The following result of indefinite integral contains hypergeometric series, but the reference answer is $-\frac{1}{2}\left(e^{-2 x} \arctan e^{x}+e^{-x}+\arctan e^{x}\right)$.
Integrate[ArcTan[E^x]/E^(2 x), x,
GeneratedParameters -> C] // FullSimplify
D[Integrate[ArcTan[E^x]/E^(2 x),
x] - (-(1/2) (E^(-2 x) ArcTan[E^x] + E^-x + ArcTan[E^x])),
x] // FullSimplify
How can I further simplify the above indefinite integral result into the form of reference answer?