3.7.51 \(\int \frac {e^x (-2+e^x)}{1+e^x} \, dx\) [651]

Optimal. Leaf size=12 \[ e^x-3 \log \left (1+e^x\right ) \]

[Out]

exp(x)-3*ln(1+exp(x))

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Rubi [A]
time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2320, 45} \begin {gather*} e^x-3 \log \left (e^x+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-2 + E^x))/(1 + E^x),x]

[Out]

E^x - 3*Log[1 + E^x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 {e^x \left (-2+e^x\right )}{1+e^x} \, dx &=\text {Subst}\left (\int \frac {-2+x}{1+x} \, dx,x,e^x\right )\\ &=\text {Subst}\left (\int \left (1-\frac {3}{1+x}\right ) \, dx,x,e^x\right )\\ &=e^x-3 \log \left (1+e^x\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} e^x-3 \log \left (1+e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x - 3*Log[1 + E^x]

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Maple [A]
time = 0.01, size = 11, normalized size = 0.92

method	result	size
derivativedivides	\({\mathrm e}^{x}-3 \ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)
default	\({\mathrm e}^{x}-3 \ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)
norman	\({\mathrm e}^{x}-3 \ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)
risch	\({\mathrm e}^{x}-3 \ln \left (1+{\mathrm e}^{x}\right )\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(-2+exp(x))/(1+exp(x)),x,method=_RETURNVERBOSE)

[Out]

exp(x)-3*ln(1+exp(x))

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Maxima [A]
time = 0.28, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - 3 \, \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x - 3*log(e^x + 1)

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Fricas [A]
time = 0.38, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - 3 \, \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x - 3*log(e^x + 1)

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Sympy [A]
time = 0.02, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - 3 \log {\left (e^{x} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x) - 3*log(exp(x) + 1)

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Giac [A]
time = 4.56, size = 10, normalized size = 0.83 \begin {gather*} e^{x} - 3 \, \log \left (e^{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x - 3*log(e^x + 1)

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Mupad [B]
time = 3.33, size = 10, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^x-3\,\ln \left ({\mathrm {e}}^x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x) - 3*log(exp(x) + 1)

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