Prove $\pi^2\int_0^\infty\frac{x\sin^4\pi x}{\cos\pi x+\cosh\pi x}dx=e^2\int_0^\infty\frac{x\sin^4ex}{\cos ex+\cosh ex}dx=\frac{176}{225}$ Marco Cantarini and Jack D'Aurizio proved hard-looking integrals (see Marco and Jack) in my recent two posts.  
This is our final hard-looking integral that yield a rational answer:

$$\pi^2\int_{0}^{\infty}\frac{x\sin^4(x\pi)}{\cos(x\pi)+\cosh(x\pi)}dx=e^2\int_{0}^{\infty}\frac{x\sin^4(xe)}{\cos(xe)+\cosh(xe)}dx=\frac{176}{225}\tag1$$

Can anyone provide us a prove of $(1)$?
 A: Taking the cue from Sophie Agnesi and doing a similar manner as my previous answer in your post, then
\begin{align}
S_A&=\int_{0}^{\infty}\frac{x\sin^4 x}{\cosh x+\cos x}\ dx\\[10pt]
&=2\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty}x\ e^{-kx}\sin^3 x\sin kx\ dx\\[10pt]
&=\frac{1}{2}\sum_{k=1}^\infty(-1)^{k-1}\left[3\int_{0}^{\infty}x\ e^{-kx}\sin x\sin kx\ dx-\int_{0}^{\infty}x\ e^{-kx}\sin3 x\sin kx\ dx\right]\\[10pt]
&=\frac{3}{4}-\frac{1}{4}\sum_{k=1}^\infty(-1)^{k-1}\left[\int_{0}^{\infty}x\ e^{-kx}\cos(k-3)x\ dx-\int_{0}^{\infty}x\ e^{-kx}\cos(k+3)x\ dx\right]\\[10pt]
&=\frac{3}{4}-\frac{1}{4}\sum_{k=1}^\infty(-1)^{k-1}\left[ \frac{\cos\left(2\tan^{-1}\left(\frac{k-3}{k}\right)\right)}{k^{2}+(k-3)^2}-\frac{\cos\left(2\tan^{-1}\left(\frac{k+3}{k}\right)\right)}{k^{2}+(k+3)^2}\right]\\[10pt]
&=\frac{3}{4}-\frac{1}{4}\left(-\frac{29}{225}\right)\\[10pt]
&=\frac{176}{225}
\end{align}
and the claim follows. I leave the latter sum to you as a brainstorming.
A: We may notice that:
$$ \frac{1}{\cos x+\cosh x}=\frac{2\,e^{ix}}{\left(1+e^{(i+1)x}\right)\left(1+e^{(i-1)x}\right)}=\frac{2\,e^x}{\left(e^x+e^{ix}\right)\left(e^x+e^{-ix}\right)}\tag{1}$$
or:
$$ \frac{1}{\cos x+\cosh x} = \frac{e^{ix}}{i\sin(x)}\cdot\frac{1}{e^x+e^{ix}}-\frac{e^{-ix}}{i\sin x}\cdot\frac{1}{e^x+e^{-ix}}\tag{1bis} $$
so that:
$$ \frac{x \sin^4(x)}{\cos x+\cosh x} = \text{Re}\left[\frac{\frac{3x}{4}-\frac{x}{4}e^{-2ix}}{e^x+e^{ix}}+\frac{\frac{3x}{4}-\frac{x}{4}e^{2ix}}{e^x+e^{-ix}}\right]\tag{1ter} $$
but:
$$ \int_{0}^{+\infty}\left(\frac{3x}{4}e^{ix}-\frac{x}{4}e^{3ix}\right) e^{-n(1+i)x}\,dx = \frac{1}{4}\left(\frac{3}{(-i+n(1+i))^2}-\frac{1}{(-3i+n(1+i))^2}\right)\tag{2} $$
so the whole problem boils down to computing:
$$ \sum_{n\geq 1}(-1)^{n-1}\left(\frac{3(2n-1)}{(n^2+(n-1)^2)^2}-\frac{6n-9}{(n^2+(n-3)^2)^2}\right)\tag{3}$$
by properly exploiting the reflection formulas for $\psi$ and $\psi'$. 
I think Marco's the expert at this point: his previous answer computes my $(3)$, too, and proves your conjecture. Anyway, I have an interesting remark: due to the particular form of the integrand function $(1)$, this kind of integrals can be directly converted into (relatively) simple series by regarding them as the integral remainder term in the Abel-Plana formula.
A: Hint:
After making substitution $\pi x\mapsto x$ for the 1st integral and $e x\mapsto x$ for the 2nd integral, one will get
\begin{equation}
\int_{0}^{\infty}\frac{x\sin^4x}{\cos x+\cosh x}\ dx
\end{equation}
Now use the following relations
\begin{equation}
 \frac{\sin x}{\cos x+\cosh x} = 2\sum_{n=1}^{\infty}\ (-1)^{n-1} e^{-nx} \sin nx
\end{equation}
and
\begin{equation}
\sin^3x=\frac{3\sin x - \sin 3x}{4}
\end{equation}
Apply the method used by user @MarcoCantarini in his answer then you should be able to get the desired result.
