Wolfram has the answer 2

The integral is $\begin{array}{rcl} \displaystyle I \; = \; \int_0^\pi x^2 \big[ \ln 2 \cos\tfrac12x\big]^2\,dx & = & \displaystyle 8\int_0^{\frac12\pi} u^2\big[ \ln(2 \cos u)\big]^2\,du \\ & = & \displaystyle -8\frac{\partial^4 F}{\partial \nu^2 \partial a^2}(1,0) \end{array}$ where $F(\nu,a) \; = \; 2^{\nu-1}\int_0^{\frac12\pi} \cos^{\nu-1}u \cos au\,du \; = \; \frac{\pi}{2 \nu B\big(\tfrac{\nu+a+1}{2},\tfrac{\nu-a+1}{2}\big)} \; = \; \frac{\pi \Gamma(\nu)}{2 \Gamma\big(\tfrac{\nu+a+1}{2}\big) \Gamma\big(\tfrac{\nu-a+1}{2}\big)} \;.$ Now $G(\nu) \; = \; -\frac{\partial^2 F}{\partial a^2}(\nu,0) \; = \; \frac{\pi \Gamma(\nu)}{4 \Gamma\big(\tfrac{\nu+1}{2}\big)^2} \psi'\big(\tfrac{\nu+1}{2}\big)$ and so $G''(1)$ can be expressed in terms of polygamma functions, and then simplified, obtaining $I \; = \; 8G''(1) \; = \; \tfrac{11}{180}\pi^5$ making the answer $11 + 180 + 5 = \boxed{196}$ .