Integral of the hypergeometric function



Barton Willis wrote:
> My Maple tells me that integrate(hypergeometric([1],[2],z),z) = z *
> hypergeometric([1, 1],[2, 2],z),
> but I don't know the general rule. Can we do better than an error?
>   
None of the references at hand had it, but fortunately I could view the 
relevant page of
Prudnikov at amazon.com.  We need 1.16.1.2 for the case when one of the 
a_i equals 1.

;; Integrals and Series: Volume 3, More Special Functions
;; Prudnikov, A. P., Brychkov, Yu A., Gould, G. G., Marichev, O.I.
;;
;;  /
;; [
;; I pFq((a_p);(b_q);c z) dz
;; ]
;; /
;;
;;     = z (p+1)F(q+1)((a_p),1;(b_q),2;c z)                 1.16.1.2
;;
;;        product((b_q - 1))
;;     =  ------------------ pFq((a_p)-1; (b_q)-1; c z)     1.16.1.3
;;        product((a_p - 1))

I have checked in a fix.