<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] \begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n}
&=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3)
\end{align}

Let $\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_3(-z)}{z}$. Then at the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
&=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{6(-1)^n}{z(z-n)^5}+\frac{6(-1)^n\zeta(2)}{z(z-n)^3}+(-1)^n\frac{(33/2)\zeta(4)+6H_n^{(4)}}{z(z-n)}\right]\\
&=6\sum^\infty_{n=1}\frac{(-1)^n}{n^5}+6\zeta(2)\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{33}{2}\zeta(4)\sum^\infty_{n=1}\frac{(-1)^n}{n}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n}\\
&=-\frac{45}{8}\zeta(5)-\frac{9}{2}\zeta(2)\zeta(3)-\frac{33}{2}\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n}
\end{align}
At zero,
$${\rm Res}(f,0)=24\zeta(5)$$
At the negative integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,-n)
&=\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n)}{n}\\
&=6\zeta(4)\ln{2}-6\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{n-1}^{(4)}}{n}\\
&=\frac{45}{8}\zeta(5)+6\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}\\
\end{align}
Since the sum of residues is zero,
\begin{align}
12\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}=-24\zeta(5)+\frac{21}{2}\zeta(4)\ln{2}+\frac{9}{2}\zeta(2)\zeta(3)\\
\end{align}
This implies that
\begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n}
&=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3)
\end{align}
Refer to this paper if you have any doubts.
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