Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$? 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.
 A: Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that
$$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2
\frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$
Integrate term-wise to get (after some regrouping)
$$\int_0^{\infty}\frac{\tanh^3x}{x^2}\,dx=\sum_{k=2}^{\infty}(-1)^k(8k^3+4k)\log k.$$
Perhaps there is some hope in view of what I see as
$$\sum_{k=2}^{\infty}(-1)^k\log k=\log\sqrt{\frac2{\pi}};$$
which is a Wallis-type formula
$$\frac23\cdot\frac45\cdot\frac67\cdots\frac{2k}{2k+1}\cdots=\sqrt{\frac2{\pi}}.$$
UPDATE. Using a divergent series approach on $\sum_k(-1)^kk^c$ and Will Sawin's comment, we can complete the solution as follows. Start with $\sum_{k\geq1}(-1)^kk^c=\zeta(-c)(2^{c+1}-1)$ to get the derivate
$$\sum_{k\geq2}(-1)^ck^c\log k=-\zeta'(-c)(2^{c+1}-1)+\zeta(-c)2^{c+1}\log2.$$
Now, apply the following facts: $\zeta(-1)=-\frac1{12},\, \zeta'(-1)=\frac1{12}-\log A,\, \zeta(-3)=\frac1{120}$ and
$$\zeta'(-3)=\frac1{120}\log(2\pi)-\frac{11}{720}+\frac1{120}\gamma-\frac{3\zeta'(4)}{4\pi^4}.$$
Next, put all these together and simplify
\begin{align}\sum_{k\geq0}(-1)^k(8k^3+4k)\log k
&=[-120\zeta'(-3)+128\zeta(-3)\log 2]+[-12\zeta'(-1)+16\zeta(-1)\log 2] \\
&=\frac56-\gamma-\log\pi-\frac{19}{15}\log2+12\log A+\frac{\zeta'(4)}{\zeta(4)}.
\end{align}
