Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tanh}^3 z}{z^2} \log(-z) \, dz , $$ where the clockwise contour tightly encircles the positive real axis, which coincides with the branch cut of the logarithm. The reason that this integral is equivalent is because the branch jump across the real line of $\frac{1}{2\pi i} \log(-z)$ is precisely $1$.
The integrand has poles at all $z=\pm i\pi(k+\frac{1}{2})$, $k=0,1,2,\ldots$. Evaluating the residues we find \begin{align*} I_{3,2} &= \sum_{k=0}^\infty \frac{8\log\pi(k+\frac{1}{2})}{\pi^2 (2k+1)^2} - \frac{96 \log\pi(k+\frac{1}{2})-80}{\pi^4 (2k+1)^4} \\ &= \frac{5}{6} - \gamma - \frac{19 \log 2}{15} + 12 \log A - \log\pi + \frac{90 \zeta'(4)}{\pi^4} \\ &= 1.1547853133231762640590704519415261475352370924508924890\ldots \end{align*} The last two lines can be checked with Wolfram Alpha, where $\gamma$ is the Euler-Mascheroni constant, and $A$ is the Glaisher constant.
Edit: Using $\gamma =12\,\log(A)-\log(2\pi)+\frac{6}{\pi^2}\,\zeta'(2)$, this can be simplified to $$I_{3,2}=\frac{5}{6} - \frac{4\log 2}{15} -\frac{6\zeta'(2)}{\pi^2}+ \frac{90 \zeta'(4)}{\pi^4},$$ that is $$\boxed{I_{3,2}=\frac{5}{6} - \frac{4\log 2}{15} -\frac{\zeta'(2)}{\zeta(2)}+ \frac{\zeta'(4)}{\zeta(4)}}. $$ And this form may hint to the existence of similar closed forms for other integrals $I_{n,m}$ with $n+m$ odd.
Edit: Indeed... Numerically, it looks like e.g. $$ I(5,2)=\frac1{3}\Bigl(\frac{8}{5}-\frac{44}{63}\log2-3\frac{\zeta'(2)}{\zeta(2)} +5 \frac{\zeta'(4)}{\zeta(4)}-2\frac{\zeta'(6)}{\zeta(6)} \Bigr)$$ $$ I(7,2)=\frac1{45}\Bigl(\frac{7943}{420}-\frac{428}{45}\log2- 45\frac{\zeta'(2)}{\zeta(2)}+ 98\frac{\zeta'(4)}{\zeta(4)}- 70\frac{\zeta'(6)}{\zeta(6)}+ 17\frac{\zeta'(8)}{\zeta(8)}\Bigr)$$ $$ I(9,2)=\frac1{315}\Bigl(\frac{71077}{630}-\frac{10196}{165}\log2- 315\frac{\zeta'(2)}{\zeta(2)}+ 818\frac{\zeta'(4)}{\zeta(4)}- 798\frac{\zeta'(6)}{\zeta(6)}+ 357\frac{\zeta'(8)}{\zeta(8)}- 62\frac{\zeta'(10)}{\zeta(10)}\Bigr)$$
(here the initial denominators are chosen such that all the coefficients of the $\frac{\zeta'(2k)}{\zeta(2k)}$ are integers.)
BTW it turns out that those coefficients always seem to sum to 0. :)
