<s>[INST] Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$ Does the following series or integral have a closed-form

\begin{equation}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx
\end{equation}

where $\Psi_3(x)$ is the polygamma function of order $3$.

Here is my attempt. Using equation (11) from Mathworld Wolfram:
\begin{equation}
\Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right)
\end{equation}
I got
\begin{equation}
\Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right)
\end{equation}
then
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=6\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\left(\zeta(4)-H_{n}^{(4)}\right)\\
&=6\zeta(4)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
&=\frac{\pi^4}{15}\ln2-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\
\end{align}
From the answers of this OP, the integral representation of the latter Euler sum is
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=\int_0^1\int_0^1\int_0^1\int_0^1\int_0^1\frac{dx_1\,dx_2\,dx_3\,dx_4\,dx_5}{(1-x_1)(1+x_1x_2x_3x_4x_5)}
\end{align}
or another simpler form
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=-\int_0^1\frac{\text{Li}_4(-x)}{x(1+x)}dx\\
&=-\int_0^1\frac{\text{Li}_4(-x)}{x}dx+\int_0^1\frac{\text{Li}_4(-x)}{1+x}dx\\
&=\text{Li}_5(-1)-\int_0^{-1}\frac{\text{Li}_4(x)}{1-x}dx\\
\end{align}
I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.

Edit :
Using the integral representation of polygamma function
\begin{equation}
\Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx
\end{equation}
then we have
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1\frac{x^{n}}{1-x}\ln^3 x\,dx\\
&=-\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n}}{n}\cdot\frac{\ln^3 x}{1-x}\,dx\\
&=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\
\end{align}
I am looking for an approach to evaluate the above integral without using residue method or double summation.
 [/INST] Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.
We can break up the integral into
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\
=&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}{\rm d}x\\
=&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x-\int^1_0\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x\\
=&\frac{15}{16}\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\frac{1}{16}\int^1_0\frac{x^{-1/2}\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\
=&\frac{15}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(1,0^{+})-\frac{1}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(0.5,0^{+})
\end{align}
After differentiating and expanding at $b=0$ (with the help of Mathematica),
\begin{align}
&\frac{\partial^4\beta}{\partial a^3 \partial b}(a,0^{+})\\
=&\left[\frac{\Gamma(a)}{\Gamma(a+b)}\left(\frac{1}{b}+\mathcal{O}(1)\right)\left(\left(-\frac{\psi_4(a)}{2}+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)\right)b+\mathcal{O}(b^2)\right)\right]_{b=0}\\
=&-\frac{1}{2}\psi_4(a)+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)
\end{align}
Therefore,
\begin{align}
-&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\
=&-\frac{15}{32}\psi_4(1)+\frac{45}{16}\psi_1(1)\psi_2(1)+\frac{1}{32}\psi_4(0.5)+\frac{1}{8}\psi_3(0.5)\ln{2}-\frac{3}{16}\psi_1(0.5)\psi_2(0.5)\\
=&-12\zeta(5)+\frac{3\pi^2}{8}\zeta(3)+\frac{\pi^4}{8}\ln{2}
\end{align}
The relation between $\psi_{m}(1)$, $\psi_m(0.5)$ and $\zeta(m+1)$ is established easily using the series representation of the polygamma function.
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