Log and exponential integral

With the substitution $u = 1 - e^{-x}$ the integral becomes $\begin{array}{rcl} \displaystyle I & = & \displaystyle \int_0^\infty \frac{x^3 \ln(e^x-1)}{e^x-1}\,dx \; = \; -\int_0^1 \frac{\ln^3(1-u)}{u}\ln\big(\tfrac{u}{1-u}\big)\,du \\ & = & \displaystyle \int_0^1 \frac{\ln^4(1-u)}{u}\,du - \int_0^1 \frac{\ln^3(1-u)\ln u}{u}\,du \; = \; \int_0^1 \frac{\ln^4u}{1-u}\,du - \int_0^1 \frac{\ln(1-u)\ln^3u}{1-u}\,du \end{array}$ Now $\int_0^1 \frac{\ln^4u}{1-u}\,du \; =\; \sum_{n=0}^\infty \int_0^1 u^n \ln^4u\,du \; = \; \sum_{n=0}^\infty \frac{24}{(n+1)^5} \; =\; 24\zeta(5) \;,$ and $\begin{array}{rcl} \displaystyle \int_0^1 \frac{\ln(1-u)\ln^3u}{1-u}\,du & = & \displaystyle -\sum_{n=1}^\infty H_n\int_0^1u^n \ln^3u\,du \\ & = & \displaystyle 6\sum_{n=1}^\infty \frac{H_n}{(n+1)^4} \; = \; 6\sum_{n=1}^\infty \left[ \frac{H_{n+1}}{(n+1)^4} - \frac{1}{(n+1)^5}\right] \\ & = & \displaystyle 6\sum_{n=1}^\infty \left[ \frac{H_n}{n^4} - \frac{1}{n^5}\right] \; = \; 6\left[(3\zeta(5) - \tfrac16\pi^2\zeta(3)) - \zeta(5)\right] \\ & = & \displaystyle 12\zeta(5) - \pi^2 \zeta(3) \;. \end{array}$ Putting these results, together $I \; = \; 24\zeta(5) - (12\zeta(5) - \pi^2\zeta(3)) \; = \; \pi^2\zeta(3) + 12\zeta(5)$ making the answer $2 + 3 + 12 + 5 = \boxed{22}$ .