convert/Whittaker converts, when possible, special functions admitting a 1F1 or 0F1 hypergeometric representation into Whittaker functions. The Whittaker functions are

FunctionAdvisor( Whittaker );

The 2 functions in the "Whittaker" class are:

$\left[\mathrm{WhittakerM}\,\mathrm{WhittakerW}\right]$

$\mathrm{AiryAi}\left(z\right)$

$\mathrm{AiryAi}\left(z\right)$

$\mathrm{convert}\left(\,\mathrm{Whittaker}\right)$

$\frac{\sqrt{3}\mathrm{WhittakerM}\left(0\,-\frac{1}{3}\,\frac{4z^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}\right)4^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{12\mathrm{\Gamma }\left(\frac{2}{3}\right){\left(z^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}-\frac{3z\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{WhittakerM}\left(0\,\frac{1}{3}\,\frac{4z^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}\right)4^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}{8\mathrm{\pi }{\left(z^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}$

$\mathrm{HermiteH}\left(a\,z\right)\mathrm{LaguerreL}\left(2\,\exp \left(z\right)\right)$

$\mathrm{HermiteH}\left(a\,z\right)\mathrm{LaguerreL}\left(2\,{\ⅇ}^z\right)$

$\mathrm{convert}\left(\&comma;\mathrm{Whittaker}\right)\mathbf{assuming}0<\mathrm{\Re }\left(z\right)$

$\frac{2^a\sqrt{\mathrm{\pi }}{\&ExponentialE;}^{\frac{z^2}{2}}\left(\frac{\mathrm{WhittakerM}\left(\frac{a}{2}+\frac{1}{4}\&comma;-\frac{1}{4}\&comma;z^2\right)}{{\left(z^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\mathrm{\Gamma }\left(\frac{1}{2}-\frac{a}{2}\right)}-\frac{2z\mathrm{WhittakerM}\left(\frac{a}{2}+\frac{1}{4}\&comma;\frac{1}{4}\&comma;z^2\right)}{{\left(z^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\mathrm{\Gamma }\left(-\frac{a}{2}\right)}\right)\mathrm{WhittakerM}\left(\frac{5}{2}\&comma;0\&comma;\frac{\mathrm{WhittakerM}\left(\mathrm{-1}\&comma;\frac{1}{2}\&comma;z\right){\&ExponentialE;}^{\frac{z}{2}}}{z}\right){\&ExponentialE;}^{\frac{\mathrm{WhittakerM}\left(\mathrm{-1}\&comma;\frac{1}{2}\&comma;z\right){\&ExponentialE;}^{\frac{z}{2}}}{2z}}}{\sqrt{\frac{\mathrm{WhittakerM}\left(\mathrm{-1}\&comma;\frac{1}{2}\&comma;z\right){\&ExponentialE;}^{\frac{z}{2}}}{z}}}$

$\frac{\left(\exp \left(z\right)\mathrm{erf}\left(z^2\right)+\mathrm{KummerU}\left(-1\&comma;\frac{1}{2}\&comma;z\right)\exp \left(\frac{1}{2}z\right)\right)\mathrm{MeijerG}\left(\left[\left[1-a\right]\&comma;\left[\right]\right]\&comma;\left[\left[0\&comma;1-b\right]\&comma;\left[\right]\right]\&comma;\frac{1}{z}\right)}{\mathrm{BesselK}\left(-3\&comma;1-z\right)}$

$\frac{\left({\&ExponentialE;}^z\mathrm{erf}\left(z^2\right)+\mathrm{KummerU}\left(\mathrm{-1}\&comma;\frac{1}{2}\&comma;z\right){\&ExponentialE;}^{\frac{z}{2}}\right)\mathrm{MeijerG}\left(\left[\left[1-a\right]\&comma;\left[\right]\right]\&comma;\left[\left[0\&comma;1-b\right]\&comma;\left[\right]\right]\&comma;\frac{1}{z}\right)}{\mathrm{BesselK}\left(3\&comma;1-z\right)}$

$\mathrm{convert}\left(\&comma;\mathrm{Whittaker}\right)$

$\frac{\left(\frac{2\mathrm{WhittakerM}\left(\mathrm{-1}\&comma;\frac{1}{2}\&comma;z\right){\&ExponentialE;}^{\frac{z}{2}}z\mathrm{WhittakerM}\left(\frac{1}{4}\&comma;\frac{1}{4}\&comma;-z^4\right)}{\sqrt{\mathrm{\pi }}{\&ExponentialE;}^{\frac{z^4}{2}}{\left(-z^4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}+\frac{2\mathrm{WhittakerW}\left(\frac{5}{4}\&comma;-\frac{1}{4}\&comma;z\right){\&ExponentialE;}^{\frac{z}{2}}\mathrm{WhittakerM}\left(\mathrm{-1}\&comma;\frac{1}{2}\&comma;\frac{z}{2}\right){\&ExponentialE;}^{\frac{z}{4}}}{z^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\right){\&ExponentialE;}^{\frac{1}{2z}}\left(\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(1-b\right)\mathrm{WhittakerM}\left(-a+\frac{b}{2}\&comma;\frac{b}{2}-\frac{1}{2}\&comma;\frac{1}{z}\right)+\mathrm{WhittakerM}\left(-a+\frac{b}{2}\&comma;\frac{1}{2}-\frac{b}{2}\&comma;\frac{1}{z}\right)\mathrm{\Gamma }\left(1+a-b\right)\mathrm{\Gamma }\left(-1+b\right)\right)\sqrt{2-2z}}{{\left(\frac{1}{z}\right)}^{\frac{b}{2}}\sqrt{\mathrm{\pi }}\mathrm{WhittakerW}\left(0\&comma;3\&comma;2-2z\right)}$