<s>[INST] Finding indefinite integral $\int{ \mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$ Could anyone help me how to solve this indefinite integral?
$$\int{\mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$$
 [/INST] Firstly let us find the anti-derivative in case $\alpha=0$.  This is pretty straightforward:
\begin{eqnarray}
\int\frac{dx}{\sqrt{\sin(x)^3+\sin(x)}}&\underbrace{=}_{y=\sin(x)}&\int\frac{dy}{\sqrt{1-y^2} \sqrt{y} \sqrt{1+y^2}}\\
&\underbrace{=}_{z=y^4}& \frac{1}{4} \int z^{-7/8} (1-z)^{-1/2} dz\\
&=& \frac{1}{4} B_{[\sin(x)]^4}(\frac{1}{8},\frac{1}{2})
\end{eqnarray}
where $B_{.}(,)$ is the incomplete beta function.
Now let us compute the derivative at $\alpha=0$. We have:
\begin{eqnarray}
-\frac{1}{2} \int \frac{\cos(x)}{(\sin(x)^3+\sin(x))^{3/2}}dx&\underbrace{=}_{y=\sin(x)}&-\frac{1}{2} \int\frac{d y}{(y^3+y)^{3/2}}\\
&\underbrace{=}_{u=-y^2}&\frac{(-1)^{3/4}}{4} B_{-\sin(x)^2}(-\frac{1}{4},-\frac{1}{2})
\end{eqnarray}
Therefore we have:
\begin{eqnarray}
\int \frac{dx}{\sqrt{\sin(x)^3+\sin(x+\alpha)}} =\frac{1}{4} B_{[\sin(x)]^4}(\frac{1}{8},\frac{1}{2})+ \alpha \frac{(-1)^{3/4}}{4} B_{-\sin(x)^2}(-\frac{1}{4},-\frac{1}{2})+O(\alpha^2)
\end{eqnarray}
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