On the Euler sum $\sum \limits_{n=1}^{\infty} \frac{H_n^{(4)} H_n^2}{n^6}$ Here is an Euler sum I ran into.
$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(4)} \mathcal{H}_n^2}{n^6}$$
where $\mathcal{H}_n^{(s)}$ is the generalised harmonic number of order $s$. I have no idea to what this evaluates neither do I have the appropriate techniques to begin cracking it.
We have seen quite plenty Euler sums but I don't think we have seen something like this before. Correct me if I am wrong!
Any help?
Edit:  Maybe this link ( inverse sin ) is helpful for an integral approach.

Update (by Editor): By stuffle relations of Multiple Zeta Values the result is: >$$S=\small -\frac{3}{20} \pi ^4 \zeta(6,2)-\frac{5}{3} \pi ^2 \zeta(8,2)+\frac{143}{4} \zeta(10,2)-6\zeta(8,2,1,1)+\frac{\zeta (3)^4}{3}-\frac{23 \pi ^6 \zeta (3)^2}{1620}-\frac{8}{45} \pi ^4 \zeta (5) \zeta (3)-\frac{31}{3} \pi ^2 \zeta (7) \zeta (3)+\frac{2531 \zeta (9) \zeta (3)}{18}-\frac{9 \pi ^2 \zeta (5)^2}{2}+\frac{1115 \zeta (5) \zeta (7)}{8}-\frac{964213 \pi ^{12}}{8756748000}$$
 A: A partial answer, designed to simplify further attempts. By Thm 4.2 of Flajolet and Salvy the Euler sum $S_{42,6}=\sum_{n\geq 1}\frac{H_n^{(4)}H_n^{(2)}}{n^6}$ can be written in terms of values of the $\zeta$ function. It follows that it is enough to evaluate
$$ \mathcal{J} = \sum_{n\geq 1}\frac{H_n^{(4)}\left(H_n^2-H_n^{(2)}\right)}{n^6}=\sum_{n\geq 1}\frac{H_n^{(4)}}{n^5}\int_{0}^{1}\log^2(1-x)x^{n-1}\,dx. $$
and in order to deal with $H_n^{(4)}$ we may steal a very interesting idea from nospoon:
$$ \arcsin(x)^6 = \frac{45}{8}\sum_{k\geq 1}\left(\left(H_{k-1}^{(2)}\right)^2-\color{red}{H_{k-1}^{(4)}}\right)\frac{4^k x^{2k}}{\color{red}{k^2}\binom{2k}{k}} $$
see also equation $(21)$ here. It might be useful to run an implementation of the PSLQ algorithm for reaching a conjectural closed form for $S_{411,6}$; in the meanwhile I'll keep digging in the literature.
The question can also be solved by evaluating
$$ \sum_{a,b,c\geq 1}\frac{(2 a+b) (b+2 c) \left(b^2+2 b c+2 c^2\right)}{a^2 b^4 (a+b)^2 c^4 (b+c)^4} $$
or
$$ \iint_{(0,1)^2}\frac{\text{Li}_4(xy)\log^4(y)\log^2(1-x)}{xy(1-xy)}\,dx\,dy ,$$
good luck. Apart from the reflection formulas for the tetralogarithm, we may notice that 
$$\iint_{(0,1)^2}\frac{\text{Li}_4(xy)\log^4(y)\log^2(1-x)}{xy}\,dx\,dy = 24\int_{0}^{1}\frac{\text{Li}_9(x)\log^2(1-x)}{x}\,dx $$
depends on the Euler sum $S_{2,10}$, which might be irreducible just like $S_{2,6}$ and $S_{2,8}$. So, unless we have some cancellation, we cannot be sure that the original Euler sum with weight $12$ can be expressed in terms of values of the $\zeta$ function only. Such impossibility for $\zeta(6,4,1,1)$ has been proved by Houches here.
