3.3.40 \(\int \frac {e^{2 x}}{2+3 e^x+e^{2 x}} \, dx\) [240]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2320, 646, 31} \begin {gather*} 2 \log \left (e^x+2\right )-\log \left (e^x+1\right ) \end {gather*}

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 31

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

Rule 646

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 2320

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]]]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 17, normalized size = 1.00 \begin {gather*} -\log \left (1+e^x\right )+2 \log \left (2+e^x\right ) \end {gather*}

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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Mathics [A]
time = 1.79, size = 17, normalized size = 1.00 \begin {gather*} -\text {Log}\left [1+E^x\right ]+2 \text {Log}\left [2+E^x\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('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 = 0.02, size = 16, normalized size = 0.94

method	result	size
default	\(-\ln \left (1+{\mathrm e}^{x}\right )+2 \ln \left (2+{\mathrm e}^{x}\right )\)	\(16\)
norman	\(-\ln \left (1+{\mathrm e}^{x}\right )+2 \ln \left (2+{\mathrm e}^{x}\right )\)	\(16\)
risch	\(-\ln \left (1+{\mathrm e}^{x}\right )+2 \ln \left (2+{\mathrm e}^{x}\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(2+3*exp(x)+exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 15, normalized size = 0.88 \begin {gather*} 2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \end {gather*}

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 = 0.36, size = 15, normalized size = 0.88 \begin {gather*} 2 \, \log \left (e^{x} + 2\right ) - \log \left (e^{x} + 1\right ) \end {gather*}

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 = 0.06, size = 14, normalized size = 0.82 \begin {gather*} - \log {\left (e^{x} + 1 \right )} + 2 \log {\left (e^{x} + 2 \right )} \end {gather*}

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 = 0.00, size = 14, normalized size = 0.82 \begin {gather*} -\ln \left (\mathrm {e}^{x}+1\right )+2 \ln \left (\mathrm {e}^{x}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(e^x + 2) - log(e^x + 1)

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Mupad [B]
time = 0.21, size = 15, normalized size = 0.88 \begin {gather*} 2\,\ln \left ({\mathrm {e}}^x+2\right )-\ln \left ({\mathrm {e}}^x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(exp(x) + 2) - log(exp(x) + 1)

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