Is there a simple expression for ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$? I have been searching through some books and also this
but I have not succeed.
I wonder if there is a simple equivalent form for ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$, in terms of elementary functions or other simpler functions.
In principle, $a>0$ but a solution for integer positive values of $a$ would be useful to me as well.
Thank you in advance.
 A: I have found a possible solution and thought I would post it here.
Applying the Euler transformation:
http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/17/02/07/0001/
the original expression ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$ can be transformed into one where the following form appears: ${}_2F_1(1,\frac{1}{2};a+1;z)$. If $a$ is a positive integner then it can be expressed in elementary (power) functions using the following relation:
http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/06/07/10/0001/
Any other ideas and suggestions are still welcome.
A: For the sake of simplicity, I'll assume $z$ is real and $0 < a \le 1/2$. First, suppose $a = 1/2$ and $z\neq 0,1$. Then
\begin{align}{}_2F_1\left(a, a+\frac{1}{2}; a + 1; z\right) &= {}_2F_1\left(\frac{1}{2},1;\frac{3}{2};z \right)\\
&= \frac{1}{i\sqrt{z}}\cdot i\sqrt{z}\,{}_2F_1\left(1,\frac{1}{2};\frac{3}{2};-(i\sqrt{z})^2\right)\\
&=\frac{1}{\sqrt{z}}\arctan(i\sqrt{z})\\
&= \frac{1}{2\sqrt{z}}\ln\frac{1 + \sqrt{z}}{1 - \sqrt{z}}.
\end{align}
Now suppose $a < 1/2$ and $z = 1$. Since $(a + 1) - a - (a + 1/2) = 1/2 - a > 0$, Gauss's summation formula gives
\begin{align}
{}_2F_1\left(a, a + \frac{1}{2};a + 1; z\right) &= 
\frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\Gamma(1)\Gamma\left(\frac{1}{2}\right)}\\
&= \frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\sqrt{\pi}}.
\end{align}
In general, since $a + 1 > a > 0$, Euler's integral representation of ${}_2F_1$ gives
\begin{align}
{}_2F_1\left(a,a+\frac{1}{2};a+1;z\right) &= \frac{\Gamma(a+1)}{\Gamma(a)\Gamma(1)} \int_0^1 t^{a-1} (1 - zt)^{-a-1/2}\, dt\\
&= a\int_0^z u^{a-1}(1 - u)^{-a-1/2}\, \frac{du}{z^a}\\
&= az^{-a}B\left(a,\frac{1}{2}-a;z\right)
\end{align}
for $z \neq 0$, where the $B$ function in the last step is the incomplete Beta function.
A: Maple can do some of these, $a$ integer or half-integer. But not, for example
$$
{}_2F_1\left(\frac{1}{3},\frac{5}{6};\frac{4}{3};z\right)
$$
In fact, 
$$
{}_2F_1\left(\frac{1}{3},\frac{5}{6};\frac{4}{3};\frac{1}{2}\right) \approx 1.1460295323744285$$
and if I put $1.1460295323744285$ into the ISC, the result is
$$
{}_2F_1\left(\frac{1}{3},\frac{5}{6};\frac{4}{3};\frac{1}{2}\right)
$$
So I am guessing that no simpler form is known.
added, based on kobe's comment 
Maple says
$$
{}_2F_1\left(a,a+\frac{1}{2};a+1;1\right) =  
\frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\sqrt{\pi}}
$$
Of course the series diverges when $a>1/2$, so I guess this is an analytic continuation in that case.   
Also, the case $a=1/2$ is
$$
{}_2F_1\left(\frac{1}{2},1;\frac{3}{2};z\right) = 
\frac{1}{2\sqrt{z}}\log\frac{1+\sqrt{z}}{1-\sqrt{z}}
$$
