For $n\geqslant m>1$, the integral $$I_{n,m}:=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx$$ converges. If $m$ and $n$ are both even or both odd, we can use the residue theorem to easily evaluate it in terms of odd zeta values, since the integrand then is a nice even function. For example, defining $e_k:=(2^k-1)\dfrac{\zeta(k) }{\pi^{k-1}}$, we have

$$ \begin{align} I_{2,2}&= 2e_3 \\ I_{4,2}&=\dfrac43(2e_3-3e_5) \\ I_{6,2}&=\dfrac2{15}(23e_3-60e_5+45e_7) \\ I_{4,4}&=\dfrac1{3}( -16e_5+60e_7) \\ I_{6,4}&=\dfrac4{15}(-23e_5+150e_7-210e_9) \\ I_{3,3}&= -e_3+6e_5 \\ I_{5,3}&= -e_3+10e_5-15e_7 \\ I_{5,5}&= e_5-25e_7 +70e_9 \\ &etc. \end{align}$$

But:

Is there a closed form for $I_{3,2}=\int\limits_0^\infty\dfrac{\tanh^3(x)}{x^2}dx$?

I am not sure at all whether nospoon's method or one of the other *ad hoc* approaches can be generalized to tackle this.

If the answer is positive, there might be chances that $I_{\frac32,\frac32}$ and the like also have closed forms.