Benchmarking the Unified Substitution Method (USM) Against Mathematica Integrate, Part 3

This post documents the updated benchmark suite for the Unified Substitution Method (USM), a branch-consistent integration framework for algebraic and half-angle integrands involving quadratic radicals. The method is described in the USM article, A Unified Substitution Method for Integration, arXiv:2505.03754, together with the supplementary appendix on half-angle identities and continuation of USM antiderivatives.

The full Wolfram benchmark code is available here: Download the Wolfram benchmark code PDF.

The key point is that USM is not merely a real-variable substitution trick. It is built on the principal complex branches inherited from the principal logarithm and the principal square root. In the USM article, the transforms are derived from identities involving exponentials of principal inverse trigonometric functions such as

e^(± i ArcCos[y])
e^(± i ArcSec[y])

These identities lead to explicit substitutions that reduce large classes of radical and half-angle integrands to rational functions of one parameter. Depending on the transform, that parameter is usually denoted by t, r, or s. These parameters act as branch-aware conformal parametrizations on the relevant analytic components of the complex plane, allowing the same rational primitive to be continued across domains with the correct principal-branch behavior.

What Type of Integrands Are Being Benchmarked?

The benchmark suite targets integrands that combine polynomial terms, quadratic radicals, and inverse-trigonometric half-angle expressions. These are precisely the types of integrands where a general-purpose symbolic integrator can suffer from branch ambiguity, expression swell, or non-termination.

Typical benchmark integrands have the form

1 / (c0 + c1*x^n0 + c2*x^n1*RadicalTerm + c3*x^n2*HalfAngleTerm)

where the radical or half-angle component belongs to one of the five USM transform geometries. For example, Transform 2 tests integrands involving

Sqrt[(x + b)^2 - a^2]

and

Sqrt[(x + b - a)/(x + b + a)]

while Transform 5 tests integrands involving

Sqrt[a^2 - (x + b)^2]

and

Sqrt[(a + b + x)/(a - b - x)]

These expressions are not artificial stress tests in the sense of being random noise. They are structured algebraic integrands generated from the exact radical geometries that the USM transforms are designed to rationalize.

The Five Geometries

Let

y = (x + b)/a

with a > 0. The topological behavior of the integrand depends on the core radical geometry.

Transform	Core Geometry	Parameter	Complex Region on the Real Axis
Transforms 1 and 2	Sqrt[y^2 - 1]	t	Complex on the interior -1 < y < 1
Transform 3	Sqrt[y^2 + 1]	s	Never complex on the real line
Transforms 4 and 5	Sqrt[1 - y^2]	r	Complex on the exteriors y > 1 and y < -1

This distinction is essential. Transforms 1 and 2 are naturally real on the exterior intervals and complex on the interior. Transforms 4 and 5 are naturally real on the bounded circular interval and complex on the exterior intervals. Transform 3 is different: since

y^2 + 1 >= 1

for every real y, the principal square root stays positive and real on the whole real axis. The parameter

s = y + Sqrt[y^2 + 1]

is real, positive, and never crosses zero. This is why Transform 3 requires no separate componentwise continuation theorem. A single universal template works globally.

Why the Benchmark Does Not Use Simplify[D[F[x], x] - I[x]]

For small examples, one might verify an antiderivative by asking Mathematica to simplify

Simplify[D[anti, x] - integrand]

to zero. That approach is not reliable for this benchmark class. The antiderivatives can contain nested radicals, branch-dependent logarithms, and large RootSum objects. In those cases, symbolic simplification can hang, return unevaluated, or spend more time proving the result than computing the integral itself.

Instead, the benchmark engine validates each antiderivative numerically at branch-safe test points. If F(x) is the returned antiderivative and I(x) is the original integrand, the engine checks whether

F'(x0) ≈ I(x0)

at carefully selected points x0 in the appropriate domain. The values are evaluated to high precision, microscopic numerical noise is removed with Chop, and the relative error must be less than

10^-4

for the branch to be marked as correct.

Why Fractional Test Points Matter

The random generator often creates examples with integer branch boundaries. Testing exactly at an integer can accidentally hit a singularity, a branch point, or an undefined value. The benchmark avoids this by stepping away from each boundary using fractional offsets.

For example, consider

I(x) = 1 / Sqrt[x^2 - 4]

The branch boundaries are x = -2 and x = 2. Instead of testing at the boundaries, the engine uses points such as:

• Right exterior: x = 2 + 2.718 = 4.718
• Interior: x = 0 + 2*0.314 = 0.628
• Left exterior: x = -2 - 2.718 = -4.718

This makes the verification branch-aware. Each returned antiderivative is tested where it is supposed to be valid, rather than at a point where the principal branch is changing.

What the Updated Tables Show

The updated benchmark tables report more than raw timing. They explicitly display:

• the generated integrand,
• the USM antiderivative,
• the interval or branch domain where the formula is used,
• the fractional test points,
• whether the derivative check passed,
• cold execution time,
• warm execution time,
• and byte count of the output expression.

The byte count is important because it measures expression bloat. A system may technically find an antiderivative but return a huge expression that is difficult to inspect, differentiate, simplify, or reuse. USM often avoids this by rationalizing the integrand first and then back-substituting through a controlled branch formula.

Where Mathematica Integrate Struggled

The screenshot below shows a native Mathematica Integrate benchmark table for a difficult family of algebraic radical integrands.

In this run, Mathematica returned several antiderivatives as unevaluated integral expressions. One case produced a large antiderivative with a byte count of 55296, and another case hit the configured 15-second timeout. The table also reports the aggregate timing:

MATH AVERAGES | Cold: 2.2397 s | Warm: 0.1952 s

This is exactly the type of behavior the benchmark is designed to expose. The issue is not that Mathematica cannot integrate simple radicals. The issue is that when polynomial factors, shifted quadratic radicals, quotient radicals, and branch-sensitive algebraic structure are mixed together, the general-purpose integrator may return an unevaluated result, a very large expression, or a timeout.

Why USM Is Different

USM attacks the structure directly. Instead of asking a general integrator to discover the right substitution, the benchmark applies the appropriate USM transform first. The radical and half-angle terms are converted into rational functions of a single parameter.

For Transforms 1 and 2, the parameter t satisfies reconstruction formulas of the form

y = (1/2) (t + 1/t)

x = (a/2) (t + 1/t) - b

dx = a*(t^2 - 1)/(2*t^2) dt

For Transforms 4 and 5, the parameter r satisfies

y = 2*r/(1 + r^2)

x = 2*a*r/(1 + r^2) - b

dx = a*2*(1 - r^2)/(1 + r^2)^2 dr

For Transform 3, the hyperbolic parameter s satisfies

y = (s - 1/s)/2

x = a*(s^2 - 1)/(2*s) - b

Sqrt[(x + b)^2 + a^2] = a*(s^2 + 1)/(2*s)

dx = a*(s^2 + 1)/(2*s^2) ds

After this substitution, the transformed integrand is rational in the parameter. Mathematica is then used for the rational integration step, which is a much more controlled problem than the original branch-heavy expression.

Dynamic Formula Merging

The updated benchmark also prevents redundant output. Some transforms require separate formulas on different real components. However, when the analytical continuation of the USM parameter aligns with Mathematica’s principal root across a boundary, the code checks whether the two returned antiderivatives are identical. If they are, the table collapses them into a single formula.

This is why the output sometimes displays one global antiderivative and sometimes displays separate formulas for the left, interior, or right branches. The goal is not to force piecewise output unnecessarily. The goal is to preserve branch correctness while avoiding redundant formulas.

Changing the Sample Size

The scripts are standalone. To run larger or smaller batches, change the final iterator count in the integrand generation block and then change the matching iterator in the results block.

For example:

SeedRandom[102];

integrands2 = Table[
  Module[
    {
      a = RandomInteger[{1, 15}],
      b = RandomInteger[{-10, 10}],
      c0, c1, c2, c3, n0, n1, n2, expr
    },

    c0 = RandomChoice[DeleteCases[Range[-5, 5], 0]];
    c1 = RandomChoice[DeleteCases[Range[-5, 5], 0]];
    c2 = RandomChoice[DeleteCases[Range[-5, 5], 0]];
    c3 = RandomChoice[DeleteCases[Range[-5, 5], 0]];

    n0 = RandomInteger[{1, 3}];
    n1 = RandomInteger[{0, 2}];
    n2 = RandomInteger[{0, 2}];

    expr =
      c0 +
      c1*x^n0 +
      c2*x^n1*Sqrt[Expand[(x + b)^2 - a^2]] +
      c3*x^n2*Sqrt[(x + b - a)/(x + b + a)];

    1/expr
  ],
  {10}
];  (* Change this number *)

Then update the matching results iterator:

results2 = Table[
  Module[
    {
      eq = integrands2[[i]],
      antiUSM, coldUSM, warmUSM, bytesUSM
    },

    (* benchmark computation here *)

  ],
  {i, 10}
];  (* Change this number to match *)

Conclusion

These benchmarks are not only measuring speed. They are measuring whether a branch-aware substitution framework can turn difficult symbolic integration problems into predictable rational integrations.

For the integrand families tested here, the central advantage of USM is structural. The method knows in advance which radical geometry is present, which parameter should be used, where the branch transitions occur, and how to back-substitute using the principal complex branches. That is why the benchmark records not only cold and warm timing, but also domains, numerical derivative verification, and expression byte count.

The attached Mathematica screenshot shows why this matters: native Integrate can return unevaluated integrals, very large expressions, or timeouts on exactly the kind of mixed algebraic radical integrands that USM is designed to rationalize.

Un problema de Igor Volzhin

Problema. Sea $\Gamma$ una circunferencia de centro $O$ y radio $R$. Desde un punto exterior $P$ se trazan las dos tangentes a $\Gamma$, que la tocan en los puntos $X$ e $Y$, siendo $X$ el punto de tangencia superior e $Y$ el punto de tangencia inferior. Sea $Q$ el primer punto de intersección de la recta $PO$ con la circunferencia $\Gamma$, al recorrer dicha recta desde $P$ hacia $O$. Se sabe que $QX=a$. La recta $XQ$ corta al segmento $PY$ en un punto $Z$ tal que $QZ=b$. Demuestre que el radio $R$ de la circunferencia puede expresarse en función de $a$ y $b$ mediante la fórmula

$$R=\frac{a\sqrt{b}}{\sqrt{3b-a}}.$$

Demostración. Note que $PQ$ es la bisectriz de $\angle{XPY}$. Por lo tanto, usando la fórmula para la longitud de la bisectriz en el triángulo $XPZ$, tenemos:

$$PQ^2=\frac{PX\cdot{PZ}}{(PX+PZ)^2}\left[(PX+PZ)^2-(a+b)^2\right].\tag{1}$$

Por el teorema de la bisectriz tambien tenemos que 

$$\frac{a}{b}=\frac{PX}{PZ} \Longrightarrow PZ=\frac{b}{a}PX.\tag{2}$$

Consecuentemente, 

$$YZ=PY-PZ=PX-\frac{b}{a}PX=PX\left(\frac{a-b}{a}\right).\tag{3}$$

Por potencia de un punto $YZ^2=ZQ\cdot{ZX}$. Sustituyendo, obtenemos 

$$PX^2\left(\frac{a-b}{a}\right)^2=b(a+b) \Longrightarrow PX^2=\frac{a^2b(a+b)}{(a-b)^2}.\tag{4}$$

Sustituyendo $(2)$ en $(1)$, simplificando y luego sustituyendo $(4)$ en $(1)$, 

$$PQ^2=\frac{ab^2(a+b)}{(a-b)^2}-ab=\frac{a^2b(3b-a)}{(a-b)^2}.\tag{5}$$

Nuevamente, por potencia de un punto, 

$$PX^2=PQ(PQ+2R) \Longrightarrow R=\frac{PX^2-PQ^2}{2PQ}.\tag{6}$$

Sustituyendo $(4)$ y $(5)$ en $(6)$ y simplificando, obtenemos

$$\boxed{R=\frac{a\sqrt{b}}{\sqrt{3b-a}}.}$$

Por la configuración, $b<a$ y necesariamente $3b>a$, de modo que la raíz de $3b−a$ tiene sentido.

Constructing an Equal-Area Triangle from a Quadrilateral

Claim. Let $ABCD$ be a convex quadrilateral, and let $O$ be an arbitrary point in the plane. Let $M$ and $N$ be the midpoints of $BO$ and $CO$, respectively. Let $A'$ be the reflection of $A$ about $N$, and let $D'$ be the reflection of $D$ about $M$. Then

$$[ABCD]=[A'D'O].$$

The area of quadrilateral ABCD is equal to the area of triangle A'D'O.

Proof. Let the diagonals $AC$ and $BD$ meet at $P$. Since $N$ is the midpoint of both $CO$ and $AA'$, the half-turn about $N$ sends $A$ to $A'$ and $C$ to $O$. Hence

$$A'O=AC \qquad \text{and} \qquad A'O \parallel AC.$$

Similarly, since $M$ is the midpoint of both $BO$ and $DD'$, the half-turn about $M$ sends $D$ to $D'$ and $B$ to $O$. Therefore

$$D'O=BD \qquad \text{and} \qquad D'O \parallel BD.$$

Let

$$\theta=\angle BPC,$$

the angle between the diagonals $BD$ and $AC$. Since $A'O \parallel AC$ and $D'O \parallel BD$, we have

$$\sin \angle A'OD'=\sin \theta.$$

Thus

$$[A'D'O]=\frac12 \cdot A'O \cdot D'O \cdot \sin \angle A'OD'=\frac12 \cdot AC \cdot BD \cdot \sin \theta.$$

On the other hand, the area of a convex quadrilateral is equal to one half the product of its diagonals times the sine of the angle between them. Hence

$$[ABCD]=\frac12 \cdot AC \cdot BD \cdot \sin \theta.$$

Therefore,

$$[A'D'O]=[ABCD],$$

as desired.

USM-style substitutions for radical equations

Introduction 

There is a tempting story one could tell about USM-style substitutions:

\[\text{radical equation} \longrightarrow \text{smart parameter} \longrightarrow \text{easy polynomial}.\]

That story is sometimes true. But when one solves the examples all the way to their actual real roots, a more precise picture emerges.

The Unified Substitution Method (USM) was developed for integration (see arXiv:2505.03754v2), not for equation solving. Still, the same branch-consistent parametrizations behind Transform 2, Transform 3, and Transform 5 give natural substitutions for equations involving

\[\sqrt{(x+b)^2-a^2}, \qquad \sqrt{(x+b)^2+a^2}, \qquad \sqrt{\frac{x+b-a}{x+b+a}}, \qquad \sqrt{a^2-(x+b)^2}, \qquad \sqrt{\frac{a+b+x}{a-b-x}}.\]

The real issue is not simply whether USM lowers the degree of the resulting polynomial. The deeper issue is whether the substitution keeps enough branch information so that, after solving the polynomial, one can still tell which roots correspond to genuine solutions of the original radical equation.

Repeated squaring usually loses sign information and therefore creates extraneous roots. A branchwise algebraic identity may keep more sign information than squaring and can sometimes outperform USM for a specially tailored problem. A USM parameter often encodes the geometry of the radical directly, so the admissible interval for the parameter itself filters out impossible roots.

The examples below make this explicit.

What USM is really good at

A USM-style substitution is strongest when one parameter rationalizes essentially all of the radicals at once.

For the circular difference case, the basic parameter is

\[x=\frac{a}{2}\left(t+\frac{1}{t}\right)-b,\]

which turns expressions of the form

\[\sqrt{(x+b)^2-a^2} \quad\text{and}\quad \sqrt{\frac{x+b-a}{x+b+a}}\]

into rational functions of \(t\). Here the admissible values of \(t\) depend on the real branch of the original equation: on the upper branch \(x\ge a-b\), one has \(0<t\le 1\), while on the lower branch \(x\le -a-b\), one has \(-1\le t<0\). Thus the parameter interval itself records which branch is being used.

For the hyperbolic sum case, the natural parameter is

\[s=\frac{x+b+\sqrt{(x+b)^2+a^2}}{a},\]

so that

\[x=\frac{a}{2}\left(s-\frac{1}{s}\right)-b, \qquad \sqrt{(x+b)^2+a^2}=\frac{a}{2}\left(s+\frac{1}{s}\right).\]

In this case the admissible parameter range is simply

\[s>0,\]

and the map \(s\mapsto x\) carries \((0,\infty)\) bijectively onto the full real line.

For the circular bounded case, one uses

\[x=\frac{2ar}{1+r^2}-b,\]

which turns

\[\sqrt{a^2-(x+b)^2} \quad\text{and}\quad \sqrt{\frac{a+b+x}{a-b-x}}\]

into rational functions of \(r\). Here the admissible parameter range is

\[-1\le r<1,\]

corresponding to the bounded real interval \(-a-b\le x<a-b\).

That is the best-case picture. The parameter respects the radical geometry instead of destroying it by squaring. But that does not mean it always wins.

Example 1 (difference form: USM is strong, but a tailored branch identity is even stronger)

Consider

\[\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}-2=0.\tag{E1}\]

The domain forced by the radicals is

\[x\in(-\infty,-1)\cup[1,\infty).\]

So the equation naturally splits into the two real branches \(x\ge 1\) and \(x < -1\).

Brute-force squaring

If one isolates a radical and squares twice, one reaches the sextic

\[x^6+2x^5-11x^4-20x^3+12x^2+48x+32=0.\]

It factors (and this could be tricky!) as

\[(x^3-x^2-4x-4)(x^3+3x^2-4x-8)=0.\]

Numerically, the real roots of this polynomial are

\[x\approx 2.875129794, \quad x\approx -3.489288572, \quad x\approx -1.289168546, \quad x\approx 1.778457118.\]

But the original radical equation does not accept all four. Direct substitution into \((E1)\) shows that only

\[\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}-2=0\quad\Longrightarrow\quad\left\{-3.489288572,\;2.875129794\right\},\]

are genuine solutions.

This is the basic weakness of repeated squaring: it removes the sign data carried by the radicals, so every real root of the polynomial must be checked back in the original equation.

The branchwise algebraic identity

Set

\[u:=\sqrt{\frac{x-1}{x+1}}\ge 0.\]

Because

\[\sqrt{x^2-1}=|x+1|\sqrt{\frac{x-1}{x+1}}=|x+1|\,u,\]

the original equation becomes branchwise simpler.

For the branch \(x\ge 1\), one has \(|x+1|=x+1\), so

\[(x+1)u-u-2=0 \quad\Longrightarrow\quad xu=2.\]

Since \(u\ge 0\) and \(x\ge 1\), the sign relation is already consistent. Squaring gives

\[x^2\frac{x-1}{x+1}=4 \quad\Longrightarrow\quad x^3-x^2-4x-4=0.\]

This cubic has one real root,

\[x\approx 2.875129794,\]

and it indeed satisfies \((E1)\).

For the branch \(x < -1\), one has \(|x+1|=-(x+1)\), so

\[-(x+1)u-u-2=0 \quad\Longrightarrow\quad (x+2)u=-2.\]

This relation is already informative \emph{before} squaring: since \(u\ge 0\), the left-hand side has the sign of \(x+2\). Therefore any valid solution on this branch must satisfy

\[x+2<0, \qquad\text{that is,}\qquad x<-2.\]

Now square:

\[(x+2)^2\frac{x-1}{x+1}=4 \quad\Longrightarrow\quad x^3+3x^2-4x-8=0.\]

This cubic has three real roots,

\[x\approx -3.489288572, \qquad x\approx -1.289168546, \qquad x\approx 1.778457118.\]

But the branch structure discards two of them immediately. The root \(x\approx 1.778457118\) is impossible because this branch assumes \(x\le -1\). The root \(x\approx -1.289168546\) is also impossible because then \(x+2>0\), so \((x+2)u\) cannot equal \(-2\). Thus the only valid root on the negative branch is

\[x\approx -3.489288572.\]

The USM route

Now use the circular difference-form parameter

\[x=\frac{1}{2}\left(t+\frac{1}{t}\right).\]

This is the Transform 2 geometry in unit radius.

For the branch \(x\ge 1\), take

\[t=x-\sqrt{x^2-1},\]

so that \(0<t\le 1\) and the map \(t\mapsto x=\frac12(t+t^{-1})\) sends \((0,1]\) monotonically onto \([1,\infty)\). The equation becomes

\[t^3+3t^2+5t-1=0.\]

This cubic has one real root,

\[t\approx 0.179509025,\]

which lies in \((0,1]\), so it is admissible. Converting back gives

\[x=\frac12\left(t+\frac1t\right)\approx 2.875129794.\]

For the branch \(x < -1\), take

\[t=x+\sqrt{x^2-1},\]

so that \(-1\le t<0\) and the same formula \(x=\frac12(t+t^{-1})\) maps \((-1,0)\) monotonically onto \((-\infty,-1)\). The equation becomes

\[t^3-t^2-7t-1=0.\]

Its real roots are approximately

\[t\approx -2.102775049, \qquad t\approx -0.146365489, \qquad t\approx 3.249140538.\]

But the parameter interval for this branch is \((-1,0)\), so only

\[t\approx -0.146365489\]

is admissible. Therefore

\[x=\frac12\left(t+\frac1t\right)\approx -3.489288572.\]

What this example proves

The full real solution set of \((E1)\) is therefore

\[\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}-2=0\quad\Longrightarrow\quad\left\{-3.489288572,\;2.875129794\right\}.\]

This example shows three different levels of structural control. Repeated squaring finds a polynomial but gives too many real roots. The special branchwise identity keeps more sign information and filters spurious roots very efficiently. USM keeps the branch information through the admissible interval for \(t\), so root validation is built into the parameter itself.

For this equation the honest ranking is

\[\text{special branchwise trick}\;>\;\text{USM}\;>\;\text{brute-force squaring}.\]

Example 2 (hyperbolic sum form: here the USM structure is genuinely better)

Now consider

\[\frac{1-\sqrt{x^2+1}}{x}=x-2, \qquad x\ne 0.\tag{E2}\]

The hyperbolic USM substitution

Set

\[s=x+\sqrt{x^2+1}.\]

Then \(s>0\) for every real \(x\), and

\[x=\frac{1}{2}\left(s-\frac{1}{s}\right), \qquad \sqrt{x^2+1}=\frac{1}{2}\left(s+\frac{1}{s}\right).\]

Substituting into \((E2)\) yields

\[s^3-s^2-7s-1=0.\]

The real roots of this cubic are

\[s\approx -2.102775049, \qquad s\approx -0.146365489, \qquad s\approx 3.249140538.\]

But the substitution itself imposes the structural condition

\[s>0.\]

So only

\[s\approx 3.249140538\]

can correspond to a real \(x\). Converting back gives

\[x=\frac12\left(s-\frac1s\right)\approx 1.470683420.\]

Direct substitution confirms that this is a genuine solution.

The classical squaring route

Starting from \((E2)\), multiply by \(x\) (allowed because \(x\ne 0\)):

\[1-\sqrt{x^2+1}=x^2-2x.\]

Hence

\[\sqrt{x^2+1}=1+2x-x^2.\]

Now the structure that matters is the sign constraint

\[1+2x-x^2\ge 0,\]

because the left-hand side is a square root. This forces

\[1-\sqrt2\le x\le 1+\sqrt2.\]

Only after recording that condition should one square:

\[(1+2x-x^2)^2=x^2+1.\]

After simplification,

\[x(x^3-4x^2+x+4)=0.\]

Since the original equation excludes \(x=0\), one solves

\[x^3-4x^2+x+4=0.\]

Its real roots are

\[x\approx -0.813606503, \qquad x\approx 1.470683420, \qquad x\approx 3.342923083.\]

Now the sign condition removes two of them immediately. The root \(x\approx -0.813606503\) fails because \(1+2x-x^2<0\). The root \(x\approx 3.342923083\) fails for the same reason. The root \(x\approx 1.470683420\) satisfies the sign condition and the original equation.

So the classical route also leads to the correct real solution set,

\[\frac{1-\sqrt{x^2+1}}{x}=x-2\quad\Longrightarrow\quad\left\{1.470683420\right\}.\]

Why the USM method is better here

In this example there is no special real factorization analogous to

\[x^2-1=(x-1)(x+1)\]

that would make a branchwise trick unusually efficient. The hyperbolic USM substitution keeps the intrinsic geometry of \(\sqrt{x^2+1}\) and provides a parameter with a built-in positivity condition \(s>0\). That single condition discards the two spurious cubic roots at once.

So here the honest ranking is

\[\text{USM}\;>\;\text{classical squaring}.\]

The reason is structural: the map \(s\mapsto x=\frac12(s-s^{-1})\) is a bijection from \((0,\infty)\) onto \(\mathbb{R}\), so an admissible positive root of the cubic corresponds to exactly one real solution of the original equation.

Example 3 (bounded circular form: USM again keeps the right branch data)

Now consider

\[\sqrt{1-x^2}-x^2\sqrt{\frac{1+x}{1-x}}=-1, \qquad -1\le x<1.\tag{E3}\]

This is naturally adapted to the bounded circular substitution.

The USM Transform 5 route

Set

\[r=\frac{x}{1+\sqrt{1-x^2}},\]

so that

\[x=\frac{2r}{1+r^2}, \qquad \sqrt{1-x^2}=\frac{1-r^2}{1+r^2}, \qquad \sqrt{\frac{1+x}{1-x}}=\frac{1+r}{1-r}.\]

For \(-1\le x<1\), this parameter satisfies

\[-1\le r<1.\]

Substituting into \((E3)\) gives

\[3r^3+r^2+r-1=0.\]

This cubic has one real root,

\[r\approx 0.469396425.\]

Since \(r\in[-1,1)\) automatically, the root is admissible. Therefore

\[x=\frac{2r}{1+r^2}\approx 0.769292354.\]

Direct substitution confirms that this solves \((E3)\).

The classical route and its sign filter

On the domain \(x<1\), one may rewrite

\[\sqrt{\frac{1+x}{1-x}}=\frac{\sqrt{1-x^2}}{1-x},\]

because \(1-x>0\). Then \((E3)\) becomes

\[\sqrt{1-x^2}\left(1-\frac{x^2}{1-x}\right)=-1,\]

that is,

\[\frac{1-x-x^2}{1-x}\sqrt{1-x^2}=-1.\]

Since \(1-x>0\) and \(\sqrt{1-x^2}\ge 0\), the left-hand side can be negative only if

\[1-x-x^2<0.\]

So before squaring one already learns that any genuine solution must satisfy

\[x>\frac{\sqrt5-1}{2}\approx 0.618033989,\]

because the other root of \(1-x-x^2=0\) lies below \(-1\) and is outside the domain.

Now square. After simplification one gets

\[x^6+2x^5-2x^4-4x^3+3x^2=0,\]

or equivalently

\[x^2(x-1)(x^3+3x^2+x-3)=0.\]

The real roots of this polynomial are

\[x=0, \qquad x=1, \qquad x\approx 0.769292354,\]

with \(x=0\) having multiplicity two in the factorization above and multiplicity three in the sextic.

Only one of these is a valid solution of \((E3)\). The root \(x=0\) is extraneous, since the original left-hand side becomes \(1\), not \(-1\). The root \(x=1\) is excluded by the domain and in any case makes \(\sqrt{\frac{1+x}{1-x}}\) blow up. The root \(x\approx 0.769292354\) satisfies the sign condition and the original equation.

Therefore the full real solution set is

\[\sqrt{1-x^2}-x^2\sqrt{\frac{1+x}{1-x}}=-1\quad\Longrightarrow\quad\left\{0.769292354\right\}.\]

Why USM helps

This example is exactly the kind of problem for which Transform 5 is designed: the same parameter rationalizes both \(\sqrt{1-x^2}\) and \(\sqrt{\frac{1+x}{1-x}}\). The admissible interval \(-1\le r<1\) keeps the bounded circular geometry visible all the way through the calculation.

By contrast, squaring destroys the sign information and inflates the algebra to a sextic before one can recover the correct solution set. So here the honest ranking is

\[\text{USM / bounded circular substitution}\;>\;\text{brute-force squaring}.\]

Final comparison (what each method tells you about correctness)

The examples above show that the real issue is not only how fast one reaches a polynomial, but how much \emph{admissibility information} survives after that step.

Repeated squaring is the least informative structurally. It often produces the right algebraic factors, but it forgets the sign relations that were originally carried by the radicals. Therefore it typically enlarges the candidate set and forces a final substitution check.

In the three examples above, squaring produced

\[\text{for \((E1)\): a sextic with four real candidates, of which only two are valid;}\]

\[\text{for \((E2)\): a cubic with three real candidates, of which only one is valid;}\]

\[\text{for \((E3)\): a sextic whose obvious real roots \(x=0\) and \(x=1\) are both extraneous.}\]

When a specially adapted identity exists, it can be even better than USM. In Example \((E1)\), the identity

\[\sqrt{x^2-1}=|x+1|\sqrt{\frac{x-1}{x+1}}\]

preserves the branch sign directly. That makes it possible to reject some extraneous roots \emph{before} solving the final cubic completely. But this advantage is highly problem-specific.

A USM parameter usually sits between those two extremes. It is more systematic than a lucky identity, and much more branch-aware than repeated squaring.

In Example \((E1)\), the branch intervals \(0<t\le 1\) and \(-1\le t<0\) immediately identify which cubic roots can represent real solutions. In Example \((E2)\), the condition \(s>0\) built into the hyperbolic substitution removes two extraneous cubic roots at once. In Example \((E3)\), the bounded interval \(-1\le r<1\) keeps the circular geometry visible and prevents the method from wandering into algebraically legal but geometrically impossible candidates.

So the real practical moral is this:

\[\text{A good substitution is not only a degree-lowering device; it is an admissibility-preserving device.}\]

Carta abierta al Presidente Abinader: Un método dominicano para el Cálculo Integral

Santo Domingo, 28 de Enero del 2026

Señor

Luis Rodolfo Abinader Corona

Presidente de la República Dominicana

C.c.: Directores y redactores de los principales medios de comunicación nacionales e internacionales

De mi consideración:

Me dirijo a usted y, por su intermedio, a los medios de comunicación del país para comunicar un avance científico que considero de interés nacional y para solicitar el apoyo institucional necesario para que la República Dominicana aproveche y difunda esta aportación.

Soy Emmanuel Antonio José García. He publicado recientemente en arXiv (Cornell) el trabajo “A Unified Substitution Method for Integration” (enlace: https://arxiv.org/abs/2505.03754), en el que presento el Método de Sustitución Unificada (USM), una propuesta matemática y metodológica destinada a simplificar y acelerar la resolución de integrales que aparecen de manera frecuente en matemáticas aplicadas, ingeniería, física y ciencias de datos.

Resumen de la contribución

El USM es un método unificado para integrar expresiones con radicales cuadráticos y composiciones trigonométricas de medio ángulo, fundamentado en identidades algebraicas explícitas para las exponenciales de funciones trigonométricas inversas principales $e^{± i \cos^{-1}(y)}$ y $e^{± i \sec^{-1}(y)}$, lo que permite derivar cinco transformaciones parametrizadas que convierten dichas integrales en formas racionales en un solo parámetro, manejando de manera coherente tanto los casos circulares como hiperbólicos. Este marco no solo subsume y generaliza técnicas clásicas, como las sustituciones de Euler (primera y segunda) y la sustitución de Weierstrass, sino que también simplifica significativamente el manejo de ramas y signos, ofrece ventajas computacionales al reducir la hinchazón de expresiones y mejora la eficiencia en la integración de estructuras mixtas.

Resultados comparativos relevantes

Para ofrecer evidencia empírica, ejecuté un benchmark con 100 integrales representativas y comparé el rendimiento del USM con la función `Integrate` de Mathematica:

• USM fue más rápido en 82 de 100 casos.
• USM produjo una antiderivada de menor tamaño (ByteCnt) en 50 de 100 casos.
• Incidencia de antiderivadas “monstruo” (≥ 10.000 bytes): USM: 5 casos vs Integrate: 24 casos.
• Máximo tamaño observado: USM: 19,840 bytes; Integrate: 150,360 bytes. En otro mini-benchmark (Ejemplo 19), el recuento de bytes de Integrate superó los 600,000 (¡la antiderivada ocupa 20 páginas!), mientras que para USM no superó los 5,000 (y la antiderivada cabe en media página).

Estos resultados se traducen en dos beneficios prácticos: ahorro de tiempo de cómputo y expresiones simbólicas más legibles y reutilizables, lo que facilita su integración en pipelines de ingeniería y en material educativo avanzado.

Vinculaciones teóricas del USM

El USM está estrechamente relacionado con conceptos matemáticos de gran utilidad. Como señaló el físico alemán Fred Hucht en MathOverflow (foro donde di a conocer la primera versión del USM): 

“The OP's relations are related to the Gudermannian...", 

lo que lo conecta con identidades elípticas y la Transformación Imaginaria de Jacobi”. La Gudermanniana es fundamental en aplicaciones como la proyección cartográfica de Mercator, mientras que la estructura paramétrica del USM se asemeja a la Transformada de Joukowsky (clave en aerodinámica para el diseño de perfiles alares) y a la Transformada de Tustin, usada en control digital para discretizar sistemas dinámicos. 

Reconocimientos y revisiones externas

El trabajo ha suscitado interés y comentarios de especialistas con trayectoria internacional:

Dr. Mohammad Alkousa (Profesor, autor de libros de cálculo e investigador vinculado al Moscow Institute of Physics and Technology - MIPT, institución conocida como el "MIT de Rusia"):

“You have done great work. I will review it and will cite your work as a reference in my calculus books. I will also mention it to my students.” (Has hecho un gran trabajo. Lo revisaré y citaré tu trabajo como referencia en mis libros de cálculo. También se lo mencionaré a mis estudiantes.)

 Relevancia: Esta validación marca un hito. Que un experto de una de las instituciones STEM más prestigiosas del mundo (el MIPT cuenta con 10 premios Nobel entre sus profesores y exalumnos) integre este método en la literatura académica confirma su superioridad pedagógica y su innegable solidez matemática.

Dr. Oleg Marichev (Wolfram Research, figura legendaria de la integración simbólica):

“I was impressed, looking on your files. I saw holes in my work, that you already found and you can fix them (even without understanding many moments). I felt that we can work.” (Quedé impresionado al ver tus archivos. Vi vacíos en mi trabajo que ya encontraste y que puedes arreglar (incluso sin entender muchos momentos). Sentí que podemos trabajar juntos.)

“You made large improvement to collecting formulas for doable Integrate situation because we with you found wide class of cases for MeijerG.” (Hiciste una gran mejora en la recopilación de fórmulas para situaciones de integración realizables, porque junto contigo encontramos una amplia clase de casos para MeijerG.)

“As I wrote, I have built collection with near 4500 cases of MeijerG. If we remove special functions we have subset of such elementary functions. There we have subset of algebraic functions. I am doing re-organization of this collection and see how important and how large subclass Fun[v ArcGun[z]]^n that you found.” (Como escribí, he construido una colección con cerca de 4500 casos de MeijerG. Si eliminamos las funciones especiales, tenemos un subconjunto de dichas funciones elementales. Allí tenemos un subconjunto de funciones algebraicas. Estoy reorganizando esta colección y veo lo importante y grande que es la subclase Fun[v ArcGun[z]]^n que encontraste.)

Dr. Sam Blake (PhD, Univ. Monash, investigador; ex-ingeniero en Wolfram Research y conocido por su participación en el descifrado del célebre Zodiac Cipher):

“That’s a very neat trick… As far as I know this is a new result.” (Ese es un truco muy ingenioso... Por lo que sé, este es un resultado nuevo.)

Daniel Lichtblau (Wolfram Research):

“You are certainly getting nice results, and we'll take a look at it.” (Ciertamente estás obteniendo buenos resultados, y le echaremos un vistazo.)

Ninad Munshi (ex-ingeniero de la NASA):

“Complexification formulas are great and it seems like this simplifies the right away.” (Las fórmulas de complejificación son geniales y parece que esto simplifica de inmediato.)

Kamila Szewczyk (programadora e investigadora matemática polaca especializada en bioinformática y algoritmos de compresión, reconocida principalmente por ser la creadora del compresor de datos de código abierto bzip3):

“One benefit of your method that I see over Rubi is that the process of applying transformation rules in USM is much clearer and more efficient to evaluate (no need to rely on transformation heuristics).” (Un beneficio de tu método que veo sobre Rubi es que el proceso de aplicar reglas de transformación en USM es mucho más claro y eficiente de evaluar (sin necesidad de depender de heurísticas de transformación).)

Importancia histórica y cultural de la integración

     

La integración no es solo una técnica matemática: es una herramienta que ha modelado el progreso científico. Lo subrayan objetos culturales oficiales (una moneda conmemorativa vinculada a la técnica de integración de Ostrogradski y una estampa postal que honra a P. L. Chebyshev) que evidencian cómo los estados y las comunidades científicas reconocen la integración como patrimonio intelectual y cultural.

Por qué esto importa para la República Dominicana

1. Innovación descentralizada: que una contribución en un área clásica como el Cálculo Integral provenga de un ingeniero dominicano demuestra que nuestro país puede generar conocimiento original en áreas matemáticas de alto impacto.

2. Aplicaciones tecnológicas: la reducción de tiempos de cómputo y la menor proliferación de expresiones simbólicas gigantescas beneficiarán desarrollos en software.

3. Potencial educativo: incorporar una metodología unificada podría simplificar la enseñanza del Cálculo Integral en bachillerato y universidad, privilegiando la comprensión sobre la memorización.

Solicitudes concretas 

Con respeto, solicito al señor Presidente y a las autoridades competentes las siguientes acciones:

1. Reconocimiento institucional y difusión oficial. Que la Presidencia y el Ministerio correspondiente (MESCYT / instituciones científicas nacionales) respalden la difusión del hallazgo y promuevan su consideración en foros académicos y tecnológicos.

2. Divulgación mediática responsable. Invito a los medios a cubrir el trabajo con rigor, entrevistando a expertos y verificando las cifras y resultados, para que el país conozca y evalúe la importancia del avance.

Ofrezco mi compromiso de colaborar estrechamente con las instituciones que lo soliciten: puedo presentar los datos del benchmark y entregar material didáctico (apuntes, ejemplos resueltos y código). 

Creo firmemente que las matemáticas pueden y deben ser un motor de desarrollo social y económico. El USM es, en mi opinión, una oportunidad para que la República Dominicana demuestre su capacidad de producir conocimiento relevante y para transformar esa producción en ventajas educativas y tecnológicas concretas.

Agradezco su atención, quedo a disposición para una reunión informativa y para coordinar las acciones que sean pertinentes.

Atentamente,

Emmanuel Antonio José García

Ingeniero 

República Dominicana

Weierstrass as a Special Case of the USM Framework

"The world's sneakiest substitution." — Michael Spivak

The classical Weierstrass substitution for integrals of the form
\[\int R(\sin\omega, \cos\omega)\,d\omega\]
is a special case of Transform 5 in the USM framework, corresponding to the circular case with parameters \(a = 1\) and \(b = 0\).

Derivation
Let
\[I = \int R(\sin\omega, \cos\omega)\,d\omega.\]
Set \(x = \sin\omega\), so that
\[\cos\omega = \sqrt{1 - x^2} \quad (\text{using the principal square root, e.g., } \cos\omega \geq 0 \text{ for } \omega \in [-\pi/2,\pi/2]),\]
and
\[d\omega = \frac{dx}{\cos\omega} = \frac{dx}{\sqrt{1 - x^2}}.\]
Hence,
\[I = \int \frac{R\!\left(x, \sqrt{1 - x^2}\right)}{\sqrt{1 - x^2}}\,dx.\]
Transform 5 handles integrals involving \(\sqrt{a^2 - (x+b)^2}\) on the domain \(|y| \leq 1\) with \(y = (x+b)/a\). Take \(a = 1\), \(b = 0\) (so \(y = x\)) and let the parameter be \(r\) (as in the paper). The transform gives:
\[x = \frac{2r}{1+r^2}, \quad \sqrt{1 - x^2} = \frac{1 - r^2}{1 + r^2}, \quad dx = \frac{2(1 - r^2)}{(1 + r^2)^2}\,dr.\]
Substitute into the integral
\[\begin{aligned} I &= \int \frac{R\!\left(x, \sqrt{1 - x^2}\right)}{\sqrt{1 - x^2}}\,dx \\ &= \int \frac{R\!\left(\frac{2r}{1+r^2}, \frac{1 - r^2}{1 + r^2}\right)}{\frac{1 - r^2}{1 + r^2}} \cdot \frac{2(1 - r^2)}{(1 + r^2)^2}\,dr \\ &= \int R\!\left(\frac{2r}{1+r^2}, \frac{1 - r^2}{1 + r^2}\right)\frac{2}{1 + r^2}\,dr.\end{aligned}\]
The final expression is exactly the Weierstrass substitution formula:
\[\int R(\sin\omega, \cos\omega)\,d\omega= \int R\!\left(\frac{2r}{1+r^2}, \frac{1 - r^2}{1 + r^2}\right) \frac{2}{1 + r^2}\,dr.\]

On the principal branch where \(\psi=\sin^{-1}(x)=\omega\), this parameter is

\[r=\tan\!\Bigl(\frac{\psi}{2}\Bigr)=\tan\!\Bigl(\frac{\omega}{2}\Bigr),\]

Thus, the Weierstrass substitution emerges naturally from Transform 5 by setting \(a = 1\), \(b = 0\) and interpreting the integrand appropriately. This demonstrates (again!) that the USM unifies and generalizes classical substitution techniques (refer to Section 6 in the paper, which details how the USM also encompasses Euler substitutions 1 and 2), including the half‑angle tangent substitution of Weierstrass. Moreover, USM creates new "Weierstrass-like" substitutions (Transforms 1 & 2) that work for hyperbolic/algebraic regions ($|y| \ge 1$) where the standard $\tan(\omega/2)$ is not typically applied.

MIT Integration Bee 2023 - Finals - Problem 3

The problem:

Solution. We assume $x > 0$.

$$\begin{aligned}I &= \int \sqrt{x^2+1+\sqrt{x^4+x^2+1}} \, dx \\ &= \int \frac{t^2-t+1}{(2t-1)^{3/2}\sqrt{t-2}} \, dt & \left(t = x^2+1+\sqrt{x^4+x^2+1}\right) \\[1em] &= 2 \int \frac{u^4+3u^2+3}{(2u^2+3)^{3/2}} \, du & \left(u^2 = t-2\right) \\[1em] &= \frac{\sqrt{2}}{2} \int \frac{3s^8+12s^6+34s^4+12s^2+3}{8s^3(s^2+1)^2} \, ds & \left(\text{USM Transform 3: } u = \sqrt{\frac{3}{2}}\frac{s^2-1}{2s}\right) \\[1em] &= \frac{\sqrt{2}}{2} \int \left( \frac{3s}{8} + \frac{3}{4s} + \frac{3}{8s^3} + \frac{2s}{(s^2+1)^2} \right) \, ds & (\text{PFD}) \\[1em] &= \frac{\sqrt{2}}{2} \left( \frac{3s^2}{16} + \frac{3}{4}\ln|s| - \frac{3}{16s^2} - \frac{1}{s^2+1} \right) + C \\[1em]
&= \frac{3\sqrt{2}}{32}\left(s^2 - \frac{1}{s^2}\right) + \frac{3\sqrt{2}}{8}\ln|s| - \frac{\sqrt{2}}{2(s^2+1)} + C
\end{aligned}$$

Where:

$$s = \sqrt{\frac{2}{3}}u + \sqrt{\frac{2}{3}u^2+1}, \quad u = \sqrt{t-2}, \quad t = x^2+1+\sqrt{x^4+x^2+1}.$$

Using the symmetry of the even function $f(x)$, we calculate $2 \int_{0}^{1/2} f(x) \, dx$:

$$\begin{aligned}\text{Limits for } s: \quad & x=0 \implies s=1 \\
& x=1/2 \implies s = \sqrt{\frac{2+\sqrt{7}}{\sqrt{3}}}\end{aligned}$$

$$\begin{aligned}\int_{-1/2}^{1/2} f(x) \, dx &= 2 \left[ F(s) \right]_{1}^{\sqrt{\frac{2+\sqrt{7}}{\sqrt{3}}}} \\[1em] &= 2 \left[ \left( \frac{\sqrt{14}}{8} - \frac{\sqrt{2}}{4} + \frac{3\sqrt{2}}{16}\ln\left(\frac{2+\sqrt{7}}{\sqrt{3}}\right) \right) - \left( -\frac{\sqrt{2}}{4} \right) \right] \\[1em] &= 2 \left[ \frac{\sqrt{14}}{8} + \frac{3\sqrt{2}}{16}\ln\left(\frac{2+\sqrt{7}}{\sqrt{3}}\right) \right] \\[1em] &= \frac{\sqrt{14}}{4} + \frac{3\sqrt{2}}{8}\ln\left(\frac{2+\sqrt{7}}{\sqrt{3}}\right) \end{aligned}$$