3.7.86 \(\int \frac {e^x}{2+3 e^x+e^{2 x}} \, dx\) [686]

Optimal. Leaf size=15 \[ \log \left (1+e^x\right )-\log \left (2+e^x\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2320, 630, 31} \begin {gather*} \log \left (e^x+1\right )-\log \left (e^x+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x/(2 + 3*E^x + E^(2*x)),x]

[Out]

Log[1 + E^x] - 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 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[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^x}{2+3 e^x+e^{2 x}} \, dx &=\text {Subst}\left (\int \frac {1}{2+3 x+x^2} \, dx,x,e^x\right )\\ &=\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^x\right )\\ &=\log \left (1+e^x\right )-\log \left (2+e^x\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 10, normalized size = 0.67 \begin {gather*} -2 \tanh ^{-1}\left (3+2 e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcTanh[3 + 2*E^x]

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Maple [A]
time = 0.02, size = 14, normalized size = 0.93

method	result	size
default	\(\ln \left (1+{\mathrm e}^{x}\right )-\ln \left (2+{\mathrm e}^{x}\right )\)	\(14\)
norman	\(\ln \left (1+{\mathrm e}^{x}\right )-\ln \left (2+{\mathrm e}^{x}\right )\)	\(14\)
risch	\(\ln \left (1+{\mathrm e}^{x}\right )-\ln \left (2+{\mathrm e}^{x}\right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 13, normalized size = 0.87 \begin {gather*} -\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(x)/(2+3*exp(x)+exp(2*x)),x, algorithm="maxima")

[Out]

-log(e^x + 2) + log(e^x + 1)

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Fricas [A]
time = 0.44, size = 13, normalized size = 0.87 \begin {gather*} -\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(x)/(2+3*exp(x)+exp(2*x)),x, algorithm="fricas")

[Out]

-log(e^x + 2) + log(e^x + 1)

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Sympy [A]
time = 0.04, size = 12, normalized size = 0.80 \begin {gather*} \log {\left (e^{x} + 1 \right )} - \log {\left (e^{x} + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(exp(x) + 1) - log(exp(x) + 2)

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Giac [A]
time = 2.27, size = 13, normalized size = 0.87 \begin {gather*} -\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(x)/(2+3*exp(x)+exp(2*x)),x, algorithm="giac")

[Out]

-log(e^x + 2) + log(e^x + 1)

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Mupad [B]
time = 3.49, size = 13, normalized size = 0.87 \begin {gather*} \ln \left ({\mathrm {e}}^x+1\right )-\ln \left ({\mathrm {e}}^x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(exp(x) + 1) - log(exp(x) + 2)

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