Finding indefinite integral $\int{ \mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$

Could anyone help me how to solve this indefinite integral?

$$\int{\mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$$

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}