2F1(1/2,1/2;3/2;z^2)=arcsin(z)/z
From specialfunctionswiki
Theorem
The following formula holds: $${}_2F_1 \left( \dfrac{1}{2}, \dfrac{1}{2}; \dfrac{3}{2}; z^2 \right) = \dfrac{\arcsin(z)}{z},$$ where ${}_2F_1$ denotes the hypergeometric 2F1 and $\arcsin$ denotes the inverse sine.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $15.1.6$