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.

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

$$ \begin{align} I_{2,2}&= 2e_3 \\ \\ I_{4,2}&= \dfrac83e_3-4e_5 \\ \\ I_{4,4}&= -\dfrac{16}{3}e_5+20e_7 \\ \\ I_{6,2}&=\dfrac{46}{15}e_3-8e_5+6e_7 \\ \\ I_{6,4}&=-\dfrac{92}{15}e_5+40e_7-56e_9 \\ \\ I_{6,6}&=\dfrac{46}{5}e_7-112e_9+252e_{11} \\ \\ I_{3,3}&= -e_3+6e_5 \\ \\ I_{5,3}&= -e_3+10e_5-15e_7 \\ \\ I_{5,5}&= e_5-25e_7 +70e_9 \\ \\ &etc. \end{align}$$

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 used here 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.

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.

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.