3.26 \(\int \frac{e^x}{2+3 e^{2 x}} \, dx\)

Optimal. Leaf size=18 \[ \frac{\tan ^{-1}\left (\sqrt{\frac{3}{2}} e^x\right )}{\sqrt{6}} \]

[Out]

ArcTan[Sqrt[3/2]*E^x]/Sqrt[6]

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Rubi [A]  time = 0.0204306, antiderivative size = 18, 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} \[ \frac{\tan ^{-1}\left (\sqrt{\frac{3}{2}} e^x\right )}{\sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Int[E^x/(2 + 3*E^(2*x)),x]

[Out]

ArcTan[Sqrt[3/2]*E^x]/Sqrt[6]

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^x}{2+3 e^{2 x}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{2+3 x^2} \, dx,x,e^x\right )\\ &=\frac{\tan ^{-1}\left (\sqrt{\frac{3}{2}} e^x\right )}{\sqrt{6}}\\ \end{align*}

Mathematica [A]  time = 0.0068214, size = 18, normalized size = 1. \[ \frac{\tan ^{-1}\left (\sqrt{\frac{3}{2}} e^x\right )}{\sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(2 + 3*E^(2*x)),x]

[Out]

ArcTan[Sqrt[3/2]*E^x]/Sqrt[6]

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Maple [A]  time = 0.004, size = 14, normalized size = 0.8 \begin{align*}{\frac{\sqrt{6}}{6}\arctan \left ({\frac{{{\rm e}^{x}}\sqrt{6}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(2+3*exp(2*x)),x)

[Out]

1/6*arctan(1/2*exp(x)*6^(1/2))*6^(1/2)

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Maxima [A]  time = 1.41214, size = 18, normalized size = 1. \begin{align*} \frac{1}{6} \, \sqrt{6} \arctan \left (\frac{1}{2} \, \sqrt{6} e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(2+3*exp(2*x)),x, algorithm="maxima")

[Out]

1/6*sqrt(6)*arctan(1/2*sqrt(6)*e^x)

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Fricas [A]  time = 1.94857, size = 50, normalized size = 2.78 \begin{align*} \frac{1}{6} \, \sqrt{6} \arctan \left (\frac{1}{2} \, \sqrt{6} e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(2+3*exp(2*x)),x, algorithm="fricas")

[Out]

1/6*sqrt(6)*arctan(1/2*sqrt(6)*e^x)

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Sympy [A]  time = 0.106956, size = 15, normalized size = 0.83 \begin{align*} \operatorname{RootSum}{\left (24 z^{2} + 1, \left ( i \mapsto i \log{\left (4 i + e^{x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(2+3*exp(2*x)),x)

[Out]

RootSum(24*_z**2 + 1, Lambda(_i, _i*log(4*_i + exp(x))))

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Giac [A]  time = 1.0934, size = 18, normalized size = 1. \begin{align*} \frac{1}{6} \, \sqrt{6} \arctan \left (\frac{1}{2} \, \sqrt{6} e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(2+3*exp(2*x)),x, algorithm="giac")

[Out]

1/6*sqrt(6)*arctan(1/2*sqrt(6)*e^x)