Help with logarithmic definite integral: $\int_0^1\frac{1}{x}\ln{(x)}\ln^3{(1-x)}\, dx$

We could use the series expansion $$\frac{\ln(1-x)}{1-x}=-\sum_{n=1}^\infty H_nx^n,$$ where $H_n=1+\frac12+\frac13+\ldots+\frac1n$ is the harmonic number. Then we get $$I=\int_0^1\frac{\ln(1-x)\ln^3x}{1-x}dx=-\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^3 x\,dx=6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4}.$$ The last series could be evaluated using Euler's formula $$\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ (see formula (20) here: http://mathworld.wolfram.com/HarmonicNumber.html): $$I=6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4}=6\sum_{n=1}^\infty \left(\frac{H_{n+1}}{(n+1)^4}-\frac{1}{(n+1)^5}\right)=6(2\zeta(5)-\zeta(2)\zeta(3)).$$

Introduce a two-parameter deformation $$\mathcal{I}(a,b)=\int_0^1 x^{a-1}\left[(1-x)^b-1\right] dx=\frac{\Gamma(a)\Gamma(b+1)}{\Gamma(a+b+1)}-\frac1a.$$ The integral we are looking for is obtained as \begin{align*} I&=\frac{\partial^3}{\partial b^3}\left[\frac{\partial\mathcal{I}(a,b)}{\partial a}\biggl|_{a=0}\right]_{b=0}=\\ &=\frac{\partial^3}{\partial b^3}\left[\frac{\gamma^2}{2}+\frac{\pi^2}{12}+\gamma\,\psi(1+b)+\frac{\psi^2(1+b)-\psi'(1+b)}{2}\right]_{b=0}=\\ &=12\zeta(5)-6\zeta(2)\zeta(3). \end{align*} At the last step, one needs to use that $\psi^{(n)}(1)=(-1)^{n+1}n!\,\zeta(n+1)$.

If you're interested in evaluating that $\psi'''(n+1)$ series you obtained in your original evaluation, a fun way to go about it is to use contours. A method popularized by Flajolet and Salvy in their famous paper on the topic.

$$\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

Begin with the kernel $$f(z)=\pi \cot(\pi z)\psi'''(-z)$$

This has poles at the positive integers, n; negative integers, -n, and 0.

The series at the positive integers is:

$$\frac{6}{(z-n)^{5}}-\frac{2\pi^{2}}{(z-n)^{3}}-\frac{\psi'''(n+1)}{z-n}+\cdot\cdot\cdot $$

Thus, the sum of the residues is:

$$Res(f,n)=\frac{6}{4!}\frac{d^{4}}{dz^{4}}\left[\frac{1}{(z-n)^{5}}\cdot \frac{1}{n}\right]-\frac{2\pi^{2}}{2!}\frac{d^{2}}{dz^{2}}\left[\frac{1}{(z-n)^{3}}\cdot \frac{1}{n}\right]-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

$$=6\sum_{n=1}^{\infty}\frac{1}{n^{5}}-2\pi^{2}\sum_{n=1}^{\infty}\frac{1}{n^{3}}-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

$$=6\zeta(5)-2\pi^{2}\zeta(3)-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}$$

the series at the negative integers has a simple pole.

$$Res(f, -n)=\sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}$$$

The residue at z=0 can be found by the Laurent expansion and noting the coefficient of the 1/z term:

$$Res(f,0)=24\zeta(5)$$

Now, put it altogether and set equal to 0 due to there being no poles in the contour:

$$6\zeta(5)-2\pi^{2}\zeta(3)+24\zeta(5)+\sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}-\sum_{n=1}^{\infty}\frac{\psi'''(n+1)}{n}=0$$

solve for the series in question. Note, since $\frac{\psi'''(n)-\psi'''(n+1)}{n}=\frac{6}{n^{5}}$, this means that $\displaystyle \sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}$ is equal to the series in question plus $6\zeta(5)$. In other words by calling our series S:

$\displaystyle \sum_{n=1}^{\infty}\frac{\psi'''(n)}{n}=S+6\zeta(5)$

$$-2S+24\zeta(5)-2\pi^{2}\zeta(3)=0$$

$$S=12\zeta(5)-\pi^{2}\zeta(3)$$