An integral of rational function with third power of cosine hyperbolic function
Brevan Ellefsen answer was an inspiration for me to solve it using Contour integration,
Consider
$$f(z) = \frac{\sinh(z)}{z \sinh^3(z-\pi/4)}$$
If we integrate around a contour of height $\pi$ and strech it to infinity we get
By taking $T \to \infty $
$$\color{red}{\int^{i\pi/2+\infty}_{-i\pi/2+\infty}f(x)\,dx}+\color{blue}{\int^{i\pi/2-\infty}_{i\pi/2+\infty}f(x)\,dx}+\color{red}{\int^{-i\pi/2-\infty}_{i\pi/2-\infty}f(x)\,dx}+ \color{blue}{\int^{-i\pi/2+\infty}_{-i\pi/2-\infty}f(x)\,dx} = 2\pi i \mathrm{Res}(f,\frac{\pi}{4})$$
Consider
$$\color{blue}{\int^{-i\pi/2+\infty}_{-i\pi/2-\infty}\frac{\sinh(x)}{x \sinh^3(x-\pi/4)}\,dx}$$
Let $x = -\pi/2i+\pi/4+y$
$$-\int^{\infty}_{-\infty}\frac{1}{-i\pi/2+\pi/4+y}\frac{\cosh(\pi/4+y)}{ \cosh^3(y)}\,dy$$
Similarly we have for
$$\color{blue}{\int^{i\pi/2-\infty}_{i\pi/2+\infty}\frac{\sinh(x)}{x \sinh^3(x-\pi/4)}\,dx}$$
By letting $x =i\pi/2+\pi/4+ y$
$$\int^{\infty}_{-\infty}\frac{1}{i\pi/2+\pi/4+y}\frac{\cosh(\pi/4+y)}{ \cosh^3(y)}\,dy$$
The other integrals go to 0 hence
$$-16 \pi i\int^{\infty}_{-\infty} \frac{1}{(5 \pi^2 + 8 \pi y + 16y^2) }\frac{\cosh\left(y+\frac{\pi}{4} \right)}{\cosh^3(y)}dy =2\pi i \mathrm{Res}(f,\frac{\pi}{4})$$
Calculating the residue we have
$$-16 \pi i\int^{\infty}_{-\infty} \frac{1}{(5 \pi^2 + 8 \pi y + 16y^2) }\frac{\cosh\left(y+\frac{\pi}{4} \right)}{\cosh^3(y)}dy = 2\pi i\frac{-(16 (π \cosh(π/4) - 4 \sinh(π/4))}{π^3}$$
Which reduces to our result
$$ \int^{\infty}_{-\infty} \frac{1}{(5 \pi^2 + 8 \pi y + 16y^2) }\frac{\cosh\left(y+\frac{\pi}{4} \right)}{\cosh^3(y)}dy=\frac{2}{\pi^3}\left(\pi \cosh\left(\frac{\pi}{4} \right)-4\sinh\left( \frac{\pi}{4}\right) \right)$$
$$I=\int_{-\infty}^{\infty}\frac{1}{(5 \pi^2 + 8 \pi x + 16x^2) }\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
$$I=\int_{-\infty}^{\infty}\frac{1}{(4x+\pi+2i\pi)(4x+\pi-2i\pi) }\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
$$I=\frac{-i}{16\pi}\int_{-\infty}^{\infty}\left(\frac{1}{x+\frac{\pi}{4}-\frac{i\pi}{2}
}-\frac{1}{x+\frac{\pi}{4}+\frac{i\pi}{2}}\right)\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
Make the substitution $x+\pi/4 \to x$
$$I=\frac{-i}{16\pi}\int_{-\infty}^{\infty}\left(\frac{1}{x-\frac{i\pi}{2}
}-\frac{1}{x+\frac{i\pi}{2}}\right)\frac{\cosh\left(x \right)}{\cosh^3(x-\pi/4)}dx$$
We now split into two integrals and investigate each
$$I_1=\int_{-\infty}^{\infty}\frac{1}{x+\frac{i\pi}{2}}\frac{\cosh\left(x \right)}{\cosh^3(x-\pi/4)}dx$$
Let $x+\frac{i\pi}{2} \to x$
$$\color{red}{I_1=\int_{\frac{i\pi}{2}-\infty}^{\frac{i\pi}{2}+\infty}\frac{\sinh(x)}{x\cosh^3(π/4 - x)} dx}$$
$$I_2=\int_{-\infty}^{\infty}\frac{1}{x-\frac{i\pi}{2}}\frac{\cosh\left(x \right)}{\cosh^3(x-\pi/4)}dx$$
Let $x-\frac{i\pi}{2} \to x$
$$\color{red}{I_2=\int_{-\frac{i\pi}{2}-\infty}^{-\frac{i\pi}{2}+\infty}\frac{\sinh(x)}{x\cosh^3(π/4 - x)} dx}$$
We now need to calculate $I_2 - I_1$ and multiply the result by $\frac{-i}{16}$, but I am stumped. There are clear symmetries in $I_1$ and $I_2$, as only the domain of integration changes; the function inside the integral stays the same. If anyone has any suggestions, I will gladly take them.