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In part of my calculations, I obtain this expression which contains the imaginary unit I but I expect that this expression might be real (from the comments below, now we are sure that it is real). How can I ask Mathematica to simplify this expression and give its explicit real form?

exp=1/96 (\[Pi]^2 + 
   12 I ArcCos[-(5/4)] ArcTanh[
     1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] + 
   20 \[Pi]^2 Cosh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] + 
   6 I ArcCos[-(5/4)] Log[2] - 
   12 ArcTanh[
     1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
     2] + 3 Log[2]^2 - 
   12 Cosh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[
     2]^2 - 12 I \[Pi] Cosh[
     1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[4] - 
   6 I ArcCos[-(5/4)] Log[9] + 
   12 I ArcCos[-(5/4)] Cosh[
     1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[16] + 
   12 ArcTanh[
     3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
     1 - I Sqrt[15]] - 
   48 I ArcCos[
     I Sqrt[2/3]
       Sinh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Cosh[
     1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[
     1 - I Sqrt[15]] - 
   12 ArcTanh[
     3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
     1 + I Sqrt[15]] + 
   48 I ArcCos[
     I Sqrt[2/3]
       Sinh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Cosh[
     1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] Log[
     1 + I Sqrt[15]] + 
   6 I ArcCos[-(5/4)] Log[I/(
     2 Sqrt[(I + Sqrt[15])/(3 I - 3 Sqrt[15])])] + 
   12 ArcTanh[
     3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[I/(
     2 Sqrt[(I + Sqrt[15])/(3 I - 3 Sqrt[15])])] - 
   12 ArcTanh[
     1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
     I/(2 Sqrt[(I + Sqrt[15])/(3 I - 3 Sqrt[15])])] + 
   6 I ArcCos[-(5/4)] Log[I/(
     2 Sqrt[(I - Sqrt[15])/(3 I + 3 Sqrt[15])])] - 
   12 ArcTanh[
     3 Coth[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[I/(
     2 Sqrt[(I - Sqrt[15])/(3 I + 3 Sqrt[15])])] + 
   12 ArcTanh[
     1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[
     I/(2 Sqrt[(I - Sqrt[15])/(3 I + 3 Sqrt[15])])] - 
   6 I ArcCos[-(5/4)] Log[(-1 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] + 
   12 ArcTanh[
     1/3 Tanh[
       1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[(-1 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] - 
   6 I ArcCos[-(5/4)] Log[(
     1 + Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] - 
   12 ArcTanh[
     1/3 Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])]] Log[(
     1 + Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] + 
   6 Log[2] Log[(
     3 + Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])] + 
   3 Log[(3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(-3 + 
      Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])]^2 - 
   48 Cosh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])] PolyLog[
     2, -(1/2)] + 
   6 PolyLog[
     2, -((-3 + 
       Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(
      2 (3 + Tanh[
          1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])))] + 
   6 PolyLog[
     2, -((3 + 
       Tanh[1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])/(
      2 (-3 + Tanh[
          1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])])))]) Sech[
  1/2 (Log[1 - I Sqrt[15]] - Log[1 + I Sqrt[15]])];
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6
  • 1
    $\begingroup$ N@exp gives 0.0889796 - 7.40149*10^-17 I and you can Chop the I part. So it looks Real. $\endgroup$
    – Syed
    Apr 11 at 14:01
  • $\begingroup$ @Syed Thanks. I do not want to use the numerics N; if possible, I like to have it in its explicit real form. $\endgroup$
    – MsMath
    Apr 11 at 14:07
  • 2
    $\begingroup$ First, use FullSimplify[exp ∈ Reals] to get $\text{Li}_2\left(\frac{i \sqrt{15}}{8}-\frac{1}{8}\right)+\text{Li}_2\left(-\frac{i \sqrt{15}}{8}-\frac{1}{8}\right)$, then apply the mirror symmetry to see that the result is indeed real. It doesn't look like Mathematica knows about the mirror symmetry, though ... $\endgroup$
    – Domen
    Apr 11 at 14:17
  • 5
    $\begingroup$ FullSimplify[Im[exp]] returns zero and that doesn't use N. $\endgroup$
    – Bill
    Apr 11 at 15:28
  • $\begingroup$ @Bill Thanks, yes, it works. And is there any hope to find the explicit real form of the function? $\endgroup$
    – MsMath
    Apr 11 at 15:40

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