3.649 \(\int \frac{e^{2 x}}{1+e^{4 x}} \, dx\)

Optimal. Leaf size=10 \[ \frac{1}{2} \tan ^{-1}\left (e^{2 x}\right ) \]

[Out]

ArcTan[E^(2*x)]/2

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Rubi [A]  time = 0.0204029, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2249, 203} \[ \frac{1}{2} \tan ^{-1}\left (e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(1 + E^(4*x)),x]

[Out]

ArcTan[E^(2*x)]/2

Rule 2249

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]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{e^{2 x}}{1+e^{4 x}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^{2 x}\right )\\ &=\frac{1}{2} \tan ^{-1}\left (e^{2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0030139, size = 10, normalized size = 1. \[ \frac{1}{2} \tan ^{-1}\left (e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(1 + E^(4*x)),x]

[Out]

ArcTan[E^(2*x)]/2

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Maple [A]  time = 0.02, size = 8, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( \left ({{\rm e}^{x}} \right ) ^{2} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(1+exp(4*x)),x)

[Out]

1/2*arctan(exp(x)^2)

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Maxima [A]  time = 1.49731, size = 9, normalized size = 0.9 \begin{align*} \frac{1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(4*x)),x, algorithm="maxima")

[Out]

1/2*arctan(e^(2*x))

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Fricas [A]  time = 0.751055, size = 28, normalized size = 2.8 \begin{align*} \frac{1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(4*x)),x, algorithm="fricas")

[Out]

1/2*arctan(e^(2*x))

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Sympy [B]  time = 0.109118, size = 17, normalized size = 1.7 \begin{align*} \operatorname{RootSum}{\left (16 z^{2} + 1, \left ( i \mapsto i \log{\left (4 i + e^{2 x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(4*x)),x)

[Out]

RootSum(16*_z**2 + 1, Lambda(_i, _i*log(4*_i + exp(2*x))))

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Giac [A]  time = 1.17685, size = 9, normalized size = 0.9 \begin{align*} \frac{1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(1+exp(4*x)),x, algorithm="giac")

[Out]

1/2*arctan(e^(2*x))