Closed form of ${_2}{F}_1\left(\frac{1}{2},s,\frac{3}{2};-\frac{1}{a^2}\right)$

Question

While solving an integral, I came acorss the term $$ \tilde{I}(a,s)= {_2}{F}_1\left(\frac{1}{2},s,\frac{3}{2};-\frac{1}{a^2}\right).$$ To be precise, it came in the following calculations \begin{align*} \int_{0}^{1} \left(1+\frac{t^2}{a^2}\right)^{-s}dt &= \int_{0}^{1} \sum_{n=0}^{\infty}\dbinom{s+n-1}{n}(-1)^n\left(\frac{t}{a}\right)^{2n}dt \\ & = \int_{0}^{1} \sum_{n=0}^{\infty}\frac{\Gamma(s+n)}{\Gamma(s)}\frac{(-1)^n}{n!}\left(\frac{t}{a}\right)^{2n}dt \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(s+n)}{\Gamma(s)}\frac{(-1)^n}{n!} \frac{1}{a^{2n}}\int_{0}^{1} t^{2n}dt \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(s+n)}{\Gamma(s)}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}\frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}\frac{\left(\frac{-1}{a^2}\right)^n}{n!} \\ &= \sum_{n=0}^{\infty} \frac{(s)_n\left(\frac{1}{2}\right)_n}{\left(\frac{3}{2}\right)_n}\frac{\left(\frac{-1}{a^2}\right)^n}{n!} = {_2}{F}_1\left(\frac{1}{2},s,\frac{3}{2};-\frac{1}{a^2}\right) \end{align*} I would like to know if this expression can be simplified more for certain values of $a>0$ and $0<s<1$. I believe there is a simpler form when $s=\frac{1}{2}$ or $a=1$.