Discrepancy wolfram alpha and hand calculation of a gaussian integral

Question

Wolfram alpha gives me the following result: $$ \int_0^{+\infty} \frac{x^2\, e^{-a x^2}}{\sqrt{x^2 +m^2}}\, dx = \frac{\sqrt{\pi}}{4 a} U\Big(\frac{1}{2},0, a m^2 \Big)$$ and it does use the usual definition $$ U(a,b,z) := \frac{1}{\Gamma(a)} \int_0^{+\infty} e^{-zt}\, t^{a-1}\, (1+t)^{b-a-1}\, dt $$ but if one does the calculus by hand, one finds instead (i use brackets to indicate a change of variable) $$ \begin{aligned} \int_0^{+\infty} \frac{x^2\, e^{-a x^2}}{\sqrt{x^2 +m^2}}\, dx &= \genfrac{[}{]}{0pt}{0}{k=\frac{x}{m}}{dk = \frac{1}{m}\, dx} = \int_0^{+\infty} \frac{m^2 k^2\, e^{-a m^2 k^2}}{\sqrt{k^2 +1}}\, dx\\ &= \genfrac{[}{]}{0pt}{0}{t=k^2}{dt = 2\hspace{.6pt} k\, dk} = \int_0^{+\infty} \frac{m^2 t e^{-\hspace{.3pt}2\hspace{.3pt} n\hspace{.3pt} m^2 t}}{\sqrt{t +1}}\, \frac{dt}{2 \sqrt{t}} = \frac{m^2}{2} \Gamma\Big(\frac{3}{2} \Big) U \Big(\frac{3}{2},2,2\hspace{.3pt} n\hspace{.3pt} m^2 \Big) \end{aligned} $$

Answer 1

Sorry to answer directly, I understood while writting...

One finds the Kummer's transformation relating both expressions... so they really are equal... no bug in the matrix...