Hypergeometric sum with two binomial factors both in the numerator and in the denominator. Let $m\ge 1$ be an integer and let $0 < z < 4(m-1)^{m-1}/m^m$ be a real number.
We consider the following sum:
\begin{equation}
{\mathcal S}^{(m)}_2(z):= \sum\limits_{i=0}^\infty \frac{\binom{m\cdot i}{i}}{\binom{2 \cdot i}{i}} \cdot \frac{z^i}{2 \cdot i+1}
\end{equation}
Now, by using the integrals represetation of the Euler' beta function and by using Yet another family of hypergeometric sums that has a closed form solution. we have found the following integral representation of our sum. We have:
\begin{equation}
{\mathcal S}^{(m)}_2(z)=\frac{1}{\sqrt{z}} \int\limits_1^x \frac{1}{\xi^{m/2} \sqrt{1-\xi+\frac{z}{4} \xi^m}} d\xi
\end{equation}
Here $x:=x(z)$ is a solution to the equation
\begin{equation}
1-x+\frac{z}{4} x^m=0
\end{equation}
which has the property that is the closest to untity.
For $m=1,2$ we get the following closed form solutions:
\begin{equation}
\left\{ {\mathcal S}^{(m)}_2(z) \right\}_{m=1}^2 = \left\{\frac{2 \pi -4 \arcsin\left(\frac{\sqrt{4-z}}{2}\right)}{\sqrt{4-z} \sqrt{z}},\frac{\log \left(\frac{\sqrt{z}+1}{\sqrt{1-z}}\right)}{\sqrt{z}}\right\}
\end{equation}
It is clear that for $m > 2$ the result will include some special functions, probably elliptic functions, as any CAS (Mathematica) will indicate. Unfortunately my knowledge about elliptic functions is very scarce and I do not want to copy and paste some enormous and ugly outputs from Mathematica without understanding how that result was actually obtained. Therefore I am asking a question how do we evaluate the integral above for $m > 2$?
 A: I am going to present a solution for $m=4$. It is clear that the integral in question has the form:
\begin{equation}
\sqrt{z} {\mathcal S}^{(4)}_2(z) = \int\limits_1^x \frac{R_1(\xi)}{w} d\xi
\end{equation}
where $R_1(\xi) = 1/\xi^2$ is a rational function  and $w:=\sqrt{1-\xi+z/4 \cdot \xi^4}$ is a square root of a quartic polynomial . Therefore following the recipe from http://mathworld.wolfram.com/EllipticIntegral.html  we can reduce the result to elementary functions and to incomplete elliptic integrals of the first, the second and the third kind $F(\phi,k)$, $E(\phi,k)$ and $\Pi(n;\phi,k)$ only. We adopt the notations as in the link above and we only state the result. 
For a given value of $0<z< 4 \cdot 27/256$ we solve the equation
\begin{equation}
1-\xi+\frac{z}{4} \xi^4=0
\end{equation}
The roots are always such that the first two are  complex conjugate  and the remaining ones are real. Now, we define:
\begin{equation}
\left\{b_1,b_2,c_1,c_2\right\}:=\left\{Re[\xi^{(1)}],-Re[\xi^{(1)}],|\xi^{(1)}|^2,\xi^{(3)} \xi^{(4)}\right\}
\end{equation}
Now, we solve a quadratic equation:
\begin{equation}
(1-\lambda)(c_1-\lambda-c_2)-(b_1-\lambda b_2)^2=0
\end{equation}
and out of the two solutions $\lambda_{1,2}$ we construct the following quantities:
\begin{eqnarray}
\left\{\alpha,\beta\right\}&:=& \left\{\sqrt{\frac{c_1-\lambda_1 c_2}{1-\lambda_1}},-\sqrt{\frac{c_1-\lambda_2 c_2}{1-\lambda_2}} \right\}\\
\left\{A_1,B_1,A_2,B_2\right\}&:=& 
\left\{
\frac{1-\lambda_1}{\lambda_2-\lambda_1},
-\frac{1-\lambda_2}{\lambda_2-\lambda_1},
\lambda_2\frac{1-\lambda_1}{\lambda_2-\lambda_1},
-\lambda_1\frac{1-\lambda_2}{\lambda_2-\lambda_1}
\right\}
\end{eqnarray}
Define:
\begin{eqnarray}
u[x]&:=&(\frac{x-\alpha}{x-\beta})^2+\frac{1}{2}(\frac{B_2}{A_2}+\frac{B_1}{A_1})\\
\kappa(x)&:=&\imath \cdot \mbox{arcsinh}(\sqrt{\frac{A_1}{B_1}} \frac{x-\alpha}{x-\beta})\\
I_1(u,\beta,B)&:=&\int \frac{1}{u-\beta}\cdot \frac{1}{\sqrt{B_1+u^2}} du = 
-\frac{2}{\sqrt{\beta^2+B}}\cdot \mbox{arctanh}\left(\frac{\sqrt{B}}{\sqrt{\beta^2+B}}\cdot \frac{u+\beta(-1+\sqrt{1+u^2/B})}{u}\right)
\\
I_2(u,\beta,B)&:=& \frac{d}{d \beta} I_1(u,\beta,B)\\
{\mathfrak I}(x)&:=&\alpha\frac{\alpha-\beta}{\beta^2}\cdot 
\left.I_2\left(u,(\frac{\alpha}{\beta})^2+\frac{1}{2}(\frac{B_2}{A_2}+\frac{B_1}{A_1}),-\frac{1}{4} (\frac{B_2}{A_2}-\frac{B_1}{A_1})^2\right)\right|_{u[x]}^{u[1]}+\\
&&
\left.I_1\left(u,(\frac{\alpha}{\beta})^2+\frac{1}{2}(\frac{B_2}{A_2}+\frac{B_1}{A_1}),-\frac{1}{4} (\frac{B_2}{A_2}-\frac{B_1}{A_1})^2\right)\right|_{u[x]}^{u[1]}
\end{eqnarray}
Now, the result reads:
\begin{eqnarray}
&&\sqrt{z} {\mathcal S}^{(4)}_2(z)=\frac{2}{\sqrt{z \cdot A_1 \cdot B_2} \cdot \beta^2}\left(\right.\\
&&\left.
\frac{\sqrt{B_2}}{\sqrt{A_2} \beta}\cdot {\mathfrak I}(x)+\right.\\
&&\left.
\frac{\imath}{\alpha-\beta}\cdot\left(F(\kappa(x),\frac{A_2 B_1}{A_1 B_2})-F(\kappa(1),\frac{A_2 B_1}{A_1 B_2})\right)+\right.\\
&&\left.
\imath\frac{3\alpha-\beta}{\alpha^2}\cdot\left(\Pi(-\frac{B_1\beta^2}{A_1\alpha^2};\kappa(x),\frac{A_2 B_1}{A_1 B_2})+\Pi(-\frac{B_1\beta^2}{A_1\alpha^2};\kappa(1),\frac{A_2 B_1}{A_1 B_2})\right)+\right.\\
&&\left.
2\frac{\alpha^2(-\alpha+\beta)}{\beta^4}\cdot \mbox{Im}
\frac{d}{d \theta} \left.\left( \Pi(-\frac{B_1}{A_1\theta};\kappa(x),\frac{A_2 B_1}{A_1 B_2})+\Pi(-\frac{B_1}{A_1\theta};\kappa(1),\frac{A_2 B_1}{A_1 B_2})\right)\right|_{\theta=\alpha^2/\beta^2}
\right)
&&
\end{eqnarray}
