Search for Ramanujan

We have $\begin{array}{rcl} \displaystyle \int_0^1 \frac{\ln x \ln^3(1-x)}{x}\,dx & = & \displaystyle \int_0^1 \frac{\ln(1-x) \ln^3x}{1-x}\,dx \\ & = & \displaystyle -\sum_{n=1}^\infty H_n \int_0^1 x^n \ln^3x\,dx \\ & = & \displaystyle 6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4} \\ & = & \displaystyle 6\sum_{n=1}^\infty \left(\frac{H_{n+1}}{(n+1)^4} - \frac{1}{(n+1)^5}\right) \\ & = & \displaystyle 6\sum_{n=2}^\infty \left( \frac{H_n}{n^4} - \frac{1}{n^5}\right) \\ & = & \displaystyle 6\sum_{n=1}^\infty \left( \frac{H_n}{n^4} - \frac{1}{n^5}\right) \\ & = & \displaystyle 6\left[ 3\zeta(5) - \tfrac16\pi^2\zeta(3) - \zeta(5)\right] \\ & = & \displaystyle 12\zeta(5) - \pi^2\zeta(3) \end{array}$ making the answer $12 + 2 + 2 + 1 + 1 = \boxed{18}$ .

Aliter :

$\displaystyle \int _{ 0 }^{ 1 }{ \dfrac { \log { \left( x \right) { \left( \log { \left( 1-x \right) } \right) }^{ 3 } } }{ x } \mathrm{d}x } = \lim_{a \to 0^+} \lim_{b \to 1} \left(\dfrac{\partial}{\partial a} \left(\dfrac{\partial^3}{\partial b^3} \operatorname{B} (a,b)\right) \right)$

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

$\implies A+B+C+D+E = \boxed{18}$