\begin{align*} I_\alpha (z) = \frac{\left(\frac{z}{2}\right)^\alpha}{\sqrt{\pi}\Gamma\left(\alpha + \frac12\right)} \int_0^\pi e^{-z\cos x} \sin^{2\alpha} x\ dx ~~~\left(\alpha > -\frac12\right) \end{align*}

と，

\begin{align*} \mathcal{L}[I_\alpha (ax)](s) = \frac{a^\alpha}{\sqrt{s^2 - a^2}(s+ \sqrt{s^2 - a^2})^\alpha}~~~(\alpha > -1) \end{align*}

を用いれば，

\begin{align*} &\int_0^\pi \frac{\sin^{2\alpha} x}{(a + b\cos x)^{\alpha + 1}} dx \\ &= \frac{1}{\Gamma(\alpha + 1)} \int_0^{\pi} \sin^{2\alpha} x \int_0^{\infty} e^{-(a+b\cos x)t} t^\alpha\ dt dx \\ &= \frac{1}{\Gamma(\alpha + 1)} \int_0^{\infty} e^{-at} t^\alpha \int_0^\pi e^{-b\cos x \cdot t}\sin^{2\alpha} x\ dx dt \\ &= \frac{\sqrt{\pi}\Gamma\left(\alpha + \frac12\right)}{\Gamma(\alpha + 1)} \left(\frac{2}{b}\right)^\alpha \int_0^\infty e^{-at} I_\alpha (bt)\ dt \\ &= \sqrt{\frac{\pi}{a^2 - b^2}} \frac{\Gamma\left(\alpha + \frac12\right)}{\Gamma(\alpha + 1)} \left(\frac{2}{a+\sqrt{a^2 - b^2}}\right)^\alpha \end{align*}