\(\int \frac {(-x-x^2) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} (x+x \log ^2(x))+(e^{-x+x \log (x)}-x \log (x)) \log (-e^{-x+x \log (x)}+x \log (x))}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx\) [3101]

Optimal result

Integrand size = 97, antiderivative size = 28 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=-e^3+x+\log (x) \log \left (-e^{-x+x \log (x)}+x \log (x)\right ) \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )\right )}{x \left (x^x-e^x x \log (x)\right )} \, dx \\ & = \int \left (-\frac {x^x}{-x^x+e^x x \log (x)}+\frac {e^x \log (x)}{-x^x+e^x x \log (x)}+\frac {e^x x \log (x)}{-x^x+e^x x \log (x)}+\frac {e^x \log ^2(x)}{-x^x+e^x x \log (x)}-\frac {x^x \log ^2(x)}{-x^x+e^x x \log (x)}-\frac {x^{-1+x} \log \left (-e^{-x} x^x+x \log (x)\right )}{-x^x+e^x x \log (x)}+\frac {e^x \log (x) \log \left (-e^{-x} x^x+x \log (x)\right )}{-x^x+e^x x \log (x)}\right ) \, dx \\ & = -\int \frac {x^x}{-x^x+e^x x \log (x)} \, dx+\int \frac {e^x \log (x)}{-x^x+e^x x \log (x)} \, dx+\int \frac {e^x x \log (x)}{-x^x+e^x x \log (x)} \, dx+\int \frac {e^x \log ^2(x)}{-x^x+e^x x \log (x)} \, dx-\int \frac {x^x \log ^2(x)}{-x^x+e^x x \log (x)} \, dx-\int \frac {x^{-1+x} \log \left (-e^{-x} x^x+x \log (x)\right )}{-x^x+e^x x \log (x)} \, dx+\int \frac {e^x \log (x) \log \left (-e^{-x} x^x+x \log (x)\right )}{-x^x+e^x x \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=x+\log (x) \log \left (-e^{-x} x^x+x \log (x)\right ) \]

[In]

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

[Out]

x + Log[x]*Log[-(x^x/E^x) + x*Log[x]]

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
risch	\(\ln \left (-x^{x} {\mathrm e}^{-x}+x \ln \left (x \right )\right ) \ln \left (x \right )+x\)	\(21\)
parallelrisch	\(\ln \left (x \right ) \ln \left (-{\mathrm e}^{\left (\ln \left (x \right )-1\right ) x}+x \ln \left (x \right )\right )+x\)	\(21\)

[In]

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

[Out]

ln(-x^x*exp(-x)+x*ln(x))*ln(x)+x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=\log \left (x \log \left (x\right ) - e^{\left (x \log \left (x\right ) - x\right )}\right ) \log \left (x\right ) + x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=x + \log {\left (x \right )} \log {\left (x \log {\left (x \right )} - e^{x \log {\left (x \right )} - x} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=-x \log \left (x\right ) + \log \left (x e^{x} \log \left (x\right ) - x^{x}\right ) \log \left (x\right ) + x \]

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {\left (-x-x^2\right ) \log (x)-x \log ^2(x)+e^{-x+x \log (x)} \left (x+x \log ^2(x)\right )+\left (e^{-x+x \log (x)}-x \log (x)\right ) \log \left (-e^{-x+x \log (x)}+x \log (x)\right )}{e^{-x+x \log (x)} x-x^2 \log (x)} \, dx=x+\ln \left (x\,\ln \left (x\right )-x^x\,{\mathrm {e}}^{-x}\right )\,\ln \left (x\right ) \]

[In]

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

[Out]

x + log(x*log(x) - x^x*exp(-x))*log(x)