Show ∫∞0log(1+x2)coshπx4sinh2πx4dx=4√2−16π+8√2πlog(√2+1)\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1)

I am trying to prove this interesting integral

I tried to write

and now using

and now introducing a parameter

But writing $I'(a)$ didn’t simplify much. The substitution $y=\sinh \pi x/4$ also was of no use because of the $x^n$ factor. So How can we prove this interesting integral? Thanks

$\ds{I\equiv\int_{0}^{\infty}\ln\pars{1 + x^{2}}\, {\cosh\pars{\pi x/4} \over \sinh^{2}\pars{\pi x/4}}\,\dd x =4\root{2} - {16 \over \pi} + {8\root{2} \over \pi}\,\ln\pars{\root{2} + 1}: \ {\large ?}}$
Zeros $\ds{\braces{x_{n}}}$ of $\ds{\sinh\pars{\pi x \over 4}}$ are given by $\quad\ds{x_{n} = 4n\ic\,,\quad n \in {\mathbb Z}\quad}$ such that
With $\ds{a \not\in {\mathbb Z}}$:
where $\ds{\Psi\pars{z}}$ is the Digamma Function $\bf\mbox{6.3.1}$.