Prove $\int_{0}^{\pi}\frac{x^2}{\sqrt{5}-2\cos{x}}dx=\frac{\pi^3}{15}+2\pi\ln^2{\left(\frac{1+\sqrt{5}}{2}\right)}$ without contour integration
How do I evaluate this integral $\int_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x$?