3.94.12 \(\int \frac {e^x (-1+x)}{e^x x+x^2} \, dx\)

Optimal. Leaf size=10 \[ \log \left (\frac {e^x+x}{x}\right ) \]

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Rubi [F]  time = 0.25, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^x (-1+x)}{e^x x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-1 + x))/(E^x*x + x^2),x]

[Out]

Defer[Int][E^x/(E^x + x), x] - Defer[Int][E^x/(x*(E^x + x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^x}{e^x+x}-\frac {e^x}{x \left (e^x+x\right )}\right ) \, dx\\ &=\int \frac {e^x}{e^x+x} \, dx-\int \frac {e^x}{x \left (e^x+x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 11, normalized size = 1.10 \begin {gather*} -\log (x)+\log \left (e^x+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[x] + Log[E^x + x]

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fricas [A]  time = 0.97, size = 10, normalized size = 1.00 \begin {gather*} \log \left (x + e^{x}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + e^x) - log(x)

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giac [A]  time = 0.12, size = 10, normalized size = 1.00 \begin {gather*} \log \left (x + e^{x}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + e^x) - log(x)

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maple [A]  time = 0.03, size = 11, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 10, normalized size = 1.00 \begin {gather*} \log \left (x + e^{x}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + e^x) - log(x)

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mupad [B]  time = 0.13, size = 10, normalized size = 1.00 \begin {gather*} \ln \left (x+{\mathrm {e}}^x\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + exp(x)) - log(x)

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sympy [A]  time = 0.09, size = 8, normalized size = 0.80 \begin {gather*} - \log {\relax (x )} + \log {\left (x + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x + exp(x))

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