Prove $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$ How to prove this advanced harmonic series of weight 6?

$$S=\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$$
  where $H_k^{(p)}=1+\frac1{2^p}+\cdots+\frac1{k^p}$ is the $k$th generalized harmonic number of order $p$.

This result can be found in the book Almost impossible integrals, sums and series page $414-419$ using pure series manipulations but can be done in different ways? 
I will post my approach soon.
 A: Using the fact that
 $$\sum_{n=1}^\infty H_n^{(3)}x^n=\frac{\operatorname{Li}_3(x)}{1-x}$$
Divide both sides by $x$ then integrate, we get
\begin{align}
\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}&=\int \frac{\operatorname{Li}_3(x)}{x(1-x)}\ dx=\int \frac{\operatorname{Li}_3(x)}{x}\ dx+\underbrace{\int \frac{\operatorname{Li}_3(x)}{1-x}\ dx}_{IBP}\\
&=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)+\int\frac{\ln(1-x)\operatorname{Li}_2(x)}{x}\ dx\\
&=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)+c
\end{align}
Set $x=0$, we get $c=0$. 
Therefore

$$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)\tag{1}$$

Now multiply both sides of $(1)$ by $-\frac{\ln(1-x)}{x}$ then integrate, we get
\begin{align}
S&=\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}\int_0^1 -x^{n-1}\ln(1-x)\ dx=\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}\left(\frac{H_n}{n}\right)=\sum_{n=1}^\infty \frac{H_n^{(3)}H_n}{n^2}\\
&=-\int_0^1\frac{\ln(1-x)\operatorname{Li}_4(x)}{x}\ dx+\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_3(x)}{x}\ dx+\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}^2_2(x)}{x}\ dx\\
&=-\sum_{n=1}^\infty\frac1{n^4}\int_0^1 x^{n-1}\ln(1-x)\ dx+\sum_{n=1}^\infty\frac1{n^3}\int_0^1 x^{n-1}\ln^2(1-x)\ dx-\frac16\operatorname{Li}^3_2(1)\\
&=-\sum_{n=1}^\infty\frac1{n^4}\left(-\frac{H_n}{n}\right)+\sum_{n=1}^\infty\frac1{n^3}\left(\frac{H^2_n}{n}+\frac{H_n^{(2)}}{n}\right)-\frac16\zeta^3(2)\\
&=\sum_{n=1}^\infty \frac{H_n}{n^5}+\sum_{n=1}^\infty \frac{H^2_n}{n^4}+\sum_{n=1}^\infty \frac{H^{(2)}_n}{n^4}-\frac{35}{48}\zeta(6)
\end{align}
By substituting:
$$\sum_{n=1}^\infty \frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$
$$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$
$$\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\zeta^2(3)-\frac13\zeta(6)$$
we get the closed form
$$\boxed{S=\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)}$$
Note: The first sum can be found using Euler identity, the second sum is evaluated here and the third sum is here.
