The definition of $\eta(s)$ is
$$\eta(s)=(1-2^{1-s})\zeta(s)$$
Take the derivative, and we get by product rule
$$\eta'(s)=2^{1-s}\ln(2)\zeta(s)+(1-2^{1-s})\zeta'(s)$$
$\eta(s)$ also has the following series representation
$$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$
Again, take the derivative
$$\eta'(s)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^s}$$
Now, we can obtain a closed form
$$\eta'(2)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}=2^{1-2}\ln(2)\zeta(2)+(1-2^{1-2})\zeta'(2)$$
Note that
$$\zeta'(2)=\frac{1}{6}\pi^2(-12\ln(A)+\ln(2\pi)+\gamma),~\zeta(2)=\frac{\pi^2}{6}$$
$$\begin{align}
\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}&=2^{-1}\ln(2)\zeta(2)+(1-2^{-1})\zeta'(2)\\
&=\frac{\pi^2}{12}\ln(2)+\frac{\pi^2}{12}(-12\ln(A)+\ln(2\pi)+\gamma)\\
&=\frac{\pi^2}{12}(\ln(4\pi)-12\ln(A)+\gamma)
\end{align}$$
We proved in this solution that
$$\mathcal{I}=\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2(2)\tag1$$
and we proved here that
$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag2$$
By $(2)$ we get
$$\mathcal{I}=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1\frac{ x^{2n}}{1+x}dx$$
Using the identity $$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
it follows that
$$\mathcal{I}=-2\ln(2)\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}+2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{2}}{2n+1}$$
$$=-2\ln(2)\mathcal{S}_1-2\mathcal{S}_2+2\mathcal{S}_3\tag3$$
For $\mathcal{S}_1$ and $\mathcal{S}_3$, we use the classical identity:
$$\sum_{n=1}^\infty(-1)^n f(2n)=\Re\sum_{n=1}^\infty i^n f(n)$$
Therefore
$$\mathcal{S}_1=\Re\sum_{n=1}^\infty i^n\frac{H_n}{n+1}=\Re\left\{\frac{\ln^2(1-i)}{i}\right\}=-\frac{\pi}{8}\ln(2)\tag4$$
where we used $\sum_{n=1}^\infty x^n\frac{H_n}{n+1}=\frac{\ln^2(1-x)}{x}$ which follows from integrating $\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$.
Similarly,
$$\mathcal{S}_3=\Re\sum_{n=1}^\infty i^n\frac{H_n^2}{n+1}$$
Using the generating function
$$\sum_{n=1}^\infty x^{n}\frac{ H_n^{2}}{n+1}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}$$
it follows that
$$\mathcal{S}_3=\Re\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3i}\right\}$$
$$=\Im\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3}\right\}$$
$$=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{16}\ln^2(2)+\frac{5}{96}\pi^3\tag5$$
Plug the results of $(4)$ and $(5)$ in $(3)$ we get
$$\mathcal{I}=4\Im\{\operatorname{Li}_3(1-i)\}+\ln(2)\ G+\frac{5\pi}{8}\ln^2(2)+\frac{5}{48}\pi^3-2\mathcal{S}_2\tag6$$
By $(1)$ and $(6)$ we get
$$\mathcal{S}_2=\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{8}\ln^2(2)+\frac{3}{64}\pi^3$$
Best Answer
This is not a pleasant result since $$S=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^6{{2n}\choose{n}}}=\frac{1}{2} \, _7F_6\left(1,1,1,1,1,1,1;\frac{3}{2},2,2,2,2,2;-\frac{1}{4}\right)$$ which is $$S=0.49746116570182992164335272848313158317029710406952\cdots$$
Similarly $$\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{{n^3}{2n\choose{n}}}t^n=\frac{1}{4} t \,\, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;-\frac{t}{4}\right)$$ which, if simplified, leads to polylogarithms.
However, if you want to compute $$I=\int\frac 1 x\sinh ^{-1}\left(\frac{\sqrt{x}}{2}\right)^2\,dx$$ $$\color{red}{x=4 \sinh^2(w)} \quad \implies \quad I=2\int w^2 \coth(w)\,dw$$ $$I=-\frac{2 }{3}w^3+2 w^2 \log \left(1-e^{2 w}\right)+2 w \text{Li}_2\left(e^{2 w}\right)-\text{Li}_3\left(e^{2 w}\right)$$ $$I=-\zeta (3)+w^2+\frac{w^4}{6}-\frac{w^6}{135}+\frac{w^8}{1890}-\frac{w^{10}}{23 625}+O\left(w^{12}\right)$$
But, using the series expansion of $\coth(w)$ and integrating termwise, $$\color{red}{I=-\zeta(3)+\sum_{n=0}^\infty \frac{2^{2 n} B_{2 n} }{(n+1) (2 n)!}w^{2( n+1)}}$$ or $$\color{red}{I=-\zeta(3)+w^2+2\sum_{n=1}^\infty(-1)^{n+1}\,\frac{ \zeta (2 n)}{(n+1)\, \pi ^{2 n}}w^{2( n+1)}}$$
As @Gary commented, the power series converges for $|w|<\pi$ (this is a serious limitation even if this corresponds to $|x| < 533$).
$$I=-\zeta (3)+w^2+\frac{w^4}{6}-\frac{w^6}{135}+\frac{w^8}{1890}-\frac{w^{10}}{23 625}+O\left(w^{12}\right)$$
Warning
From a numerical point of view, you will face problems when $n$ is "large" (because of the order of operations). For this case, use the asymptotics $$ \frac{2^{2 n} B_{2 n} }{(n+1) (2 n)!}\sim (-1)^{n+1}\, \frac{2 \,\pi ^{-2 n}}{n+1}$$
(which is in a relative error smaller than $0.0001$% as soon as $n>9$) could be useful.
So, since we face an alternating series, we know in advance how many $p$ terms have to added if we write $$I=-\zeta(3)+\sum_{n=0}^p \frac{2^{2 n} B_{2 n} }{(n+1) (2 n)!}w^{2( n+1)}+\sum_{n=p+1}^\infty \frac{2^{2 n} B_{2 n} }{(n+1) (2 n)!}w^{2( n+1)}$$ $$\frac{2 \pi ^{-2 (p+1)} }{p+2}w^{2 (p+2)} ~\leq ~\epsilon \quad \implies \quad p \sim -2-\frac{1}{2 \log \left(\frac{w}{\pi }\right)}W\left(-\frac{4 \pi ^2 }{\epsilon }\log \left(\frac{w}{\pi }\right)\right)$$
As a consequence of $|w|<\pi$, $p$ varies extremely fast since, close to the limit, $$p=2\left(\frac{ \pi ^2}{\epsilon }-1\right)+\frac{8 \pi ^3 }{\epsilon ^2}(w-\pi)+O\left((w-\pi )^2\right)$$