Procedure for evaluating the hypergeometric series $_2F_1\left\{\frac{v+2}{2},\frac{v+3}{2};v+1;z\right\}$ I'm trying to work out the procedure to get the following hypergeometric series into a simpler form, for all postive integer $v$:
$$ _2F_1\left\{\frac{v+2}{2},\frac{v+3}{2};v+1;z\right\}$$
For example, plugging this into Wolfram Alpha gives for $v$ = 1,
$$\frac{1}{(1-z)^{3/2}}$$
for $v$ = 2,
$$\frac{4 (2 \sqrt{1-z} \,(z-1)-3 z+2)}{3 \sqrt{1-z}\, (z-1) z^2}$$
for $v$ = 3,
$$-\frac{2 (3 z^2+4 (2 \sqrt{1-z}-3) z-8 \sqrt{1-z}+8)}{(1-z)^{3/2} z^3}$$
and so on.
I'm guessing a transformation is repeatedly applied until a terminating form of the hypergeometric series is obtained. For $v$ = 1, applying Euler's transformation, 
$$_2F_1 (a,b;c;z) =
(1-z)^{c-a-b}{}_2F_1 (c-a, c-b;c ; z)$$
gives the correct form; however, I cant work out what is used for $v$ = 2 and higher.
 A: Answering my own question with this post:
Tables of Hypergeometric Functions
There is a link to a paper that describes a method for performing the transformations. The method is quite complicated and not worth rewriting here. Apparently the python package sympy has a function "hyperexpand" that can work it out.
Update:
Thanks again to Antonio for the hint to start from a known form. I've worked out the process:
Substituting $a = v/2$ we get
$$ _2F_1\left\{\frac{v+2}{2},\frac{v+3}{2};v+1;z\right\} = \,_2F_1\left\{a+1,a+\frac{3}{2};2a+1;z\right\} $$
Since [1],
$$ _2F_1\left\{a,a+\frac{1}{2};2a+1;z\right\} = 2^{2a}(1 + \sqrt{1-z})^{-2a} $$
and [2],
$$ _2F_1\{a_1 + 1,a_2;b;z\} = \left(\frac{z}{a_1}\frac{d}{dz} + 1\right) \,_2F_1\{a_1, a_2;b;z\}, $$
$$ _2F_1\{a_1,a_2+1;b;z\} = \left(\frac{z}{a_2}\frac{d}{dz} + 1\right) \,_2F_1\{a_1, a_2;b;z\}, $$
then,
$$ _2F_1\left\{a+1,a+\frac{3}{2};2a+1;z\right\}  = \left(\frac{z}{a_1}\frac{d}{dz} + 1\right)\left(\frac{z}{a_2}\frac{d}{dz} + 1\right)\left(2^{2a}(1 + \sqrt{1-z})^{-2a}\right).$$
The rest follows by doing the differentiation and substuting back $v/2 = a$, and gives the same result as described in Antonio's answer.
