\(\int \frac {e^x}{1+e^{2 x}} \, dx\) [38]

Optimal result

Integrand size = 13, antiderivative size = 4 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^x\right ) \]

[Out]

arctan(exp(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2281, 209} \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^x\right ) \]

[In]

Int[E^x/(1 + E^(2*x)),x]

[Out]

ArcTan[E^x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2281

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[d*e*(Log[F]/(g*h*Log[G]))]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Denominator[m]))], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right ) \\ & = \tan ^{-1}\left (e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^x\right ) \]

[In]

Integrate[E^x/(1 + E^(2*x)),x]

[Out]

ArcTan[E^x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00

method	result	size
default	\(\arctan \left ({\mathrm e}^{x}\right )\)	\(4\)
risch	\(\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2}\)	\(20\)

[In]

int(exp(x)/(1+exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

arctan(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^{x}\right ) \]

[In]

integrate(exp(x)/(1+exp(2*x)),x, algorithm="fricas")

[Out]

arctan(e^x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (3) = 6\).

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\operatorname {RootSum} {\left (4 z^{2} + 1, \left ( i \mapsto i \log {\left (2 i + e^{x} \right )} \right )\right )} \]

[In]

integrate(exp(x)/(1+exp(2*x)),x)

[Out]

RootSum(4*_z**2 + 1, Lambda(_i, _i*log(2*_i + exp(x))))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^{x}\right ) \]

[In]

integrate(exp(x)/(1+exp(2*x)),x, algorithm="maxima")

[Out]

arctan(e^x)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\arctan \left (e^{x}\right ) \]

[In]

integrate(exp(x)/(1+exp(2*x)),x, algorithm="giac")

[Out]

arctan(e^x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {e^x}{1+e^{2 x}} \, dx=\mathrm {atan}\left ({\mathrm {e}}^x\right ) \]

[In]

int(exp(x)/(exp(2*x) + 1),x)

[Out]

atan(exp(x))