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evaluate (x ArcTanh[(Sqrt[2] Sqrt[-x^2] Sin[u])/Sqrt[1 - 2 x^2 + Cos[2 u]]] Sqrt[1 - 2 x^2 + Cos[2 u]] Sec[u])/(8 Sqrt[2] Pi Sqrt[-x^2] Sqrt[1 - x^2 Sec[u]^2]) with u=arccos(x)
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