Hi I integrated a function yesterday and found the below output in terms of Bessels function

In[131]:= ClearAll["Global`*"];
U = Integrate[
  2 ((1 + r \[Beta]^2)/(
     M0 \[Beta]^4) + ((-1 - \[Beta]^2 - \[Alpha] \[Beta]^2) BesselI[0,
        2 Sqrt[r] \[Beta]])/(
     M0 \[Beta]^4 (BesselI[0, 
         2 \[Beta]] + \[Alpha] \[Beta] BesselI[1, 2 \[Beta]]))) r, {r,
    0, 1}]

Out[132]= (\[Beta] (6 + (9 + 6 \[Alpha]) \[Beta]^2 + 
    2 \[Beta]^4) BesselI[0, 
   2 \[Beta]] + (-6 - 6 (2 + \[Alpha]) \[Beta]^2 - 
    3 (2 + \[Alpha]) \[Beta]^4 + 2 \[Alpha] \[Beta]^6) BesselI[1, 
   2 \[Beta]])/(3 M0 \[Beta]^7 (BesselI[0, 
    2 \[Beta]] + \[Alpha] \[Beta] BesselI[1, 2 \[Beta]]))

But now when i tried it again i am getting a different solution with hypergeometric function. It seems both the solution are same. As i need the function in simplest form for futher calculations, mathematica is not producing same solution which it produced yesterday. Is there any way to convert the hypergeometeric function to bessles one again or can i get the same solution cause it looks simple to read. Second solution is given below:

In[9]:= ClearAll["Global`*"];
U = Integrate[
  2 ((1 + r \[Beta]^2)/(
     M0 \[Beta]^4) + ((-1 - \[Beta]^2 - \[Alpha] \[Beta]^2) BesselI[0,
        2 Sqrt[r] \[Beta]])/(
     M0 \[Beta]^4 (BesselI[0, 
         2 \[Beta]] + \[Alpha] \[Beta] BesselI[1, 2 \[Beta]]))) r, {r,
    0, 1}]

Out[10]= (3 + 2 \[Beta]^2 - (
 6 (1 + (1 + \[Alpha]) \[Beta]^2) (Hypergeometric0F1Regularized[
     3, \[Beta]^2] + \[Beta]^2 Hypergeometric0F1Regularized[
      4, \[Beta]^2]))/(
 BesselI[0, 
   2 \[Beta]] + \[Alpha] \[Beta] BesselI[1, 
    2 \[Beta]]))/(3 M0 \[Beta]^4)