# Inverse VS Reduce

I got a solution I want to get inverse but it is not working in Mathematica or Rubi:

f[x_] := (a*ArcTanh[(a*x)/Sqrt[-d + b*x^2]])/(a^2 - b)(Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[-d + b*x^2]])/(a^2 - b) + (a*Log[d + (a^2 - b)*x^2])/(2*(a^2 - b));

InverseFunction[(a*ArcTanh[(a*x)/Sqrt[-d + b*x^2]])/(a^2 - b) -(Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[-d + b*x^2]])/(a^2 - b) + (a*Log[d + (a^2 - b)*x^2])/(2*(a^2 - b)), x]

Reduce[y == f[x], x]


All parameters are positive.

• Probably you consider a real function? If yes, FunctionDomain[f[x], x, Reals] shows several necessary conditions which must be fullfilled. Without further restriction Mathematica cannot evaluate the  InverseFunction Nov 10, 2021 at 21:29
• Duplicates: (257680), (257851) -- You make the same syntax error in InverseFunction above as was corrected for you in (257680). Nov 11, 2021 at 18:40
• What makes you think this equation/inverse can be solved symbolically? Nov 11, 2021 at 18:41