Showing $\frac{1}{e^{i\alpha_2}} + \frac{1}{e^{i\beta_2}} + \frac{1}{e^{i\gamma_2}} = \frac{1}{e^{i(\alpha_2+\beta_2+\gamma_2)}}$

Let $x$ be any of $\alpha_1$, $\beta_1$, or $\gamma_1$ and suppose $\alpha_1+\beta_1+\gamma_1=\pi$. Then

$$e^{i(\alpha_2+\beta_2)}+e^{i(\alpha_2+\gamma_2)}+e^{i(\beta_2+\gamma_2)}=1\qquad \left(x \in \mathbb{R} : \frac{x + \pi}{2\pi} \not\in \mathbb{Z}\right)\tag{1}$$

or

$$\frac{1}{e^{i\alpha_2}} + \frac{1}{e^{i\beta_2}} + \frac{1}{e^{i\gamma_2}} = \frac{1}{e^{i(\alpha_2+\beta_2+\gamma_2)}},\qquad \left(x = 2\pi n + \pi, \quad n \in \mathbb{Z}\right) \tag{2}$$

where complex $\alpha_2=\cos^{-1}(\csc(\alpha_1))$ and similarly for $\beta_2$ and $\gamma_2$.

The identities $(1-2)$ are consequences of Theorem 1 in this blog post and the trigonometric identities 3 and 4 (counting from top to bottom) in this list.

Addendum: generalize $(1-2)$.

Integrals of the form $\int f\left(x,\frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$

We give a general approach for integrals of the form $\int f\left(x,\frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$.

Let $b-a=m$ and $b+a=n$, where $a$ and $b$ are real numbers. Then the following identities hold:

$$\frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}=\tan{\left(\frac12\sec^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}=\frac{\sqrt{x+m}}{\sqrt{x+n}},\tag{1}$$

$$e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b - \sqrt{(x + b)^2 - a^2}}{a}\tag{2}$$

These identities are generalizations of identities $(1-2)$ and $(9-10)$ at "Integration using Euler-like identities" and can be derived similarly as shown in the linked blog. The general identities $(1)$ and $(2)$ lead us to the following general transformation formula: 

$$\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{3}$$

Where $\alpha=\cos^{-1}\left(\frac{x+b}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x+b}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x+b}{a}\right).$ Use the upper sign for the alternating signs $\mp$ when $\frac{x + b}{a} \geq 1$ and the lower sign when $0 \leq \frac{x + b}{a} \leq 1$.

To illustrate its usefulness, let's look at the following example taken from the YouTube channel Prime Newtons.

Example 1. Evaluate 

$$\int \frac{\sqrt{x+1}}{\sqrt{x+2}}\,dx$$

Solution. First, we obtain the values of $a$ and $b$ by solving the following (simple!) system of equations:

$$\left. b-a=1\atop a+b=2\right\}$$

The solutions are $a = \frac{1}{2}$ and $b = \frac{3}{2}$. Applying formula $(3)$ for $x\geq-1$, the integral becomes

$$\begin{aligned}\int \frac{\sqrt{x+1}}{\sqrt{x+2}}\,dx&= -\frac{i}{4}\int \frac{(1 - e^{i\alpha}) (e^{-i\alpha} - e^{i\alpha})}{(e^{i \alpha} + 1)} \, d\alpha\\&=-\frac{i}{4}\int \frac{e^{-i\alpha} (e^{i\alpha} - 1) (e^{2i\alpha} - 1)}{e^{i\alpha} + 1} \, dx\\&=-\frac{i}{4}\int e^{-i\alpha}(e^{i\alpha}-1)^2\,d\alpha\\&=-\frac{i}{4}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=\frac14(2i\alpha+e^{-i\alpha}-e^{i\alpha})+C\end{aligned}$$

To switch to real numbers, replace from $(2)$, then replace the values of $a$ and $b$ and simplify, obtaining

$$=\sqrt{x^2+3x+2} + \frac{1}{2} \log\left(2x - 2\sqrt{x^2+3x+2} + 3\right)+C$$

On the Prime Newtons YouTube channel, they first perform a variable change, followed by a trigonometric substitution, reducing the integral to integrals involving the secant and secant cubed. This integral calculator uses $u^2 = \frac{x+1}{x+2}$ and then (a complicated) partial fraction decomposition. This can be even more complicated for integral like this one:

$$\int \frac{x\sqrt{x+1}}{\sqrt{x+2}}\,dx$$

where as our method only requires a few lines of algebra and basic calculus, as in Example 1.