Alternating second power Euler sum $\sum_{k\geq 1} \frac{(H'_k)^2}{k^2}$

This is going to be almost the full solution because I lack time know. We start from the following identity: \begin{equation} H_k^{'} = (-\log(2)) (1-(-1)^k) + \int\limits_0^{-1} Li_1(\xi) k \xi^{k-1} d\xi \end{equation} Now we square the above and divide by $k^2$ and sum from one to infinity. We end up with three terms. The first one is trivial the second one is a one dimensional integral which is doable because it contains only a product of logs and the last term is a two dimensional integral. We have: \begin{eqnarray} &&\sum\limits_{k=1}^\infty \frac{[H_k^{'}]^2}{k^2} = \\ &&\log(2)^2 \frac{\pi^2}{2} - 2 \log(2) \frac{7}{8} \zeta(3) + \int\limits_{[0,-1]^2} \frac{\log(1-\xi_1) \log(1-\xi_2) }{1-\xi_1 \xi_2} d\xi_1 d\xi_2=\\ &&3 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{43 \pi ^4}{1440}+\frac{\log ^4(2)}{8}+\frac{3}{8} \pi ^2 \log ^2(2)-\int\limits_0^{-1} \frac{\log(1-\xi_1) Li_2(\frac{2 \xi_1}{-1+\xi_1})}{\xi_1} d\xi_1 =\\ &&3 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{43 \pi ^4}{1440}+\frac{\log ^4(2)}{8}+\frac{3}{8} \pi ^2 \log ^2(2)+\\ &&\int\limits_0^1 \log(1-\frac{u}{2}) Li_2(u) \left(\frac{1}{2-u} + \frac{1}{u}\right)d u=\\ &&3 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{7 \pi ^4}{160}+\frac{\log ^4(2)}{8}+\frac{3}{8} \pi ^2 \log ^2(2) + \sum\limits_{m=1}^\infty \frac{H_m^2}{m^2} \frac{1}{2^m} \end{eqnarray} Now in the third line from the top we integrated over $\xi_2$ and then used Landen's identity and then carried out the doable integrals. In the fourth line from the top we substituted for $2 \xi_1/(-1+\xi_1)$ and in the last line we evaluated the remaining integral by expanding the integrand in a series and integrating term by term. Now we use my answer to this question Double harmonic sum $\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$ . We have: \begin{eqnarray} \sum\limits_{m=1}^\infty \frac{H_m^2}{m^2} \frac{1}{2^m} &=& -\frac{1}{3} \int\limits_0^{\frac{1}{2}} \frac{[\log(1-\xi)]^3}{\xi}d\xi + \frac{1}{2} [Li_2(\frac{1}{2})]^2 + Li_4(\frac{1}{2}) \\ &=&-\frac{1}{3} \int\limits_{\frac{1}{2}}^1 \frac{Li_0(\xi)}{\xi} \log(\xi)^3d\xi + \frac{1}{2} [Li_2(\frac{1}{2})]^2 + Li_4(\frac{1}{2}) \\ &=& -\text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{4} \zeta (3) \log (2)+\frac{37 \pi ^4}{1440}-\frac{\log ^4(2)}{24}+\frac{1}{24} \pi ^2 \log ^2(2) \end{eqnarray} Bringing everything together we have: \begin{eqnarray} \sum\limits_{k=1}^\infty \frac{[H_k^{'}]^2}{k^2} =2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{13 \pi ^4}{720}+\frac{\log ^4(2)}{12}+\frac{5}{12} \pi ^2 \log ^2(2) \end{eqnarray} as expected.

Note: As usual in those cases the consecutive terms correspond to decompositions of the weight (in this case number four) into sums of integers. Apart from the first term on the rhs we have $(4)$, $(1,1,1,1)$ and $(2,1,1)$. Oddly enough the term $(3,1)$ has canceled out in this case. Why?!

Numerically, this is $$ -\frac{13}{8}\zeta(4)+\frac5{2}\zeta(2)\log^22+\frac1{12}\log^42+2\mathrm{Li}_4({\textstyle\frac12}), $$ from "Experimental Evaluation of Euler Sums" by Bailey, Borwein and Girgensohn.

This can be found with an integer relation algorithm applied to the sum evaluated to high precision using this integral: $$ \int_0^1\frac{\log(\frac1z)dz}{z(1-z)}\left(-\zeta(2)+\log^22+2\log(1-z)\log\left(\frac{1+z}{2}\right)+2\mathrm{Li}_2\left(\frac{1-z}{2}\right)+\mathrm{Li}_2(z)\right). $$