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

Optimal. Leaf size=17 \[ 2 \log \left (e^x+2\right )-\log \left (e^x+1\right ) \]

[Out]

-Log[1 + E^x] + 2*Log[2 + E^x]

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Rubi [A]  time = 0.0318288, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2282, 632, 31} \[ 2 \log \left (e^x+2\right )-\log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(2 + 3*E^x + E^(2*x)),x]

[Out]

-Log[1 + E^x] + 2*Log[2 + E^x]

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 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0134487, size = 17, normalized size = 1. \[ 2 \log \left (e^x+2\right )-\log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(2 + 3*E^x + E^(2*x)),x]

[Out]

-Log[1 + E^x] + 2*Log[2 + E^x]

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Maple [A]  time = 0.007, size = 16, normalized size = 0.9 \begin{align*} -\ln \left ( 1+{{\rm e}^{x}} \right ) +2\,\ln \left ( 2+{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(1+exp(x))+2*ln(2+exp(x))

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Maxima [A]  time = 0.949909, size = 20, normalized size = 1.18 \begin{align*} 2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)

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Fricas [A]  time = 2.25475, size = 42, normalized size = 2.47 \begin{align*} 2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)

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Sympy [A]  time = 0.112225, size = 14, normalized size = 0.82 \begin{align*} - \log{\left (e^{x} + 1 \right )} + 2 \log{\left (e^{x} + 2 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(exp(x) + 1) + 2*log(exp(x) + 2)

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Giac [A]  time = 1.0665, size = 20, normalized size = 1.18 \begin{align*} 2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)