Roaring Integral!

$\begin{aligned} I & = \int_0^\pi \frac{1}{(5-3\cos x)^3} dx \quad \quad \small \color{#3D99F6}{\text{Let } t = \tan \frac{x}{2} \space \Rightarrow dx = \frac{2}{1+t^2}dt \space \Rightarrow \cos x = \frac{1-t^2}{1+t^2}} \\ & = \int_0^\infty \frac{(1+t^2)^2}{4(1+4t^2)^3} dt \\ & = \frac{1}{4} \int_0^\infty \frac{1+2t^2+t^4}{(1+4t^2)^3} dt \quad \quad \small \color{#3D99F6}{\text{Let }u = 4t^2 \space \Rightarrow dt = \frac{1}{4\sqrt{u}}du} \\ & = \frac{1}{16} \int_0^\infty \left(1+\frac{1}{2}u+\frac{1}{16}u^2\right)u^{-\frac{1}{2}}(1+u)^{-3} du \\ & = \int_0^\infty \frac {u^{-\frac 12}}{16(1+u)^3} + \frac {u^{\frac 12}}{32(1+u)^3} + \frac {u^{\frac 32}}{256(1+u)^3} du \quad \quad \small \color{#3D99F6}\text{Beta function }B(m+1, n-m-1) = \int_0^\infty \frac {u^m}{(1+u)^n} du \\ & = \frac{B\left(\frac{1}{2}, \frac{5}{2} \right)}{16} + \frac{B\left(\frac{3}{2}, \frac{3}{2} \right)}{32} + \frac{B\left(\frac{5}{2}, \frac{1}{2} \right)}{256} \\ & = \frac{17\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{5}{2} \right)}{256\Gamma\left(3\right)} + \frac{ \Gamma^2 \left(\frac{3}{2}\right)}{32\Gamma\left(3\right)} \quad \quad \small \color{#3D99F6} \Gamma(s) \text{ is gamma function} \\ & = \frac{17 \left(\sqrt{\pi}\right) \left(\frac{3}{4}\sqrt{\pi}\right)}{256(2!)} + \frac{\left(\frac{1}{2}\sqrt{\pi}\right)^2}{32(2!)} \\ & = \frac{59\pi}{2048} \end{aligned}$

$\Rightarrow \alpha + \beta = 59 + 2048 = \boxed{2107}$

References:

• Beta function
• Gamma function

The complex substitution $z = e^{ix}$ gives $I \; = \; \int_0^\pi \frac{dx}{(5 - 3\cos x)^3} \; =\; \tfrac12\int_{-\pi}^\pi \frac{dx}{(5 - 3\cos x)^3} \; = \; \tfrac12\int_{|z|=1} \frac{1}{(5 - \frac32(z+z^{-1}))^3} \frac{dz}{iz} \; = \; \frac{4}{i} \int_{|z|=1} \frac{z^2\,dz}{(3z-1)^3(3-z)^3}$ Contour integration then gives $I \; = \; \frac{8\pi}{27} \mathrm{Res}_{z=\frac13}\frac{z^2}{(z-\frac13)^3(3-z)^3} \; = \; \frac{4\pi}{27} \frac{d^2}{dz^2}\frac{z^2}{(3-z)^3} \Bigg|_{z=\frac13} \; = \; \frac{59\pi}{2048}$ making the answer $\boxed{2107}$ .