Calculate $\int_0^\infty \frac1{(1+z)^{2+\gamma}} \ _2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)\,dz$ On page $46$ of this paper, the authors say that

By writing the Gauss Hypergeometric function in series, integrating and summing again we find the following expression
$$\int_0^\infty \frac1{(1+z)^{2+\gamma}} \ _2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)\,dz=\frac{\Gamma\left(-\frac\gamma2\right)\Gamma\left(\frac{\gamma-1}2\right)}{\Gamma\left(\frac{1-\gamma}2\right)\Gamma\left(\frac\gamma2\right)}.$$

I can do the first two steps: writing the Gauss Hypergeometric function in series and integrating, but I don't know how to "sum again" to get the neat result. Let me first explain the notations. Here $\gamma\in(0,1)$ is a given parameter and $\Gamma$ is the Gamma function. The Gauss Hypergeometric function is defined for $|z|<1$ by the series
$$_2F_1(a,b;c;z)=\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!},$$
where the (rising) Pochhammer symbol $(x)_n$ is defined by
$$(x)_n=\begin{cases} 1, & n=0,\\ x(x+1)\cdots(x+n-1), & n\geq1.\end{cases}$$
It follows from definition that $(x)_n=\Gamma(x+n)/\Gamma(x)$ for $x>0$.
My calculation. By definition,
\begin{align*}
_2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)&=\sum_{n=0}^\infty \frac{\left(1+\frac\gamma2\right)_n\left(\frac32\right)_n}{(3)_n}\frac1{n!}\left(\frac{4z}{(1+z)^2}\right)^n\\
&=\sum_{n=0}^\infty \frac{\Gamma\left(1+\frac\gamma2+n\right)}{\Gamma\left(1+\frac\gamma2\right)}\frac{\Gamma\left(\frac32+n\right)}{\Gamma\left(\frac32\right)}\frac{\Gamma\left(3\right)}{\Gamma\left(3+n\right)}\frac{4^n}{n!}\frac{z^n}{(1+z)^{2n}}.
\end{align*}
Now, we do the integration. Change the variable $z=\tan^2\theta$ gives that $\frac{dz}{d\theta}=2\tan\theta\frac1{\cos^2\theta}$ and thus
\begin{align*}
\int_0^\infty \frac{z^n}{(1+z)^{2+\gamma+2n}}\,dz&=2\int_0^{\frac\pi2}\frac{\tan^{2n}\theta}{(1+\tan^2\theta)^{2+\gamma+2n}}\tan\theta\frac1{\cos^2\theta}\,d\theta\\
&=2\int_0^{\frac\pi2}\cos^{2n+2\gamma+1}\theta\sin^{2n+1}\theta\,d\theta\\
&=B(n+\gamma+1,n+1)=\frac{n!\Gamma(n+\gamma+1)}{\Gamma(2n+\gamma+2)}.
\end{align*}
Therefore,
$$\int_0^\infty \frac1{(1+z)^{2+\gamma}} \ _2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)\,dz=\sum_{n=0}^\infty \frac{\Gamma\left(1+\frac\gamma2+n\right)}{\Gamma\left(1+\frac\gamma2\right)}\frac{\Gamma\left(\frac32+n\right)}{\Gamma\left(\frac32\right)}\frac{2\cdot 4^n}{(n+2)!}\frac{\Gamma(n+\gamma+1)}{\Gamma(2n+\gamma+2)}.$$
I can do a little more cancellation between the the numerator and the denominator. But I still don't know how to calculate the sum even after the simplication.
Any help would be appreciated!
 A: It works
Using
$$\ _2F_1\left(1+\frac\gamma2,\frac32;3;t\right)=\, _2F_1\left(\frac{3}{2},1+\frac{\gamma }{2};3;t\right)$$ then
$$\, _2F_1\left(\frac{3}{2},1+\frac{\gamma }{2};3;t\right)=\sum_{n=0}^\infty \frac{4}{\sqrt{\pi }}\frac{\left(n+\frac{1}{2}\right)! \left(n+\frac{\gamma }{2}\right)!}{\frac{\gamma}{2}!\, n!\, (n+2)!}\,t^n$$
$$I=\int_0^\infty \frac1{(1+z)^{2+\gamma}} \ _2F_1\left(1+\frac\gamma2,\frac32;3;\frac{4z}{(1+z)^2}\right)\,dz$$ $$I=\sum_{n=0}^\infty \frac{4^{n+1} \left(n+\frac{1}{2}\right)! \left(\frac{\gamma
   }{2}+n\right)!}{\sqrt{\pi } \,\frac{\gamma }{2}!\, n! \,(n+2)!}\int_0^\infty z^n (z+1)^{-\gamma -2 n-2}\,dz$$
$$\int_0^\infty z^n (z+1)^{-\gamma -2 n-2}\,dz=\frac{\Gamma (n+1) \Gamma (n+\gamma +1)}{\Gamma (2 n+\gamma +2)}$$
$$I=\sum_{n=0}^\infty \frac{2^{1-\gamma }\, \Gamma \left(n+\frac{3}{2}\right) \,\Gamma (n+\gamma +1)}{\Gamma
   \left(1+\frac{\gamma }{2}\right)\, \Gamma (n+3)\, \Gamma \left(n+\frac{\gamma
   +3}{2}\right)}=-\frac{\Gamma \left(\frac{2-\gamma }{2}\right) \Gamma \left(\frac{\gamma
   -1}{2}\right)}{\Gamma \left(\frac{1-\gamma }{2}\right) \Gamma
   \left(\frac{\gamma +2}{2}\right)}$$ which is the same as the rhs.
