Let $n\ge 0$ be an integer and let $m\in(0,1)$ be a real number. Consider a following sum: \begin{equation} {\mathfrak I}^{(n)}_m := \frac{\pi}{2} (-1)^n \sum\limits_{j=0}^\infty \binom{-1/2}{j} \binom{-1/2}{j+n} m^j =\frac{\pi}{2} (-1)^n\binom{-\frac{1}{2}}{n} \, _2F_1\left(\frac{1}{2},n+\frac{1}{2};n+1;m\right). \end{equation} It is not hard to see that this sum has a following integral representation: \begin{equation} {\mathfrak I}^{(n)}_m = \int\limits_0^{\pi/2} \frac{\cos(u)^{2 n}}{\sqrt{1-m \cos(u)^2}} du = \int\limits_0^1 \frac{y^{2 n}}{\sqrt{1-m y^2}}\cdot \frac{1}{\sqrt{1-y^2}}dy. \end{equation} The integrals above can be always reduced to elliptic integrals by using the procedure in http://mathworld.wolfram.com/EllipticIntegral.html . We have actually done this and then we found that the quantities above satisfy the following recurrence relations: \begin{eqnarray} {\mathfrak I}^{(n)}_m - \sum\limits_{q=0}^{n-1}(1+m^{q-n}) {\mathfrak I}^{(q)}_m = \frac{\sqrt{m} E\left(\sin ^{-1}\left(\sqrt{m}\right)|\frac{1}{m}\right)-m^{-n} E(\pi/2|m)}{1-m} \end{eqnarray} where $E(\phi|m)$ is the elliptic integral of the second kind.
Now, my question is the usual one. Is it possible to find a closed form expression for the quantities in question?
I agree with the comment of reuns, but I also would like to point out that $m=\frac{1}{2}$ is one of such special values. In such a case
$$\mathfrak{I}^{(n)}=\int_{0}^{\pi/2}\frac{\sin(\theta)^{2n}\,d\theta}{\sqrt{1-\frac{1}{2}\sin^2\theta}}=\frac{\pi}{2}\sum_{h\geq 0}\frac{\binom{2h}{h}\binom{2h+2n}{h+n}}{16^h\cdot 4^{h+n}} $$ is always given by a linear combination with rational coefficients of $E\left(\frac{1}{2}\right)$ and $K\left(\frac{1}{2}\right)$, where $$ K\left(\tfrac{1}{2}\right) = \tfrac{1}{4\sqrt{\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2,\qquad E\left(\tfrac{1}{2}\right) = \frac{\pi\sqrt{\pi}}{\Gamma\left(\frac{1}{4}\right)^2}+\frac{\Gamma\left(\frac{1}{4}\right)^2}{8\sqrt{\pi}} $$ by Clausen's formula (${}_2 F_1(\ldots)^2 = {}_3 F_2(\ldots)$) and Legendre's identity.