If $a^2>b^2$ prove that $\int\limits_0^{\pi} \frac{dx}{(a+b\cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.

It is well-known that $$\int_0^\pi\frac{dx}{t+\cos x}=\frac{\pi}{\sqrt{t^2-1}}$$ for $t>1$. Differentiating gives $$-\int_0^\pi\frac{dx}{(t+\cos x)^2}=-\frac{\pi t}{(t^2-1)^{3/2}}.$$ Differentiating again gives $$2\int_0^\pi\frac{dx}{(t+\cos x)^3} %=-\frac{\pi(t^2-1)}{(t^2-1)^{5/2}}+\frac{3\pi t^2}{(t^2-1)^{5/2}} =\frac{\pi(2t^2+1)}{(t^2-1)^{5/2}}.$$ Homogenising this gives your formula.