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][1]  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][2] 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.


  [1]: https://math.stackexchange.com/a/1583085
  [2]: https://math.stackexchange.com/a/1584566