Re: An exact simplification challenge - 57 (arctangents) - Proud

Vladimir Bondarenko schrieb:
> Indeed, Derive 6.1 handles this the way you have shown.
> 
> Actually, by a mistake I quoted the result produced by
> Derive 6.0. 8-(
> 
> I was wondering if someone can get this nice result
> using Maple or Mathematica?
> 
Using Mathematica 5.2 :
expr = (-((1/120)*(30 - 6*5^(1/2))^(1/2) + (1/120)*5^(1/2) - 1/40))*
     ArcTan[(8*(30 - 6*5^(1/2))^(1/2) - 8*5^(1/2) - 8)/
       (8*(10 - 2*5^(1/2))^(1/2) + 8*3^(1/2)*(5^(1/2) + 1))] +
    ((1/120)*(30 - 6*5^(1/2))^(1/2) - (1/120)*5^(1/2) + 1/40)*
     ArcTan[(8*(30 - 6*5^(1/2))^(1/2) + 8*5^(1/2) + 8)/
       (8*(10 - 2*5^(1/2))^(1/2) - 8*3^(1/2)*(5^(1/2) + 1))] -
    ((1/120)*(30 + 6*5^(1/2))^(1/2) + (1/120)*5^(1/2) + 1/40)*
     ArcTan[(8*(30 + 6*5^(1/2))^(1/2) - 8*5^(1/2) + 8)/
       (8*(10 + 2*5^(1/2))^(1/2) + 8*3^(1/2)*(5^(1/2) - 1))] -
    ((1/120)*(30 + 6*5^(1/2))^(1/2) - (1/120)*5^(1/2) - 1/40)*
     ArcTan[(8*(30 + 6*5^(1/2))^(1/2) + 8*5^(1/2) - 8)/
       (8*(10 + 2*5^(1/2))^(1/2) - 8*3^(1/2)*(5^(1/2) - 1))] +
    ((1/240)*(30 + 6*5^(1/2))^(1/2) + (1/240)*5^(1/2) + 1/80)*
     ArcTan[((10 - 2*5^(1/2))^(1/2) + 3^(1/2)*(5^(1/2) + 1))/
       ((30 - 6*5^(1/2))^(1/2) - 5^(1/2) - 1)] +
    ((1/240)*(30 + 6*5^(1/2))^(1/2) - (1/240)*5^(1/2) - 1/80)*
     ArcTan[((10 - 2*5^(1/2))^(1/2) - 3^(1/2)*(5^(1/2) + 1))/
       ((30 - 6*5^(1/2))^(1/2) + 5^(1/2) + 1)] -
    ((1/240)*(30 - 6*5^(1/2))^(1/2) - (1/240)*5^(1/2) + 1/80)*
     ArcTan[((10 + 2*5^(1/2))^(1/2) + 3^(1/2)*(5^(1/2) - 1))/
       ((30 + 6*5^(1/2))^(1/2) - 5^(1/2) + 1)] +
    ((1/240)*(30 - 6*5^(1/2))^(1/2) + (1/240)*5^(1/2) - 1/80)*
     ArcTan[((10 + 2*5^(1/2))^(1/2) - 3^(1/2)*(5^(1/2) - 1))/
       ((30 + 6*5^(1/2))^(1/2) + 5^(1/2) - 1)] +
    Pi*((1/24)*(15 + 6*5^(1/2))^(1/2) + (7/120)*5^(1/2) + 1/10);
Timing[FullSimplify[expr]]
--> {1.776731*Second,
      (1/30)*(3 + 2*Sqrt[5] + Sqrt[15 + 6*Sqrt[5]])*Pi}