Integral of the hypergeometric function

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?

Additional Maple results:

> int(hypergeom([1,1,a],[b,c],z),z);

           z hypergeom([1, 1, 1, a], [2, b, c], z)

> int(hypergeom([1,1,1,a],[b,c,d],z),z);

            z hypergeom([1, 1, 1, 1, a], [2, b, c, d], z)

Barton

-----maxima-bounces at math.utexas.edu wrote: -----


>The?following?code?allows?maxima?to?integrate?the?hypergeometric?function.
>hypergeometric([a1,...,ap],[b1,...,bq],z).??It?is?added?to
>share/hypergeometric/hypergeometric.lisp.
>
>At?present?it?returns?a?"Division?by?0"?error?if?any?of?ai?are?equal?to
>one.
>Is?this?too?harsh???I?really?don't?see?the?value?in?giving?a?different
>error?message.
>
>######
>
>(defun?hyp-integral-3?(a?b?z)
>??"Integral?of?hypergeometric(a,b,z)?wrt?z"
>??(let?((a-1?(add?a?-1))
>????????(b-1?(add?b?-1)))
>????(mul?(reduce?#'mul?(margs?b-1))
>?????????(inv?(reduce?#'mul?(margs?a-1)))
>?????????(take?'($hypergeometric)?a-1?b-1?z))))
>
>;;Integral?of?hypergeometric(a,b,z).??Only?defined?for?third?arg?z.
>(putprop?'$hypergeometric?`((a?b?z)?nil?nil?,#'hyp-integral-3)?'integral)
>
>#####
>
>(%i1)?integrate(hypergeometric([],[2],z),z);
>(%o1)?????????????????????hypergeometric([],?[1],?z)
>(%i2)?integrate(hypergeometric([i,j],[3],z),z);
>???????????????????2?hypergeometric([i?-?1,?j?-?1],?[2],?z)
>(%o2)??????????????----------------------------------------
>???????????????????????????????(i?-?1)?(j?-?1)
>(%i3)?integrate(hypergeometric([1],[2],z),z);
>
>Division?by?0
>?--?an?error.?To?debug?this?try:?debugmode(true);
>
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