Solution for $\int_0^1 y^{(1 + b)} \left(\frac{c + y}{1 + c y}\right)^b dy$, where $(0Please help me in solving integral 
$\int_0^1 y^{(1 + b)} \left(\frac{c + y}{1 + c y}\right)^b dy$,  where   $(0<b<1)$, $(0<c<1)$ 
I want the final solution in a simplified form not in terms of hypergeometric forms like Appell $F_1$ and $_2F_1$. Approximate solution. will also work.
Thanks
 A: Mathematica gives
$$\frac{c^b \text{AppellF1}\left[2+b,-b,b,3+b,-\frac{1}{c},-c\right]}{2+b}$$
I don't think it has a simpler form other than that. To do an approximation, maybe you need to specify some conditions on $b$ and $c$, such as $b \approx 0$ or $c \approx 0$.
A: The interpolating polynomial gives a good approximation taking $x=0$,$x=1/2$ and $x=1$.
$$f(x)={{{c}}^b}\left( {x - 1} \right)\left( {2x - 1} \right) - 4x\left( {x - 1} \right){\left( {\frac{{2c + 1}}{{c + 2}}} \right)^b} + x\left( {2x - 1} \right)$$
so you can evaluate
$$\int_0^1x^{1+b}f(x)dx$$
and get an approximation. For $b$ fixed large and $c$ any value, the approximation is very good. Similarily for the other symmetric situation. For $c=0$ and $b<0.1$ the approximation is very bad, but for $c\neq 0$ we can push the values to $.9$ and things look good, for $c<.2;.3$ we get a not too good approximation.
Here you can see some images. $p$ is blue, the original function in red.
Bad $c=0$ case. $b$ ranges $0$ to $1$ in tenths of unity.

Good $b=1$ case. $c$ ranges $0$ to $1$ in tenths of unity.

Not that bad $c=0.1$ case. $b$ ranges $0$ to $1$ in tenths of unity.

Good $c=0.6$ case. $c$ ranges $0$ to $1$ in tenths of unity.

Good $b=0.62$ case. $c$ ranges $0$ to $1$ in tenths of unity.

