According to mathematica, for $\alpha,z>0$ we have \begin{multline} G^{3,1}_{2,3}\left(z\bigg|{0,1\atop 0,0,1-\alpha}\right)=% -\frac{z}{\alpha}\Gamma(1-\alpha){_2F_2}\left({1,1\atop 2,\alpha+1};z\right)\\% \qquad\qquad\qquad-\pi\sin(\pi\alpha)\csc^2(\pi(\alpha-1))\Gamma(1-\alpha)(-z)^\alpha z^{-\alpha}\\% \qquad\qquad\qquad\qquad\qquad\qquad+\pi\sin(\pi\alpha)\csc^2(\pi(\alpha-1))(-z)^\alpha z^{-\alpha}\Gamma(1-\alpha,-z)\\% -\Gamma(1-\alpha)\log z% +\Gamma(1-\alpha)\psi(1-\alpha). \end{multline}
The Wolfram functions website at http://functions.wolfram.com/07.34.03.0987.01 provides a way to write $G^{3,1}_{2,3}()$ as a sum of $_2F_2()$'s but only when the bottom parameters do not differ by integer ammounts. Everywhere else I have looked has also required this restriction. Is there a property of the G function that would allow me to write this G-function such that the bottom parameters do not differ by an integer? How else might I reduce this if not?
It seems like if I were able to get rid of one of the zero's in the bottom argument and assume $\alpha$ is not an integer, then I would be able to apply the relation on the wolfram functions site.
As pointed out by @Maxim, this expression is due to the residue series expansion of the $G$ function. Using the residue expansion we write \begin{align} G^{3,1}_{2,3}\left(z\bigg|{0,1\atop 0,0,1-\alpha}\right)% =& \sum_{k=0}^{\infty}\operatorname*{Res}_{s=-k}\left(\frac{\Gamma(s)\Gamma(s+1-\alpha)\Gamma(1-s)}{s}z^{-s}\right)\\% &\quad+\sum_{k=0}^{\infty}\operatorname*{Res}_{s=\alpha-1-k}\left(\frac{\Gamma(s)\Gamma(s+1-\alpha)\Gamma(1-s)}{s}z^{-s}\right). \end{align} As pointed out, the residue at the double pole for $s=0$ yields logarithmic and polygamma terms. The sums of the residues at the poles for $s=-k$ for $k>0$ and $s=\alpha-1-k$ yield hypergeometric and incomplete gamma terms. In total we find
\begin{align} G^{3,1}_{2,3}\left(z\bigg|{0,1\atop 0,0,1-\alpha}\right)=% &-\Gamma(1-\alpha)\biggl(% \log z-\psi(1-\alpha)% +\frac{z}{\alpha}{_2F_2}\left({1,1\atop 1+\alpha,2};z\right)\\% &\quad+\pi\csc(\pi\alpha)(-1)^\alpha\left(1-\frac{\Gamma(1-\alpha,-z)}{\Gamma(1-\alpha)}\right)% \biggr). \end{align} With a bit of algebra one can show that this expression is the same as in the original post above.