Log-sine

Note that $I \; = \; \int_0^{2\pi} x^2 \big[ \ln(2\sin\tfrac12x)\big]^2\,dx \; = \; 8\int_0^\pi x^2\ln^2(2\sin x)\,dx \; = \; -8\frac{\partial^4F}{\partial a^2 \partial b^2}(0,0)$ where $\begin{array}{rcl} \displaystyle F(a,b) & = & \displaystyle 2^a \int_0^\pi \sin^ax \cos bx\,dx \; = \; \frac{\pi \cos\frac12b\pi}{(a+1)B\big(\frac12(a+b+2),\frac12(a-b+2)\big)} \\ & = & \displaystyle \frac{\pi \Gamma(a+1) \cos \frac12b\pi}{\Gamma\big(\frac12(a+b+2)\big) \Gamma\big(\frac12(a-b+2)\big)} \;. \end{array}$ We first obtain that $-\frac{\partial^2F}{\partial b^2}(a,0) \; = \; \frac{\Gamma(a + 1)\big(\pi^3 + 2 \pi \psi^{(1)}\big(1 + \frac12a\big)\big)}{ 4 \Gamma\big(1 + \frac12a\big)^2}$ and hence, differentiating two more times,. we obtain $I \; = \; -8\frac{\partial^4F}{\partial a^2 \partial b^2}(0,0) \; = \; \tfrac{13}{45}\pi^5$ giving the answer $13 + 5 + 45 = \boxed{63}$ .