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.
 A: We can have a nice generalization,
From
$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_n x^n$$
We have 
$$I_m=\int_0^1\frac{\ln(1+x)\ln^{m-1}x}{1-x}\ dx=\sum_{n=1}^\infty \overline{H}_n \int_0^1 x^n\ln^{m-1}x\ dx$$
$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_n}{(n+1)^m}$$
$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_{n-1}}{n^m}$$
$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_n+\frac{(-1)^n}{n}}{n^m}$$
$$=(-1)^{m-1}(m-1)!\left[\sum_{n=1}^\infty \frac{\overline{H}_n}{n^m}-\eta(m+1)\right]$$
Substitute
$$\sum_{n = 1}^\infty \frac{\overline H_n}{n^m} = \ln 2\zeta (m)  - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1)$$
We get

$$I_m=(-1)^{m}(m-1)!\left[\frac{1}{2} m \zeta (m + 1)-\ln 2\zeta (m)    - \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1)\right]$$


The generalization $\displaystyle \small \sum_{n = 1}^\infty \frac{\overline H_n}{n^m}$ can be found here (see Theorem 3.5 on page 9).
A: 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.
A: \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.
A: Computation of $\displaystyle U=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx$
\begin{align*}
U&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3 t}{1-t}\,dt\right)\,dx\\
&=-6\zeta(4)\ln 2+\int_0^1\int_0^1 \left(\frac{\ln^3(tx)}{(1+t)(1+x)}-\frac{\ln^3(tx)}{(1+t)(1-tx)}\right)\,dt\,dx\\
&=-6\zeta(4)\ln 2+6\left(\int_0^1\frac{\ln^2 t}{1+t}\,dt\right)\left(\int_0^1\frac{\ln x}{1+x}\,dx\right)+\\
&2\left(\int_0^1\frac{\ln^3 t}{1+t}\,dt\right)\left(\int_0^1\frac{1}{1+x}\,dx\right)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\
&=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\
&\overset{\text{IBP}}=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\
&=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\
&=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+24\zeta(5)-U\\
U&=\boxed{-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)}
\end{align*}
A: The following proof is very simple and short but it works for only the odd powers of $\ln(x):$
By integrating
$$\operatorname{Li}_2(-x)+\operatorname{Li}_2\left(-\frac1x\right)=-\frac{\ln^2(x)}{2}-2\eta(2)$$
repeatedly, we get
$$\operatorname{Li}_{2a}(-x)+\operatorname{Li}_{2a}\left(-\frac1x\right)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(x)$$
Using the integral form of the polylogarithm function, we have
\begin{gather*}
\operatorname{Li}_{2a}(-x)+\operatorname{Li}_{2a}\left(-\frac1x\right)\\
=\frac{-1}{(2a-1)!}\int_0^1\frac{-x\ln^{2a-1}(y)}{1+xy}\mathrm{d}y-\frac1{(2a-1)!}\int_0^1\frac{-\frac1{x}\ln^{2a-1}(y)}{1+\frac{y}{x}}\mathrm{d}y\\
=\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\frac{x}{1+xy}+\frac{1}{x+y}\right)\mathrm{d}y\\
=\frac{1}{(2a-1)!}\int_0^1\frac{(1+2xy+x^2)\ln^{2a-1}(y)}{(1+xy)(x+y)}\mathrm{d}y.
\end{gather*}
and so
\begin{equation}
\int_0^1\frac{(1+2xy+x^2)\ln^{2a-1}(y)}{(1+xy)(x+y)}\mathrm{d}y=-2(2a-1)!\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(x).
\end{equation}
Divide the latter identity by $1+x$ using $\int_0^1\frac{\ln^s(x)}{1+x}dx=(-1)^s s!\eta(s+1),$ we get
$$-2(2a-1)!\sum_{k=0}^a \eta(2a-2k)\eta(2k+1)=\int_0^1\ln^{2a-1}(y)\left(\int_0^1\frac{1+2xy+x^2}{(1+xy)(x+y)(1+x)}\mathrm{d}x\right)\mathrm{d}y$$
$$=\int_0^1\ln^{2a-1}(y)\left(\frac{2\ln(1+y)}{1-y}+\frac{\ln(1+y)}{y}-\frac{\ln(y)}{1-y}-\frac{2\ln(2)}{1-y}\right)\mathrm{d}y$$
$$=2\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1-x}\mathrm{d}x-(2a-1)!\eta(2a+1)-(2a)!\zeta(2a+1)$$
$$+2(2a-1)!\ln(2)\zeta(2a)$$
$$\Longrightarrow \int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1-x}\mathrm{d}x$$
$$=(2a-1)!\left(\frac12(2a-4^{-a}+1)\zeta(2a+1)-\ln(2)\zeta(2a)-\sum_{k=0}^a \eta(2a-2k)\eta(2k+1)\right).$$
