I would calculate the following integral \begin{equation} I_x = \int_{0}^{1} y^{b+\mu-1} (1-y)^{\nu-1}\, _2F_1(a,b+\nu +\mu;c; xy) \, dy. \end{equation} Such that $\quad \Re a,\Re b,\Re \mu, \Re \nu >0$ and $ -1<x<1$.
Someone can help me!!! Thanks in advance
One method is the following: \begin{align} I_{x} &= \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b+\nu + \mu)_{n}}{n! \, (c)_{n}} \, x^{n} \, J_{n} \end{align} where $$J_{n} = \int_{0}^{1} y^{n + b + \mu -1} \, (1-y)^{\nu-1} \, dy.$$ Now, \begin{align} J_{n} &= B(n + b + \mu, \nu) = B(b+\mu, \nu) \, \frac{(b+\mu)_{n}}{(b+\mu + \nu)_{n}} \end{align} and leads to \begin{align} I_{x} &= B(b+\mu, \nu) \, {}_{2}F_{1}(a, b+\mu; c; x), \end{align} or \begin{align} \int_{0}^{1} y^{b+ \mu -1} \, (1-y)^{\nu -1} \, {}_{2}F_{1}(a, b + \nu + \mu; c; xy) \, dy = B(b+\mu, \nu) \, {}_{2}F_{1}(a, b+\mu; c; x). \end{align}