Radial integral involving WhittakerM and Hypergeometric0F1

I am trying to find the definite integral of $$I=\int_{0}^{\infty}dx\left(\frac{\sqrt{n^{2l-1}}M_{n,l+\frac{1}{2}}(\frac{2x}{n})}{\Gamma(2l+2)}\right)x\left(\frac{(2r)^{l_f+1} {}_0F_1(2l_f+2;-2r)}{\sqrt{2}\Gamma(2l_f+2)}\right)$$.

Where $$M_{n,l+\frac{1}{2}}(\frac{2x}{n})$$ is the WhittakerM function, $${}_0F_1(2l_f+1;-2r)$$ is the Hypergeometric0F1 function, $$r=a_0*x$$, where $$a_0$$ is the bohr radius in Rydberg units ($$a_0=\frac{h^2}{4\pi^2m_ee_0^2}$$). I am trying to compute this integral for when $$n=2,l=0$$ and $$l_f=1$$ and my input is this:

h=UnitConvert[Quantity["PlanckConstant"]];
m=UnitConvert[Quantity["ElectronMass"]];
q=UnitConvert[Quantity["ElectronCharge"]];
a=h^2/(4*Pi^2*m*q^2)
Integrate[
x (Sqrt[2^(2*0 - 1)]*WhittakerM[2, 0 + 1/2, (2*x)/2])/
Gamma[2*0 + 2] ((2*a*x)^(1 + 1) Hypergeometric0F1[2*1 + 2, -2*a*x])/(
Sqrt[2] Gamma[2*1 + 2]), {x, 0, \[Infinity]}]


But I get

Integrate:Integrate was unable to determine the units of quantities that appear
in the input"


And if I omit the bohr radius the output I get is the input, any ideas on what's going wrong?

• Could you please update your title? I think it'd be better to write something like "Integral involving WhittakerM and Hypergeometric0F1". I mean more beneficial for other users as well :-)
– bmf
Mar 26, 2022 at 18:37
• @bmf I hope it's fine that I used your suggestion for the new title? Mar 26, 2022 at 18:39
• Of course it is. Perfectly fine. Thanks a lot!
– bmf
Mar 26, 2022 at 18:40
• Question: did you try to input only the numerical values for the constants without their units? The units can be restored at the end by means of dimensional analysis
– bmf
Mar 26, 2022 at 18:42
• Sort of? Might have been lazy but I tried just replacing $a$ with an undefined $c$ and adding $Assuming[c>0,....]$ which gave me the same result as if I omitted $a$. Mar 26, 2022 at 18:45

Integrate[x (Sqrt[2^(2*0 - 1)]*WhittakerM[2, 0 + 1/2, (2*x)/2])/