I will open with the main theorem.
Let $s$ be a positive real number. Then the following equality holds.
$$\frac{\pi^2}{4}\int_0^{\infty} \dfrac{\tanh(x)\,\tanh(x s)}{x^2}\,dx = s \int_0^1 \ln\left(\frac{1-x}{1+x}\right) \ln\left(\frac{1-x^s}{1+x^s}\right) \,\frac{dx}{x}.\tag{1}$$
Proof.
First, loagrithmically differentiate the Weierstrass-form product of the hyperbolic cosine,
to obtain $\displaystyle \, \frac{1}{8x}\tanh(x)=\sum_{n=0}^{\infty} \frac1{\pi^2 (2n+1)^2+(2x)^2}\,.$
Also, since (elementarily) we have $\displaystyle \,\,\int_0^{\infty} \frac1{a^2+x^2}\,dx=\frac{\pi}{2a},\,\,$ it follows that
$\displaystyle \,\int_0^{\infty} \frac1{(a^2+x^2)(b^2+x^2)}\,dx=\frac1{b^2-a^2}\int_0^{\infty}\left(\frac1{a^2+x^2}-\frac1{b^2+x^2}\right)dx=\frac{\pi}{2}\,\frac1{ab(a+b)}.$
So,
$$\begin{align*}
\int_0^{\infty} \frac{\tanh(x)\,\tanh(x s)}{64 x^2 s}\,dx\\
&=\int_0^{\infty} \sum_{n,m=0}^{\infty} \dfrac1{(\pi^2(2n+1)^2+(2x)^2)(\pi^2(2m+1)^2+(2xs)^2)}\,\,dx\\
&=\frac1{2^2 (2s)^2} \sum_{n,m=0}^{\infty} \int_0^{\infty} \dfrac1{x^2+\left(\frac{\pi}{2}(2n+1)\right)^2}\,\dfrac1{x^2+\left(\frac{\pi}{2s}(2m+1)\right)^2}\,dx\\
&=\frac1{4\pi^2} \sum_{n,m=0}^{\infty} \dfrac1{(2n+1)\,(2m+1)\,\,((2n+1)+s(2m+1))}\\
&=\frac1{4\pi^2} \sum_{n,m=0}^{\infty} \dfrac1{(2n+1)(2m+1)} \int_0^1 x^{(2n+1)+s(2m+1)}\,\frac{dx}{x}\\
&=\frac1{16\pi^2} \int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^s}{1+x^s}\right)\,\frac{dx}{x}.
\end{align*}$$
Example no. 1
Set $s=1$. With the substitution $x \mapsto \frac{1-x}{1+x},$ we have
$$\begin{align*}
\frac{\pi^2}{4}\int_0^{\infty} \frac{\tanh^2 x}{x^2}\,dx\\
&=\int_0^1 \ln^2\left(\frac{1-x}{1+x}\right)\,\frac{dx}{x}\\
&=2\int_0^1 \frac{\ln^2 x}{1-x^2}\,dx\\
&=2\int_0^1 \ln^2 x \sum_{n=0}^{\infty} x^{2n} \,dx\\
&=\sum_{n=0}^{\infty} \frac{4}{(2n+1)^3}=\frac{7}{2}\zeta(3).
\end{align*}$$
Therefore,
$$ \,\,\int_0^{\infty} \frac{\tanh^2 x}{x^2}\,dx=\frac{14\zeta(3)}{\pi^2}.$$
Example no. 2
Set $s=3$. Employing the same substitution, we have
$$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^3}{1+x^3}\right)\,\frac{dx}{x}=2\int_0^1 \frac{\ln x}{1-x^2} \ln\left(\frac{x(x^2+3)}{3x^2+1}\right)dx$$
Now, I was able to obtain the following:
$$ f(a)=\int_0^1 \dfrac{\ln x \,\ln(a^2+x^2)}{1-x^2}dx=-\frac{\pi^2}{8}\ln(1+a^2)+\frac14 F\left(-\frac1{a^2}\right)-2\operatorname{Re} F\left(\frac{i}{a}\right),$$
where
$$ F(z)=\text{Li}_3(z)+2\text{Li}_3(1-z)-\ln(1-z)\text{Li}_2(1-z)-\frac{\pi^2}{6}\ln(1-z)-2\zeta(3).$$
It comes from the fact that whenever $|z|<1$, $\displaystyle \,\, F(z)=\sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n}\,z^n.$
Using that notation,
$\displaystyle \,\,2\int_0^1 \frac{\ln x}{1-x^2}\ln\left(\frac{x(x^2+3)}{3x^2+1}\right)dx=\frac{7}{2}\zeta(3)+\frac{\pi^2}{4}\ln3+2f(\sqrt{3})-2f\left(\frac1{\sqrt{3}}\right).$
The trouble was simplifying the hideous, monstrous expression. After several hours of painful simplification by hand, I finally obtained
$$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\,\ln\left(\frac{1-x^3}{1+x^3}\right)\frac{dx}{x}=\frac{\pi^2}{18}\ln3-\frac{2\pi^2}{3}\ln2+\frac{8}{3}\ln^3 2-\frac{7}{2}\zeta(3)-2\text{Li}_3\left(\frac14\right)\\+16\operatorname{Re} \text{Li}_3(1-i\sqrt{3})-\frac{2\pi}{3}\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).$$
(can it be simplified more?)
Or,
$$\int_0^{\infty} \frac{\tanh(x)\tanh(3x)}{x^2}\,dx=\frac23\ln3-8\ln2+\frac{32\ln^3 2}{\pi^2}-\frac{42\zeta(3)}{\pi^2}-\frac{24\text{Li}_3\left(\frac14\right)}{\pi^2}\\+\frac{192}{\pi^2}\operatorname{Re}\text{Li}_3(1-i\sqrt{3})-\frac{8}{\pi}\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).$$
If we set $s=4$ and again substitute $x \mapsto \frac{1-x}{1+x},\,$ we find
$$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\,\ln\left(\frac{1-x^4}{1+x^4}\right)\frac{dx}{x}=2\int_0^1 \frac{\ln x}{1-x^2} \ln\left(\frac{4x(1+x^2)}{x^4+6x^2+1}\right)\,\frac{dx}{x}
\\=-\frac{\pi^2}{2}\ln2+\frac{7}{2}\zeta(3)+2f(1)-2f(\sqrt{2}+1)-2f(\sqrt{2}-1)$$
I am not courageous enough to even begin simplifying that.
It seems that a closed form may exist for each natural $s$ (and therfore, for each $1/n$ where $n$ is natural).
For example, under the subsitution $x \mapsto \frac{1-x}{1+x}$, the expression $\displaystyle \frac{1-x^5}{1+x^5}$ turns into $\displaystyle \frac{x(x^4+10x^2+5)}{5x^4+10x^2+1}$.
Factorizing $x^4+10x^2+5=(x^2+5+2\sqrt{5})(x^2+5-2\sqrt{5})$ and $\displaystyle 5x^4+10x^2+1=5\left(x^2+\frac1{5+2\sqrt{5}}\right)\left(x^2+\frac1{5-2\sqrt{5}}\right)$,
it follows that
$$ \int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^5}{1+x^5}\right)\frac{dx}{x}=\frac72\zeta(3)+\frac{\pi^2}{4}\ln5+2f\left(\sqrt{5+2\sqrt{5}}\right)+2f\left(\sqrt{5-2\sqrt{5}}\right)-2f\left(\frac1{\sqrt{5+2\sqrt{5}}}\right)-2f\left(\frac1{\sqrt{5-2\sqrt{5}}}\right).$$
Similarly,
$$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^6}{1+x^6}\right)\frac{dx}{x}=\frac72\zeta(3)-\frac{\pi^2}{4}\ln6+2f(\sqrt{3})+2f\left(\frac1{\sqrt{3}}\right)-2f(1)-2f(2+\sqrt{3})-2f(2-\sqrt{3}).$$
A pattern can be seen.
It seems that we can always factorize the expression that results by applying $x\mapsto \frac{1-x}{1+x}$ to $\frac{1-x^n}{1+x^n}$ into factors of the form $x$ and $x^2+r^2$ ($r \in \mathbb{C}$), implying that there is indeed a closed form in terms of logarithms and polylogarithms
for $\int_0^{\infty} \tanh(x)\tanh(xn)/x^2\,\,dx$.
In fact, the roots $r$ in the factorization seem to always be $\tan(q \pi)$ (up to multiplication by an imaginary unit), where q is a rational number (whose denominator is $n$ when $n$ is odd and $2n$ when it is even) . For example, $\displaystyle \sqrt{2}-1=\tan\left(\frac{\pi}{8}\right),\,\,\,\,\sqrt{5-2\sqrt{5}}=\tan\left(\frac{\pi}{5}\right),\,\,\,\,\,2-\sqrt{3}=\tan\left(\frac{\pi}{12}\right)\,\,$etc.
I am too lazy to write a general formula that works for every natural number.
$\Large \mathbf {EDIT}$
For the sake of completeness, I add below the general formula, which works for every natural number.
Theorem 2.$\,$
Let $n$ be a positive integer. Define the function $F$ by
$$ F(z)=\text{Li}_3(z) + 2\text{Li}_3(1-z) - \ln(1-z)\text{Li}_2(1-z) -\frac{\pi^2}{6}\ln(1-z)-2\zeta(3),$$
where we assume the principal value of the logarithm.
Now, define the function $g$ by
$$g(z)=F(-z^2)-8 \operatorname{Re} F(i z).$$
Then
$$
\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^n}{1+x^n}\right)\frac{dx}{x}
=\frac{7}{2}\zeta(3)+\frac12 \sum_{k=1}^{n-1} (-1)^k g\left(\cot\left(\frac{k \pi}{2 n}\right)\right) .$$
This result, with the help of theorem $(1)$, can easily be seen to establish a closed form
for $\mathcal A(n) = \int_0^{\infty} \frac{\tanh(x) \tanh(n x)}{x^2}dx $ for each natural $n$.
Best Answer
Plot of the integrand for $a=8$.
As the integrand is even, $I(a) = 2 \int_{0}^\infty \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy$
We can break the limits of integration into three intervals, $0 < y< a(1-\delta)$, $a(1-\delta) < y< a(1+\delta)$, and $a(1+\delta)<y<\infty$, where $\delta$ is chosen such that $\frac{1}{a} \ll \delta \ll 1$, so that $\tanh{(a \delta)} \approx 1$.
Then, the first part is, $I_1(a) = 2 \int_{0}^{a-a\delta} \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy \approx 2 \int_{0}^{a-a\delta} \frac{(-1)}{y^2-a^2} dy = \frac{1}{a} \log\frac{2-\delta}{\delta} \approx \frac{1}{a} \log\frac{2}{\delta}$.
The second part is, $I_2(a) = 2 \int_{a-a\delta}^{a+a\delta} \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy \approx 2 \int_{a-a\delta}^{a+a\delta} \frac{\tanh{(2a(y-a))}}{2a(y-a)} dy = \frac{1}{a} \int_{-2a^2 \delta}^{2a^2 \delta} \frac{\tanh{z}}{z} dz$.
Here I use the known result $\int_0^b \frac{\tanh x}{x} \approx C + \log b$, where $C = \gamma + \log(\frac{4}{\pi})$. Here $\gamma$ is the Euler's constant.
Then, $I_2(a) \approx \frac{2}{a} (C + \log(2 a^2 \delta))$.
The third part is, $I_3(a) = 2 \int_{a+a\delta}^{\infty} \frac{\tanh{(y^2-a^2)}}{y^2-a^2} dy \approx 2 \int_{a+a\delta}^{\infty} \frac{1}{y^2-a^2} dy = \frac{1}{a} \log\frac{2+\delta}{\delta} \approx \frac{1}{a} \log\frac{2}{\delta}$.
After adding everything, $\delta$ cancels, and we get,
$$I(a) = I_1 (a) + I_2(a) + I_3 (a) \approx \frac{2}{a} (C + 2 \log 2 + 2 \log a) = \frac{2\gamma + 2 \log(16/\pi) + 4 \log a}{a}.$$
The numerical value of $\boxed{c_1 = 2 \gamma + 2\log(16/\pi)}$ is approximately $4.41015$.