The equation is $\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi ^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},% \frac{1}{2};2;\Pi ^{2})$, where $\Pi \le 1$. How to reduce the left integral into the right hypergeometric fucntion? Is there anyone can solve this problem? Thank you.

Thank you for @T. Amdeberhan. I learn general binomial expansion from your answer. Your result is more general. After some modifications, the result is \begin{eqnarray*} &&\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi ^{2}-2\Pi \sqrt{1-y^{2}}}\right) \\ &=&\frac{\pi }{2}\frac{\Pi }{\sqrt{1+\Pi ^{2}}}\ _{2}F_{1}\left( \frac{1}{4},% \frac{3}{4};2;\left( \frac{2\Pi }{1+\Pi ^{2}}\right) ^{2}\right) , \end{eqnarray*} which is valid for any value of $\Pi $. But for $\Pi \leq 1$, how to transform the result into

$ \int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi ^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},% \frac{1}{2};2;\Pi ^{2}) $?

It is seems that we should expand the integrad with respect to $\Pi $. Should we use multinomial expansion? But how to deal with the double summation? Could we prove the equality of the two hypergeometrical functions directly for $\Pi \leq 1$?