# How to modify this integral to bring it in the form of an interal representation of Hypergeometric function?

I want to find a simple expression for the following integral :

$F(x)=\displaystyle \int_0^x \dfrac{ay^{a\alpha -1}\left[ 1-(1-c)y^a \right]^{\beta -1}}{B(\alpha,\beta)\left[1+cy^a\right]^{\alpha +\beta}}dy$

If I put $(1-c)y^a=t$, then I get

$F(x)=\dfrac{1}{(1-c)^{\alpha}B(\alpha,\beta)}\displaystyle \int_0^{(1-c)x^a}t^{\alpha-1}(1-t)^{\gamma - \alpha -1}\left[ 1-\dfrac{c}{1-c}t \right]^{-\gamma}dt$

where $\gamma=\alpha +\beta$.

My aim is to put the integral in the RHS in the form of a hypergeometric function as it looks similar to the same :

${}_2F_1 (a,b,c;z)=\dfrac{1}{B(b,c-b)}\displaystyle \int_0^1x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx$

This is where I am stuck. Any ideas ?

I change my answer to a hint: You could start by doing the substitution $$u=\frac{(1-c)(1-t)}{1-c-ct}.$$ It will transform the integral $$\int t^{\alpha-1}(1-t)^{\gamma-\alpha-1}\Bigl[1-\frac{c}{1-c}t\Bigr]^{-\gamma}\,dt$$ into (modulo a multiplicative constant, I hope I did the calculations correctly) $$\int (1-u)^{a-1}u^{\alpha+\gamma-1}\,du.$$ This will lead you to an incomplete beta function, and then you can use equation (3) at this site to transform it into the hypergeometric function you want.