3.1.73 \(\int \frac {1}{1+e^x} \, dx\) [73]

Optimal. Leaf size=10 \[ x-\log \left (1+e^x\right ) \]

[Out]

x-ln(1+exp(x))

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Rubi [A]
time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {2320, 36, 29, 31} \begin {gather*} x-\log \left (e^x+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^x)^(-1),x]

[Out]

x - Log[1 + E^x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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

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Mathematica [A]
time = 0.01, size = 10, normalized size = 1.00 \begin {gather*} -2 \tanh ^{-1}\left (1+2 e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^x)^(-1),x]

[Out]

-2*ArcTanh[1 + 2*E^x]

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Maple [A]
time = 0.01, size = 12, normalized size = 1.20

method	result	size
norman	\(x -\ln \left (1+{\mathrm e}^{x}\right )\)	\(10\)
risch	\(x -\ln \left (1+{\mathrm e}^{x}\right )\)	\(10\)
derivativedivides	\(-\ln \left (1+{\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}\right )\)	\(12\)
default	\(-\ln \left (1+{\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}\right )\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 1.47, size = 9, normalized size = 0.90 \begin {gather*} x - \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(e^x + 1)

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Fricas [A]
time = 0.85, size = 9, normalized size = 0.90 \begin {gather*} x - \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(e^x + 1)

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Sympy [A]
time = 0.02, size = 7, normalized size = 0.70 \begin {gather*} x - \log {\left (e^{x} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(exp(x) + 1)

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Giac [A]
time = 0.47, size = 9, normalized size = 0.90 \begin {gather*} x - \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(e^x + 1)

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Mupad [B]
time = 0.04, size = 9, normalized size = 0.90 \begin {gather*} x-\ln \left ({\mathrm {e}}^x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(exp(x) + 1),x)

[Out]

x - log(exp(x) + 1)

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