A ln Li integral

Since $\mathrm{Li}_2(-x) \; = \; \sum_{n=0}^\infty \frac{(-x)^{n+1}}{(n+1)^2}$ we deduce that the integral is $\begin{aligned} I & = \; \int_0^1 \frac{\ln x}{x(1-x^2)}\mathrm{Li}_2(-x)\,dx \; = \; \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(n+1)^2}\int_0^1 \frac{\ln x \, x^n}{1-x^2}\,dx \\ & = \; \sum_{m,n=0}^\infty \frac{(-1)^{n+1}}{(n+1)^2}\int_0^1 \ln x \,x^{n+2m}\,dx \; = \; \sum_{m,n=0}^\infty \frac{(-1)^n}{(n+1)^2(n+2m+1)^2} \\ & = \; \sum_{m,n=0}^\infty \frac{(-1)^{n+2m}}{(n+1)^2(n+2m+1)^2} \; = \; \sum_{N=1}^\infty \frac{(-1)^{N+1}}{N^2}\sum_{{1 \le u \le N} \atop {u \equiv N \pmod{2}}}\frac{1}{u^2} \\ & = \; \frac12\sum_{N=1}^\infty\frac{(-1)^{N+1}}{N^2}\sum_{u=1}^N \frac{1 + (-1)^{N-u}}{u^2} \; = \; \frac12\sum_{N=1}^\infty \frac{(-1)^{N+1}}{N^2}\sum_{u=1}^N \frac{1}{u^2} + \frac12\sum_{N=1}^\infty \frac{1}{N^2}\sum_{u=1}^N \frac{(-1)^{u+1}}{u^2} \\ & = \; \frac12\sum_{N=1}^\infty \frac{(-1)^{N+1}}{N^2}\sum_{u=1}^N\frac{1}{u^2} + \frac12\sum_{u=1}^\infty \frac{(-1)^{u+1}}{u^2}\sum_{N=u}^\infty \frac{1}{N^2} \; = \; \frac12\sum_{N=1}^\infty \frac{(-1)^{N+1}}{N^2}\left(\sum_{u=1}^\infty \frac{1}{u^2} + \frac{1}{N^2}\right) \\ & = \; \tfrac1{12}\pi^2\sum_{N=1}^\infty \frac{(-1)^{N+1}}{N^2} + \tfrac12 \sum_{N=1}^\infty \frac{(-1)^{N+1}}{N^4} \; = \; \tfrac1{12}\pi^2 \times \tfrac1{12}\pi^2 + \tfrac12 \times \tfrac{7}{720}\pi^4 \\ & = \; \tfrac{17}{1440}\pi^4 \end{aligned}$ where the key variable substitution in the third line is $N = n + 2m + 1$ and $u = n+1$ . This makes the answer $17 + 4 + 1440 = \boxed{1461}$ .