\(\int \frac {e^x (1-x)+(-e^{-1+2 x} x+e^{-1+x} x^2) \log (\frac {x}{e^x-x})}{(e^x x-x^2) \log (\frac {x}{e^x-x})} \, dx\) [2304]

Optimal result

Integrand size = 71, antiderivative size = 24 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx=2-e^{-1+x}+\log \left (15 \log \left (\frac {x}{e^x-x}\right )\right ) \]

[Out]

2-exp(-1+x)+ln(15*ln(x/(exp(x)-x)))

Rubi [F]

\[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx=\int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx=-e^{-1+x}+\log \left (\log \left (\frac {x}{e^x-x}\right )\right ) \]

[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)]]

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(\ln \left (\ln \left (\frac {x}{{\mathrm e}^{x}-x}\right )\right )-{\mathrm e}^{-1+x}\)	\(20\)
norman	\(-{\mathrm e}^{-1} {\mathrm e}^{x}+\ln \left (\ln \left (\frac {x}{{\mathrm e}^{x}-x}\right )\right )\)	\(22\)
risch	\(-{\mathrm e}^{-1+x}+\ln \left (\ln \left (x -{\mathrm e}^{x}\right )-\frac {i \left (-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{x}-x}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \pi -2 i \ln \left (x \right )\right )}{2}\right )\)	\(150\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx={\left (e \log \left (\log \left (-\frac {x}{x - e^{x}}\right )\right ) - e^{x}\right )} e^{\left (-1\right )} \]

[In]

integrate(((-x*exp(-1+x)*exp(x)+x^2*exp(-1+x))*log(x/(exp(x)-x))+(1-x)*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)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx=- \frac {e^{x}}{e} + \log {\left (\log {\left (\frac {x}{- x + e^{x}} \right )} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx=-e^{\left (x - 1\right )} + \log \left (-\log \left (x\right ) + \log \left (-x + e^{x}\right )\right ) \]

[In]

integrate(((-x*exp(-1+x)*exp(x)+x^2*exp(-1+x))*log(x/(exp(x)-x))+(1-x)*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))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx={\left (e \log \left (\log \left (-\frac {x}{x - e^{x}}\right )\right ) - e^{x}\right )} e^{\left (-1\right )} \]

[In]

integrate(((-x*exp(-1+x)*exp(x)+x^2*exp(-1+x))*log(x/(exp(x)-x))+(1-x)*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)

Mupad [B] (verification not implemented)

Time = 8.52 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^x (1-x)+\left (-e^{-1+2 x} x+e^{-1+x} x^2\right ) \log \left (\frac {x}{e^x-x}\right )}{\left (e^x x-x^2\right ) \log \left (\frac {x}{e^x-x}\right )} \, dx=\ln \left (\ln \left (-\frac {x}{x-{\mathrm {e}}^x}\right )\right )-{\mathrm {e}}^{x-1} \]

[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)