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

Optimal. Leaf size=24 \[ 2-e^{-1+x}+\log \left (15 \log \left (\frac {x}{e^x-x}\right )\right ) \]

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Rubi [F]  time = 1.03, 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)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-E^(-1 + x) - Defer[Int][Log[x/(E^x - x)]^(-1), x] + Defer[Int][1/((E^x - x)*Log[x/(E^x - x)]), x] + Defer[Int
][1/(x*Log[x/(E^x - x)]), x] - Defer[Int][x/((E^x - x)*Log[x/(E^x - x)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{-1+x}+\frac {1-x}{x \log \left (\frac {x}{e^x-x}\right )}+\frac {-1+x}{\left (-e^x+x\right ) \log \left (\frac {x}{e^x-x}\right )}\right ) \, dx\\ &=-\int e^{-1+x} \, dx+\int \frac {1-x}{x \log \left (\frac {x}{e^x-x}\right )} \, dx+\int \frac {-1+x}{\left (-e^x+x\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx\\ &=-e^{-1+x}+\int \left (-\frac {1}{\log \left (\frac {x}{e^x-x}\right )}+\frac {1}{x \log \left (\frac {x}{e^x-x}\right )}\right ) \, dx+\int \left (\frac {1}{\left (e^x-x\right ) \log \left (\frac {x}{e^x-x}\right )}-\frac {x}{\left (e^x-x\right ) \log \left (\frac {x}{e^x-x}\right )}\right ) \, dx\\ &=-e^{-1+x}-\int \frac {1}{\log \left (\frac {x}{e^x-x}\right )} \, dx+\int \frac {1}{\left (e^x-x\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx+\int \frac {1}{x \log \left (\frac {x}{e^x-x}\right )} \, dx-\int \frac {x}{\left (e^x-x\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 21, normalized size = 0.88 \begin {gather*} -e^{-1+x}+\log \left (\log \left (\frac {x}{e^x-x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(-1 + x) + Log[Log[x/(E^x - x)]]

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fricas [A]  time = 1.09, size = 24, normalized size = 1.00 \begin {gather*} {\left (e \log \left (\log \left (-\frac {x}{x - e^{x}}\right )\right ) - e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*log(log(-x/(x - e^x))) - e^x)*e^(-1)

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giac [A]  time = 0.19, size = 24, normalized size = 1.00 \begin {gather*} {\left (e \log \left (\log \left (-\frac {x}{x - e^{x}}\right )\right ) - e^{x}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*log(log(-x/(x - e^x))) - e^x)*e^(-1)

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maple [A]  time = 0.28, size = 22, normalized size = 0.92

method	result	size
norman	\(-{\mathrm e}^{x} {\mathrm e}^{-1}+\ln \left (\ln \left (\frac {x}{{\mathrm e}^{x}-x}\right )\right )\)	\(22\)
risch	\(-{\mathrm e}^{x -1}+\ln \left (\ln \left (x -{\mathrm e}^{x}\right )-\frac {i \left (-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right ) \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{3}-2 \pi \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{2} \mathrm {csgn}\left (i x \right )+2 \pi -2 i \ln \relax (x )\right )}{2}\right )\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 20, normalized size = 0.83 \begin {gather*} -e^{\left (x - 1\right )} + \log \left (-\log \relax (x) + \log \left (-x + e^{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(x - 1) + log(-log(x) + log(-x + e^x))

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mupad [B]  time = 1.70, size = 20, normalized size = 0.83 \begin {gather*} \ln \left (\ln \left (-\frac {x}{x-{\mathrm {e}}^x}\right )\right )-{\mathrm {e}}^{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x - 1) - log(-x/(x - exp(x)))*(x^2*exp(x - 1) - x*exp(x - 1)*exp(x)))/(log(-x/(x - exp(x)))*(x*e
xp(x) - x^2)),x)

[Out]

log(log(-x/(x - exp(x)))) - exp(x - 1)

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sympy [A]  time = 0.35, size = 15, normalized size = 0.62 \begin {gather*} - \frac {e^{x}}{e} + \log {\left (\log {\left (\frac {x}{- x + e^{x}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(-1)*exp(x) + log(log(x/(-x + exp(x))))

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