3.643 \(\int \frac{e^x}{\sqrt{1+e^x+e^{2 x}}} \, dx\)

Optimal. Leaf size=14 \[ \sinh ^{-1}\left (\frac{2 e^x+1}{\sqrt{3}}\right ) \]

[Out]

ArcSinh[(1 + 2*E^x)/Sqrt[3]]

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Rubi [A]  time = 0.0362688, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2282, 619, 215} \[ \sinh ^{-1}\left (\frac{2 e^x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/Sqrt[1 + E^x + E^(2*x)],x]

[Out]

ArcSinh[(1 + 2*E^x)/Sqrt[3]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.0095732, size = 14, normalized size = 1. \[ \sinh ^{-1}\left (\frac{2 e^x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/Sqrt[1 + E^x + E^(2*x)],x]

[Out]

ArcSinh[(1 + 2*E^x)/Sqrt[3]]

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Maple [A]  time = 0.062, size = 11, normalized size = 0.8 \begin{align*}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ({{\rm e}^{x}}+{\frac{1}{2}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(2/3*3^(1/2)*(exp(x)+1/2))

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Maxima [A]  time = 1.44264, size = 16, normalized size = 1.14 \begin{align*} \operatorname{arsinh}\left (\frac{1}{3} \, \sqrt{3}{\left (2 \, e^{x} + 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(1/3*sqrt(3)*(2*e^x + 1))

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Fricas [A]  time = 0.881631, size = 61, normalized size = 4.36 \begin{align*} -\log \left (2 \, \sqrt{e^{\left (2 \, x\right )} + e^{x} + 1} - 2 \, e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(2*sqrt(e^(2*x) + e^x + 1) - 2*e^x - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{x}}{\sqrt{e^{2 x} + e^{x} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)/sqrt(exp(2*x) + exp(x) + 1), x)

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Giac [A]  time = 1.28356, size = 28, normalized size = 2. \begin{align*} -\log \left (2 \, \sqrt{e^{\left (2 \, x\right )} + e^{x} + 1} - 2 \, e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(2*sqrt(e^(2*x) + e^x + 1) - 2*e^x - 1)