A curious family of integrals that give rational multiples of $\pi$

 I have noticed experimentally that:

$$\int_0^1 \frac{\color{red}{x}}{x^4+2x^3+2x^2-2x+1} \,dx=\color{blue}{\frac{\pi}{8}},\tag{1}$$

$$\int_0^1 \frac{\color{red}{1-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx=\color{blue}{\frac{\pi}{4}},\tag{2}$$

$$\int_0^1 \frac{\color{red}{1+x-x^2}}{x^4+2x^3+2x^2-2x+1} \, dx =\color{blue}{\frac{3\pi}{8}}.\tag{3}$$

So slight variations in the numerator always seem to produce something like $n\pi$, where $n$ is a rational number.

At MathSE I have asked what the exact relationship is between $n$ and the numerator of the integrand, to which Quanto has responded with a general formula:

$$\int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} \, dx=\frac\pi8\color{green}{(c+b-a)}+\frac\pi{3\sqrt3}\color{green}{(a+c)}.\tag{4}$$

However, is it necessary for the denominator to remain fixed? Not really. The following integral is formula $(34)$ in this list of $\pi$ formulas:

$$\int_0^1 \frac{\color{red}{16x-16}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\pi}.\tag{5}$$

Notice that the denominator is different. But again, a slight variation in the numerator and it still produces something like $n\pi$:

$$\int_0^1 \frac{\color{red}{x^2-x-1}}{x^4-2x^3+4x-4}\,dx=\color{blue}{\frac{3\pi}{16}}.\tag{6}$$

More generally, 

$$\int_{0}^{1} \frac{ax^2+bx+c}{x^4-2x^3+4x-4}\,dx=\frac{\pi}{16}\color{green}{(2a-c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt32}\color{green}{(b+c)}+\frac{\ln{(3-\sqrt8)}}{\sqrt8}\color{green}{a}.\tag{7}$$

From $(7)$, we can deduce that the integral will yield rational multiples of $\pi$ when $b$ and $c$ are opposite to each other and $a=0$. However, this formula is still far from being the ultimate generalization since it does not take into account the coefficients of the denominator. The fact that the denominator is not the same in $(4)$ and $(7)$ suggests that a further generalization is possible.

A more intimidating integral

Doing some arithmetic with the integrands, we can obtain more intimidating integrands that still yield $\pi$. The following one comes from adding integrands (2) and (5) with different denominators:

$$\int_{0}^{1} \frac{2(1-x)(x^5-5x^4-10x^3-4x^2+8x-8)}{x^8-2x^6-2x^5+9x^4-2x^3-16x^2+12x-4}=\pi.$$

Addendum. I wonder if it is possible to characterize the integrand $\frac{P(x)}{Q(x)}$ in such a way that by simple inspection we can say $I=n\pi$? Or in other words, what should be the relationship between the coefficients of the numerator and the denominator for the integral to yield $n\pi$?