<s>[INST] 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
$$\begin{align}&\mathcal B(4)\stackrel{\color{gray}?}=-\frac{496\,\zeta(5)}{3\,\pi^4}+\frac{2540\,\zeta(7)}{\pi^6}\\
&\mathcal B(5)\stackrel{\color{gray}?}=\frac{31\,\zeta(5)}{\pi^4}-\frac{3175\,\zeta(7)}{\pi^6}+\frac{35770\,\zeta(9)}{\pi^8}\\
&\mathcal B(6)\stackrel{\color{gray}?}=\frac{5842\,\zeta(7)}{5\,\pi^6}-\frac{57232\,\zeta(9)}{\pi^8}+\frac{515844\,\zeta(11)}{\pi^{10}}\end{align}\tag6$$
 [/INST] $$\mathcal{A}(m,n)=\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n}\text{d}z\qquad m+1>n>1$$
We have
$$\begin{aligned}
&\mathcal{A}\left(1,p+1\right)
=\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)\\
&\mathcal{A}\left(2,p+1\right)=
\frac{(p+1)(2^{p+2}-1)}{\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)\\
&\mathcal{A}\left(3,p+1\right)=
\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)
-\frac{(p+1)(p+2)(2^{p+3}-1)}{2\pi^{p+2}\cos\frac{p\pi}{2} }\zeta(p+3)\\
&\mathcal{A}\left(4,p+1\right)=
\frac{4(p+1)(2^{p+2}-1)}{3\cdot\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)
-\frac{(p+1)(p+2)(p+3)(2^{p+4}-1)}{6\cdot\pi^{p+3}\sin\frac{p\pi}{2} }\zeta(p+4)\\
&\mathcal{A}\left(5,p+1\right)=
\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)
-\frac{5(p+1)(p+2)(2^{p+3}-1)}{6\cdot\pi^{p+2}\cos\frac{p\pi}{2} } \zeta(p+3)
+\frac{(p+1)(p+2)(p+3)(p+4)(2^{p+5}-1)}{24\cdot\pi^{p+4}\cos\frac{p\pi}{2} }\zeta(p+5)\\
&\mathcal{A}\left(6,p+1\right)=
\frac{23(p+1)(2^{p+2}-1)}{15\cdot\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)
-\frac{(p+1)(p+2)(p+3)(2^{p+4}-1)}{3\cdot\pi^{p+3}\sin\frac{p\pi}{2} } \zeta(p+4)
+\frac{(p+1)(p+2)(p+3)(p+4)(p+5)(2^{p+6}-1)}{120\cdot\pi^{p+5}\sin\frac{p\pi}{2} }\zeta(p+6)
\end{aligned}$$
And substitute different p(can be a non-integer),your cases will done.

Proof:
For $n+m=2k$,consider the function
$$
f(z)=\frac{\tanh(z)^m}{z^n}
$$
And I get
$$
\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n} \text{d}z
=-\pi i\sum_{k=-\infty}^{\infty} \text{Res}(f,\frac{\pi}{2}(2k+1)i )
$$
For $n+m=2k+1$,consider
$$
f(z)=\frac{\tanh(z)^m}{z^n}\ln z
$$
And I get
$$
\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n} \text{d}z
=-\sum_{k=-\infty}^{\infty} \text{Res}(f,\frac{\pi}{2}(2k+1)i )
$$
I dont have time to write a full answer.Thanks.
(I will have exams for three days tomorrow.)

If we consider the function
$$
f(z)=\frac{\tanh z}{z^{1+p}}
$$
And use the keyhole contour,we can obtain
$$
(1-e^{-2p\pi i})\int_{0}^{\infty} \frac{\tanh z}{z^{1+p}}\text{d}z
= 2\pi i\sum_{k=-\infty}^{\infty}\text{Res}\left (\frac{\tanh z}{z^{1+p}}, \frac{\pi}{2}(2k+1)i  \right )
$$

A few examples
$$
\begin{aligned}
&\int_{0}^{\infty} \frac{\tanh z}{z^{3/2}} \text{d}z
=\frac{4-\sqrt{2} }{\sqrt{\pi} } \zeta\left ( \frac{3}{2}  \right )\\
&\int_{0}^{\infty} \frac{\tanh z}{z^{5/3}} \text{d}z
=\frac{2(2^{5/3}-1)}{\pi^{2/3}}\zeta\left ( \frac{5}{3}  \right )\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{2}} \text{d}z
= \frac{14}{\pi^2}\zeta(3)\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{3/2}} \text{d}z
= \frac{24-3\sqrt{2} }{2\pi^{3/2}} \zeta\left ( \frac{5}{2}  \right )\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{5/2}} \text{d}z
= \frac{80-5\sqrt{2} }{2\pi^{5/2}} \zeta\left ( \frac{7}{2}  \right )\\
&
\end{aligned}
$$
</s>