I need a closed form for the Hypergeometric series $$_2F_1(a,a;1+a;-1)$$ where $a<1$
I tried the following integral representation $$_2F_1(p,q;r;z)=\frac{\Gamma(r)}{\Gamma(q)\Gamma(r-q)} \int_{0}^{1} t^{q-1} (1-t)^{r-q-1} (1-tz)^{-p}\ dt $$ So setting $p=a, q=a,r=1+a,z=-1$ we obtain $$_2F_1(a,a;1+a;-1)=\frac{\Gamma(1+a)}{\Gamma(a)} \int_{0}^{1} t^{a-1} (1-t)^{1+a-a-1} (1+t)^{-a}\ dt $$ So we get using $\Gamma(1+a)=a\Gamma(a)$ $$_2F_1(a,a;1+a;-1)=a\int_{0}^{1} t^{a-1}(1+t)^{-a}\ dt $$
Now substitute $t=\cos x$ so that $dt=-\sin x \ dx$ and we get $$_2F_1(a,a;1+a;-1)=a\int_{0}^{\pi/2} \cos^{a-1} x \ (1+\cos x)^{-a}\ \sin x\ dx $$ Using $1+\cos x=2\cos^2(x/2)$ we get $$_2F_1(a,a;1+a;-1)=a\ 2^{-a}\int_{0}^{\pi/2} \cos^{a-1} x \ \cos^{-2a}\left(\frac{x}{2}\right)\ \sin x\ dx $$
I request for a closed form expression of $_2F_1(a,a;1+a;-1)$ where $a<1$
As pointed out by @TymaGaidash your particular hypergeometric function is related to the incomplete beta function via $$ F\left({a,a\atop a+1};-1\right)=a(-1)^{-a}\operatorname{B}_{-1}(a,1-a). $$ Without any further restrictions on $a$, it is unlikely that your $F(-1)$ reduces to any elementary functions.
A comprehensive search of special cases for rational $a$ yields the special cases $$ F\left({1/2,1/2\atop 3/2};-1\right)=\operatorname{arcsinh}(1), $$ $$ F\left({-1/2,-1/2\atop 1/2};-1\right)=\sqrt 2-\operatorname{arcsinh}(1), $$ $$ F\left({-3/2,-3/2\atop -1/2};-1\right)=5\sqrt 2+3\operatorname{arcsinh}(1). $$ Further special cases for $a=-5/2,-7/2,-9/2,-11/2$ as well as rational $a$ with denominator $3$ and $4$ can also be found here.