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I solve an easy example to check that Mathematica can understand the difference between ArcTan and ArcTanh but as you see here I can't get Arctanh??!! Even I don't know why the inverse function is not working! thanks for helping

g = Integrate[1/((-(3/2))*x^2 + (3/2)*a*x+ b/2), x, Assumptions -> {a > 0, b > 0}]
Assuming[{a > 0, b >0},InverseFunction[g][x]]

a,b are positive.

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  • $\begingroup$ g isn't a function; it's a symbolic expression. Try g[x_] = Integrate[...] instead (after Clear[g]); $\endgroup$
    – Sjoerd Smit
    Nov 18 '21 at 14:42
  • $\begingroup$ My problem is the answer is Tanh[] but it gives me Tan[] !!! tan(ix) = i tanh x $\endgroup$
    – Mathecis
    Nov 18 '21 at 15:12
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You'll get the result in terms of the hyperbolic functions if you use FullSimplify:

Clear[g]
g[x_] = Assuming[
  a > 0 && b > 0,
  FullSimplify@Integrate[1/((-(3/2))*x^2 + (3/2)*a*x + b/2), x]
  ]
Assuming[
 a > 0 && b > 0,
 FullSimplify[InverseFunction[g][x]]
 ] // Normal

-((4 ArcTanh[(a - 2 x)/Sqrt[a^2 + (4 b)/3]])/Sqrt[9 a^2 + 12 b])

1/6 (3 a + Sqrt[9 a^2 + 12 b] Tanh[1/4 Sqrt[9 a^2 + 12 b] x])

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  • $\begingroup$ Thank you, so assuming will work here $\endgroup$
    – Mathecis
    Nov 19 '21 at 15:51

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