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.
$$ I(5,2)=\frac1{3}\Bigl(\frac{8}{5}-\frac{44}{63}\log2-3\frac{\zeta'(2)}{\zeta(2)} +5 \frac{\zeta'(4)}{\zeta(4)}-2\frac{\zeta'(6)}{\zeta(6)} \Bigr)$$
$$ I(7,2)=\frac1{45}\Bigl(\frac{7943}{420}-\frac{428}{45}\log2- 45\frac{\zeta'(2)}{\zeta(2)}+ 98\frac{\zeta'(4)}{\zeta(4)}- 70\frac{\zeta'(6)}{\zeta(6)}+ 17\frac{\zeta'(8)}{\zeta(8)}\Bigr)$$
$$ I(9,2)=\frac1{315}\Bigl(\frac{71077}{630}-\frac{10196}{165}\log2- 315\frac{\zeta'(2)}{\zeta(2)}+ 818\frac{\zeta'(4)}{\zeta(4)}- 798\frac{\zeta'(6)}{\zeta(6)}+ 357\frac{\zeta'(8)}{\zeta(8)}- 62\frac{\zeta'(10)}{\zeta(10)}\Bigr)$$
$$I(11,2)=\frac1{14175}\Bigl(\frac{2230796}{495}-\frac{10719068}{4095}\log2- 14175\frac{\zeta'(2)}{\zeta(2)}+ 41877\frac{\zeta'(4)}{\zeta(4)}- 50270\frac{\zeta'(6)}{\zeta(6)}+ 31416\frac{\zeta'(8)}{\zeta(8)}- 10230\frac{\zeta'(10)}{\zeta(10)}+1382\frac{\zeta'(12)}{\zeta(12)}\Bigr)$$
$$I(13,2)=\frac1{467775}\Bigl(\frac{270932553}{2002}-\frac{25865068}{315}\log2- 467775\frac{\zeta'(2)}{\zeta(2)}+ 1528371\frac{\zeta'(4)}{\zeta(4)}- 2137564\frac{\zeta'(6)}{\zeta(6)}+ 1672528\frac{\zeta'(8)}{\zeta(8)}- 771342\frac{\zeta'(10)}{\zeta(10)}+197626\frac{\zeta'(12) }{\zeta(12)}-21844\frac{\zeta'(14) }{\zeta(14)}\Bigr)$$

Here the initial denominators are chosen as the least common denominators of the $\frac{\zeta'(2k)}{\zeta(2k)}$ terms, such that inside the big parentheses, we have integer coefficients here. It turns out that the sequence of those denominators, viz. $1, 3, 45, 315, 14175, 467775$, coincides up to signs with A117972, the numerators of the rational numbers $\frac{\pi^{2n}\;\zeta'(-2n)}{\zeta(2n+1)}$.

Moreover, note that the coefficients of the $\frac{\zeta'(2k)}{\zeta(2k)}$ have alternating signs and always sum to 0, meaning that the closed forms can be decomposed into terms $\frac{\zeta'(2k)}{\zeta(2k)}-\frac{\zeta'(2k-2)}{\zeta(2k-2)}$, which are thus "exponential periods".

Further, the numerators of the $\frac{\zeta'(2n)}{\zeta(2n)}$ term of $I_{2n-1,2}$, viz. $1, -2, 17, -62, 1382, -21844$, coincide with A002430, the numerators of the Taylor series for $\tanh(x)$, which are also closely related to the rational values $\zeta(1-2n)$. All this seems rather interesting.

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