<s>[INST] Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve 

$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$ 

through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration. 

Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$ 

Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$
 [/INST] Decompose the integrand to express the integral as
\begin{align}
I=&\int_{-\infty}^\infty \frac1{1+x^8}dx \\
=& \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2+x^2}{x^4+\sqrt2 x^2+1}dx
 + \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2-x^2}{x^4+\sqrt2 x^2+1}dx\\
\overset{x\to \frac1x}=&
 \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{(\sqrt2+1)x^2}{x^4+\sqrt2 x^2+1}dx
 + \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{(\sqrt2-1)x^2}{x^4+\sqrt2 x^2+1}dx\\
 =& \frac{\sqrt2+1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{(x-\frac1x)^2+2+\sqrt2}dx
 + \frac{\sqrt2-1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{(x-\frac1x)^2+2-\sqrt2}dx\\
\end{align}
Then, apply the Glasser master theorem  $\int_{-\infty}^\infty f(x-\frac1x)dx= \int_{-\infty}^\infty f(x)dx$
\begin{align}
I =& \frac{\sqrt2+1}{2\sqrt2}\int_{-\infty}^\infty \frac1{x^2+(2+\sqrt2)}dx
 + \frac{\sqrt2-1}{2\sqrt2}\int_{-\infty}^\infty \frac{1}{x^2+(2-\sqrt2)}dx\\
=& \frac{\sqrt2+1}{2\sqrt2}\cdot \frac{\pi}{\sqrt{2+\sqrt2}}
 + \frac{\sqrt2-1}{2\sqrt2}\cdot \frac{\pi}{\sqrt{2-\sqrt2}}
=\frac\pi2 \sqrt{1+\frac1{\sqrt2}}=\frac\pi4 \csc\frac\pi8
\end{align}
</s>