Mystery Zeta

Recalling the alternating Euler sum $a_h(1,2) \; =\; \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n+1)^2}H_n \; = \; \tfrac18\zeta(3)$ we observe that $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_n & = & \displaystyle -1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n+1)^2}H_{n+1} \\ & = & \displaystyle -1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n+1)^2}H_n + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n+1)^3} \; = \; a_h(1,2) + \sum_{n=1}^\infty \frac{(-1)^n}{n^3} \\ & = & \displaystyle \tfrac18\zeta(3) - \tfrac34\zeta(3) \; = \; -\tfrac58\zeta(3) \end{array}$ Since $\int_0^1 x^{n-1}\ln(1-x)\,dx \; = \; -\sum_{m=1}^\infty \frac{1}{m}\int_0^1 x^{n+m-1}\,dx \; = \; -\sum_{m=1}^\infty \frac{1}{m(m+n)} \; = \; -\tfrac{1}{n}H_n$ for any $n \ge 1$ , we see that $\int_0^1 \frac{\ln(1+x)\ln(1-x)}{x}\,dx \; = \; =\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\int_0^1 x^{n-1}\ln(1-x)\,dx \; = \; \sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_n \; = \; -\tfrac58\zeta(3)$ making the answer $5+8+3 = \boxed{16}$ .