I'm attempting to solve the indefinite integral
$$S\left(v\right) = 2a\sqrt{\alpha E}\int\!v^{-1/2}\left(\frac{Q^2}{2E}-v^2\right)^{1/2}\left(v^3+\alpha^3\right)^{-1/2}\left(v^3+2\alpha^3\right)^{-1/6}\,\mathrm dv$$
Because this looks like it could be a Hypergeometric integral I have defined $x\equiv\frac{2E}{Q^2}v^2$, $z\equiv -\frac{Q^2}{4E\alpha^3}v$, $a = \frac 16$, $b = \frac 23$, and $c = \frac 94$ so that it can be rewritten in the form
$$S\left(x\right) \sim \int\!\left(\frac{x}{1+\beta x^{3/2}}\right)^{1/2}x^{b-1}\left(1-x\right)^{c-b-1}\left(1-zx\right)^{-a}\,\mathrm dx$$
Clearly the last three terms in this integral look like a hypergeometric integral, but I have a leading term which messes things up.
Note the integral form of the Hypergeometric function I find on Wikipedia is defined as
$$B(b,c-b)\,{}_2F_1(a,b;c;z) = \int_0^1 x^{b-1} (1-x)^{c-b-1}(1-zx)^{-a} \, dx \qquad Re(c) > Re(b) > 0$$
Also, my integral is not bounded. Does anyone have any suggestions for how to solve this (or even relevant transformations which could be useful)?