simplification

[dimitris]

Hello to all.

I would appreciate your suggestions about simplifying further the
following lengthy expression.

ArcTan[(3*(Sqrt[3] + I*x - I*x^2)*(1 - x + x^2))/(-3 + I*Sqrt[3] - 3*x
- 3*I*Sqrt[3]*x + 3*x^2 + I*Sqrt[3]*x^2 - 6*x^3 -
2*I*Sqrt[3]*x^4 - I*Sqrt[3*(1 - I*Sqrt[3])]*Sqrt[1 - x + x^2] +
I*Sqrt[3*(1 - I*Sqrt[3])]*x*Sqrt[1 - x + x^2] +
I*Sqrt[3*(1 - I*Sqrt[3])]*x^2*Sqrt[1 - x + x^2] + 2*I*Sqrt[3*(1
- I*Sqrt[3])]*x^3*Sqrt[1 - x + x^2])]/
Sqrt[3*(1 - I*Sqrt[3])] - ArcTan[(3*(Sqrt[3] - I*x + I*x^2)*(1 - x
+ x^2))/(3 + I*Sqrt[3] + 3*x - 3*I*Sqrt[3]*x - 3*x^2 +
I*Sqrt[3]*x^2 + 6*x^3 - 2*I*Sqrt[3]*x^4 - I*Sqrt[3*(1 +
I*Sqrt[3])]*Sqrt[1 - x + x^2] +
I*Sqrt[3*(1 + I*Sqrt[3])]*x*Sqrt[1 - x + x^2] + I*Sqrt[3*(1 +
I*Sqrt[3])]*x^2*Sqrt[1 - x + x^2] +
2*I*Sqrt[3*(1 + I*Sqrt[3])]*x^3*Sqrt[1 - x + x^2])]/Sqrt[3*(1 +
I*Sqrt[3])] -
(I*Log[(-I + Sqrt[3] - 2*I*x)^2*(I + Sqrt[3] + 2*I*x)^2])/
(2*Sqrt[3*(1 - I*Sqrt[3])]) +
(I*Log[(-I + Sqrt[3] - 2*I*x)^2*(I + Sqrt[3] + 2*I*x)^2])/
(2*Sqrt[3*(1 + I*Sqrt[3])]) +
(I*Log[(1 + x + x^2)*(11*I + 4*Sqrt[3] - 17*I*x - 4*Sqrt[3]*x +
11*I*x^2 + 4*Sqrt[3]*x^2 +
10*I*Sqrt[1 - I*Sqrt[3]]*Sqrt[1 - x + x^2] - 8*I*Sqrt[1 -
I*Sqrt[3]]*x*Sqrt[1 - x + x^2])])/(2*Sqrt[3*(1 - I*Sqrt[3])]) -
(I*Log[(1 + x + x^2)*(-11*I + 4*Sqrt[3] + 17*I*x - 4*Sqrt[3]*x -
11*I*x^2 + 4*Sqrt[3]*x^2 -
10*I*Sqrt[1 + I*Sqrt[3]]*Sqrt[1 - x + x^2] + 8*I*Sqrt[1 +
I*Sqrt[3]]*x*Sqrt[1 - x + x^2])])/(2*Sqrt[3*(1 + I*Sqrt[3])])

Thanks in advance.
Dimitris

[dimitris]

I would like to add some background for the expression.
This expression arose when I tried to evaluate with Mathematica the
following indefinite integral

intMath=Integrate[1/((x^2 + x + 1)*Sqrt[x^2 - x + 1]), x]

and Mathematica produced above lengthy expression.
This integral appeared in another forum.
What it is very interesting is that the person who mentioned the
integral gave an expression
for the antiderivative which is just:

intUser=ArcTan[(Sqrt[2]*(1 + x))/Sqrt[1 - x + x^2]]/Sqrt[2] +
ArcTanh[(Sqrt[2/3]*(-1 + x))/Sqrt[1 - x + x^2]]/Sqrt[6];

As you can see this is a correct antiderivative since

D[intUser, x] - 1/((x^2 + x + 1) Sqrt[x^2 - x + 1]) // FullSimplify
0

User's antiderivative and Mathematica's differ just a single constant:

Chop[Table[intUser - intMath /. x -> RandomReal[{-1000, 0}], {20}]]
Chop[Table[intUser - intMath /. x -> RandomReal[{0, 1000}], {20}]]
(*output omitted*)

I have tried with a lot of ways to simplify Mathematica's lengthy
expression, just by curiosity, trying every way
I know (and I remember! It has been a lot time since I dealt with
procedures like this.)
But I thought it will be an interesting thread for many people in this
forum!

Dimitris

P.S. It doesn't have a connection with my original query, but just for
the record:
User's antiderivative is continuous in the whole real axis.
Mathematica is not.