Integration of the HyperGeometric function DLMF 15.5.E8 The goal is to have an understandable proof of DLMF 15.5.E8 that can be generalized, or not, to Generalized HyperGeometric functions. This result is also in Wolfram http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/20/02/05/
and Slater “Generalized Hypergeometric Functions” 1.4.1.8
Rewriting DLMF http://dlmf.nist.gov/15.5.E8:
$\left((1-z)\frac{\mathrm{d}}{\mathrm{d}z}(1-z)\right)^{n}(z^{c-1}(1-z)^{b-c}F(a,b;c;z))$
$=(c-1)_{n} z^{c-1-n}(1-z)^{b-c+n} F(a-1,b;c-1;z)$
with $n=1$ and, if you want, a an integer $<0$
$(\frac{\mathrm{d}}{\mathrm{d}z}\left(z^{c-1}(1-z)^{b-c+1}F\left(a,b;c;z\right)\right)$
$=(c-1)z^{c-2}(1-z)^{b-c} F(a-1,b;c-1;z)$
 Now for a particular power k we have:
$(\Sigma_{k}\frac{d}{dz}(z^{c-1}(1-z)^{b-c+1}\frac{(a)_{k},(b)_{k}}{(c)_{k}} \frac{z^{k}}{k!})$
$=\Sigma_{k} \frac{z^{c + k-2}  (1 - z )^{b-c}  (a)_{k}  (b)_{n}  ( - b\cdot z + c - k\cdot z + k - 1)}{ k! (c)_{k} }$
$ = (c-n)_{n} z^{(c-n-1)}(1-z)^{(b-c+n)} F(a-n,b;c-n;z) $
I will settle for the case $n=1$
Maxima/wxMaxima code for examining terms
t:z^(c+k-1)(1-z)^(b-c+1)/k!;
tt:diff(t,z);
factor(tt);
s:z^(c-1)(1-z)^(b-c+1)*hypergeometric([a,b],[c],z);
ss:diff(s,z);
factor(ss);  
I will try to give bonus points for a proof using the results in (or similar to) DLMF http://dlmf.nist.gov/16 
My underlying goal is the raising of a in F(a-1,b,c,x) by
$h(x,b,c)\int g(x,b,c)F(a-1,b,c,x)dx$
with h(),g() independent of a.
Perhaps it should be noted that when a is a negative integer, this a result on polynomials.  
Progress: After thinking I realized that, using DLMF numbers, applying 15.8.1 and then 15.5.4 and then 15.8.1 in reverse will prove, at least "formally", 15.5.9.  So I am presuming that a similar sequence might work for 15.5.8; although it gets a little confusing/complicated.
As far as Generalized HyperGeometric goes these seem to point to solving via. the Mellin transform, or not.
 A: This is a lot simpler than it looks:


*

*Beat up an identity until it is in the form you want and is differentiable

*Differentiate

*Reverse the identity  
Constructive comments welcomed.

Using http://dlmf.nist.gov/15.5.E4 and the chain rule
$\frac{d}{dz}\left(\left(\frac{-z}{1-z}\right)^{c-1}\cdot F\left(a,b;c;\frac{-z}{1-z}\right)\right) [diff]$
$=\left(\left(c-1\right)\cdot\left(\frac{-z}{1-z}\right)^{c-2}\cdot F\left(a,b;c-1;\frac{-z}{1-z}\right)\right)\cdot\left(\frac{d}{dz}\left(\frac{-z}{1-z}\right)\right)$
We have:
$\frac{d}{dz}\left(\left(\frac{-z}{1-z}\right)^{c-1}\cdot F\left(a,b;c;\frac{-z}{1-z}\right)\right)  [xform]$ 
$=\left(\left(c-1\right)\cdot\left(\frac{-z}{1-z}\right)^{c-2}\cdot F\left(a,b;c-1;\frac{-z}{1-z}\right)\right)\cdot\frac{-1}{\left(1-z\right)^{2}}$ 

Constructing the LHS of http://dlmf.nist.gov/15.5.E8
$\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot F\left(a,b;c;z\right)$
$=\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot\left(1-z\right)^{-b}\cdot F\left(c-a,b;c;\frac{-z}{1-z}\right)
 $
$=\left(\frac{-z}{1-z}\right)^{c-1}\cdot\left(F\left(c-a,b;c;\frac{-z}{1-z}\right)\right)$  
We have from [eq:diff]:
$\frac{d}{dz}\left(\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot F\left(a,b;c;z\right)\right)$
$=\frac{d}{dz}\left(\left(\frac{-z}{1-z}\right)^{c-1}\cdot\left(F\left(c-a,b;c;\frac{-z}{1-z}\right)\right)\right)$
$=\left(\left(c-1\right)\cdot\left(\frac{-z}{1-z}\right)^{c-2}\cdot F\left(c-a,b;c-1;\frac{-z}{1-z}\right)\right)\cdot\frac{-1}{\left(1-z\right)^{2}}
 $   

Reversing [eq:xform]
$\frac{d}{dz}\left(\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot F\left(a,b;c;z\right)\right)$
 $=\left(c-1\right)\cdot z^{c-2}\cdot\left(1-z\right)^{b-\left(c-2\right)}\cdot\left(1-z\right)^{-2}\cdot F\left(a-1,b;c-1;z\right)$
QED

And the indefinite integral
${\displaystyle \int}\left(c-1\right)\cdot z^{c-2}\cdot\left(1-z\right)^{b-c}\cdot F\left(a-1,b;c-1;z\right)$
 $=\left(z^{c-1}\cdot\left(1-z\right)^{b-c+1}\right)\cdot F\left(a,b;c;z\right)$

Missing: detailed evaluation of regions of convergence
Generalization thoughts:  


*

*http://dlmf.nist.gov/15.5.E4 is generalizable as http://dlmf.nist.gov/16.3.E4

*The process is dependent on the existence of an identity which, as far as my research goes, is specialized. I plan to work on that.

