where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(3)}=1+\frac1{2^3}+\frac1{3^3}+...+\frac1{n^3}$
I managed to prove $\quad\displaystyle\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)\quad$ using logarithmic integral and couple harmonic identities. other approaches are much appreciated.
you can find a different solution by Cornel in his book (Almost impossible integrals, sum and series).
using the fact that $\displaystyle\sum_{n=1}^\infty z^nH_n^{(3)}=\frac{\operatorname{Li}_3(z)}{1-z}\ $ divide both sides by $z$ then integrate from $z=0$ to $x$, we get $$\sum_{n=1}^\infty \frac{x^nH_n^{(3)}}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)\tag{1}$$
replace $x$ with $-x$ in(1), then divide both sides by $x$ and integrate from $0$ to $1$, we get
\begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}&=\operatorname{Li}_5(-1)-\underbrace{\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}\ dx}_{IBP}-\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ &=\operatorname{Li}_5(-1)+\operatorname{Li}_2(-1)\operatorname{Li}_3(-1)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ &=\frac38\zeta(2)\zeta(3)-\frac{15}{16}\zeta(5)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ \end{align} I proved here \begin{align} \int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx=\frac34\zeta(2)\zeta(3)-\frac{17}{16}\zeta(5) \end{align} which follows that
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$
BONUS:
By setting $x=-1$ in (1) we have
$$\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$
