Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$ It appears that
$$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$
(so far I have about $1000$ decimal digits to confirm that).
After changing variable $x=-\tfrac12\ln z$, it takes an equivalent form
$$\int_0^1\frac{(1-z)^2}{z\,(1+z)^2 \ln^2z}dz\stackrel{\color{gray}?}=\frac{7\,\zeta(3)}{\pi^2}.\tag2$$
Quick lookup in Gradshteyn—Ryzhik and Prudnikov et al. did not find this integral, and it also is returned unevaluated by Mathematica and Maple. How can we prove this result? Am I overlooking anything trivial?
Further questions: Is it possible to generalize it and find a closed form of 
$$\mathcal A(a)=\int_0^\infty\frac{\tanh(x)\tanh(ax)}{x^2}dx,\tag3$$
or at least of a particular case with $a=2$?
Can we generalize it to higher powers
$$\mathcal B(n)=\int_0^\infty\left(\frac{\tanh(x)}x\right)^ndx?\tag4$$

Thanks to nospoon's comment below, we know that 
$$\mathcal B(3)=\frac{186\,\zeta(5)}{\pi^4}-\frac{7\,\zeta(3)}{\pi^2}\tag5$$
I checked higher powers for this pattern, and, indeed, it appears that
$$\begin{align}&\mathcal B(4)\stackrel{\color{gray}?}=-\frac{496\,\zeta(5)}{3\,\pi^4}+\frac{2540\,\zeta(7)}{\pi^6}\\
&\mathcal B(5)\stackrel{\color{gray}?}=\frac{31\,\zeta(5)}{\pi^4}-\frac{3175\,\zeta(7)}{\pi^6}+\frac{35770\,\zeta(9)}{\pi^8}\\
&\mathcal B(6)\stackrel{\color{gray}?}=\frac{5842\,\zeta(7)}{5\,\pi^6}-\frac{57232\,\zeta(9)}{\pi^8}+\frac{515844\,\zeta(11)}{\pi^{10}}\end{align}\tag6$$

Let $\psi_{1}(z)$ be the trigamma function.
Similar to the answer here, we can integrate the function $$g(z) = \psi_{1}\left(\frac{z}{\pi i } \right) \tanh^{2}(z) $$ around an rectangular contour with vertices at $\pm R, \pm R + \pi i $.
The function $\psi_{1}\left(\frac{z}{\pi i} \right) $ has a double pole at the origin, but it's canceled by the double zero of $\tanh^{2}(z)$.
Also, as $\Re(z) \to \pm \infty$, the magnitude of $\tanh^{2}(z)$ tends to $1$, while the magnitude of $\psi_{1}\left(\frac{z}{\pi i} \right) $ decays like $\frac{\pi}{z}$.   This, coupled with the fact the height of the contour is fixed, causes the integral to vanish on the vertical sides of the contour as $R \to \infty$.
We therefore get
$$\begin{align}\lim_{R \to \infty}\oint g(z) \, \mathrm dz &=\int_{-\infty}^{\infty} \psi_{1}\left(\frac{x}{\pi i } \right) \tanh^{2}(x) \, \mathrm dx - \int_{-\infty}^{\infty}\psi_{1}\left(\frac{x}{\pi i} + 1 \right) \tanh^{2}(x) \, \mathrm dx \\ &= - \pi^{2} \int_{-\infty}^{\infty} \frac{\tanh^{2}(x)}{x^{2}} \, \mathrm dx \tag{1} \\ &= 2 \pi i \operatorname{Res}\left[g(z), \frac{\pi i}{2} \right] \\ &= 2 \pi i \left(\frac{\psi_{2} \left(\frac{1}{2} \right)}{\pi i}\right)\\& = -28 \zeta(3) \tag{2}.\end{align}$$
The result then follows.

The fastest way to determine the residue at $z = \frac{\pi i}{2}$ is to multiply the Laurent series expansion $$\tanh^{2}(z) = \frac{1}{\left(z-\frac{\pi  i}{2}\right)^{2}} + \mathcal{O}(1)$$ with the Taylor expansion of $\psi_{1} \left(\frac{z}{\pi i} \right)$ at $z= \frac{\pi i}{2}$.

$(1)$ https://en.wikipedia.org/wiki/Polygamma_function#Recurrence_relation
$(2)$ https://mathworld.wolfram.com/PolygammaFunction.html (16)

To evaluate the integral $$\mathcal A(3)= \int_{0}^{\infty} \frac{\tanh (x) \tanh(3x)}{x^{2}} \, \mathrm dx,$$ we can integrate the function $$g(z) = \psi_{1} \left(\frac{z}{\pi i} \right) \tanh (z) \tanh(3z) $$ around the same contour.
We get $$\begin{align} \mathcal{A}(3) &= \frac{1}{\pi i } \left(\operatorname{Res}\left[g(z), \frac{\pi i }{6}\right] + \operatorname{Res}\left[g(z), \frac{5 \pi i}{6}\right] + \operatorname{Res}\left[g(z), \frac{\pi i}{2} \right]\right)  \\ &= \frac{1}{\pi i} \left(\frac{i}{3} \,  \psi_{1} \left(\frac{1}{6} \right) \tan \left(\frac{\pi}{6} \right)  + \frac{i}{3} \, \psi_{1} \left(\frac{5 }{6} \right) \tan \left(\frac{5 \pi }{6} \right)- \frac{i}{3 \pi} \, \psi_{2} \left(\frac{1}{2} \right)\right) \\ &= \frac{1}{3 \sqrt{3} \, \pi} \left(\psi_{1} \left(\frac{1}{6} \right) - \psi_{1} \left(\frac{5}{6} \right) + \frac{14\sqrt{3}}{\pi} \,  \zeta(3) \right). \end{align} $$

To show that $$\operatorname{Res} \left[g(z), \frac{\pi i}{2} \right] = \frac{1}{3 \pi i} \, \psi_{2} \left(\frac{1}{2} \right),$$ we can use the formula for the residue of a double pole derived here.
$$\begin{align} \operatorname{Res} \left[g(z), \frac{\pi i}{2} \right] &= \operatorname{Res} \left[\frac{\psi_{1} \left(\frac{z}{\pi i }\right) \sinh(z) \sinh(3z)}{\cosh(z) \cosh(3z)}, \frac{\pi i}{2}  \right] \\ &=  \operatorname{Res} \left[\frac{\psi_{1} \left(\frac{z}{\pi i }\right) \left(\cosh(4z) - \cosh(2z) \right)}{\cosh(2z) +\cosh(4z)}, \frac{\pi i}{2}  \right]  \\ &= \frac{6 \left( \frac{2}{\pi i} \, \psi_{2} \left(\frac{1}{2} \right)(12) \right) - 2 \left( 2 \,  \psi_{1} \left(\frac{1}{2} \right)(0) \right)}{3(12)^{2}} \\ &= \frac{1}{3 \pi i} \, \psi_{2} \left(\frac{1}{2} \right) \end{align}$$
