How to prove the following equality
$$\mathcal S=\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}\\=672\zeta(9)-240\zeta(2)\zeta(7)-105\zeta(3)\zeta(6)-168\zeta(4)\zeta(5)+24\zeta^3(3)$$
Where $H_n^{(r)}=\sum_{k=1}^n\frac1{k^r}$ is the harmonic number and $\zeta$ is The Riemann zeta function.
Here is my approach and would like to see different ways.
From here we have
$$\frac{\ln^4(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}\right)x^n$$
Multiply both sides by $\frac{\ln^4x}{4!x}$ then integrate from $x=0$ to $1$
and use the fact that $\frac1{4!}\int_0^1 x^{n-1}\ln^4x\ dx=\frac1{n^5}$ to have
\begin{align} \mathcal S&=\frac1{4!}\int_0^1\frac{\ln^4(1-x)\ln^4x}{x(1-x)}\ dx\\ &=\frac1{4!}\int_0^1\frac{\ln^4(1-x)\ln^4x}{x} dx+\frac1{4!}\underbrace{\int_0^1\frac{\ln^4(1-x)\ln^4x}{1-x}dx}_{1-x\mapsto x}\\ &=\frac2{4!}\int_0^1\frac{\ln^4(1-x)\ln^4x}{x}dx\overset{IBP}{=}\frac1{15}\int_0^1\frac{\ln^3(1-x)\ln^5x}{1-x}dx\tag1 \end{align}
The interesting part in this solution is that we can calculate the last integral without using the derivative of beta function:
We proved here
$$\int_0^1\frac{x^n\ln^m(x)\ln^3(1-x)}{1-x}dx=\frac1{4}\frac{\partial^m}{\partial n^m}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)$$
Set $m=5$ then let $n$ approach $0$ we get
$$\int_0^1\frac{\ln^3(1-x)\ln^5x}{1-x}\ dx\\=10080\zeta(9)-3600\zeta(2)\zeta(7)-1575\zeta(3)\zeta(6)-2520\zeta(4)\zeta(5)+360\zeta^3(3)$$
Substitute this result in $(1)$ we get the closed form of $\mathcal S.$
