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if $0<a\leq1$, then canwe get a closed form of $$I(a)=\int_0^\infty\frac{x}{\tanh x}\frac{1}{\cosh^2(ax)}dx.$$ In fact,if $a=1$,$I(a=1)=\pi^2/8$.

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If you are interested in particular cases: $$\begin{align} I\left(\frac14\right) & = \pi^2 + 2 \\ I\left(\frac13\right) & = -12\operatorname{Li}_2\left( \frac{2}{i\sqrt{3}-1} \right) -12\operatorname{Li}_2\left( -\frac{2}{i\sqrt{3}+1} \right) - \frac{5\pi^2}{8} \\ I\left(\frac12\right) & = \frac{\pi^2}{4} + 1 \\ I\left(\frac34\right) & = -\frac{13\pi^2}{243} + \frac{32\pi\sqrt{3}}{81} + \frac29\\ I\left(1\right) & = \frac{\pi^2}{8} \\ \end{align}$$

There are also some closed-form for $a>1$:

$$\begin{align} I\left(2\right) & = \frac{3\pi^2}{32} - \frac{\pi}{8} \\ I\left(3\right) & = \frac{49\pi^2}{648} - \frac{2\pi \sqrt{3}}{27} \end{align}$$

I've got them by using CAS and some manual simplification.

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