\(\int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x (-x+e^x (x^2+x^3)+(1+e^x (-x-x^2)) \log (x)+(-2+2 x) \log (-x+\log (x)))}{-x^2+x \log (x)} \, dx\) [6260]

Optimal result

Integrand size = 80, antiderivative size = 22 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=e^{-e^x x-\log ^2(-x+\log (x))} x \]

[Out]

exp(-ln(ln(x)-x)^2+ln(x)-exp(x)*x)

Rubi [F]

\[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=\int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx \]

[In]

Int[(E^(-(E^x*x) - Log[-x + Log[x]]^2)*x*(-x + E^x*(x^2 + x^3) + (1 + E^x*(-x - x^2))*Log[x] + (-2 + 2*x)*Log[
-x + Log[x]]))/(-x^2 + x*Log[x]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x+\log (x)} \, dx \\ & = \int \left (-e^{x-e^x x-\log ^2(-x+\log (x))} x (1+x)+\frac {e^{-e^x x-\log ^2(-x+\log (x))} (x-\log (x)+2 \log (-x+\log (x))-2 x \log (-x+\log (x)))}{x-\log (x)}\right ) \, dx \\ & = -\int e^{x-e^x x-\log ^2(-x+\log (x))} x (1+x) \, dx+\int \frac {e^{-e^x x-\log ^2(-x+\log (x))} (x-\log (x)+2 \log (-x+\log (x))-2 x \log (-x+\log (x)))}{x-\log (x)} \, dx \\ & = -\int \left (e^{x-e^x x-\log ^2(-x+\log (x))} x+e^{x-e^x x-\log ^2(-x+\log (x))} x^2\right ) \, dx+\int \frac {e^{-e^x x-\log ^2(-x+\log (x))} (x-\log (x)-2 (-1+x) \log (-x+\log (x)))}{x-\log (x)} \, dx \\ & = -\int e^{x-e^x x-\log ^2(-x+\log (x))} x \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x^2 \, dx+\int \left (e^{-e^x x-\log ^2(-x+\log (x))}-\frac {2 e^{-e^x x-\log ^2(-x+\log (x))} (-1+x) \log (-x+\log (x))}{x-\log (x)}\right ) \, dx \\ & = -\left (2 \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} (-1+x) \log (-x+\log (x))}{x-\log (x)} \, dx\right )+\int e^{-e^x x-\log ^2(-x+\log (x))} \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x^2 \, dx \\ & = -\left (2 \int \left (-\frac {e^{-e^x x-\log ^2(-x+\log (x))} \log (-x+\log (x))}{x-\log (x)}+\frac {e^{-e^x x-\log ^2(-x+\log (x))} x \log (-x+\log (x))}{x-\log (x)}\right ) \, dx\right )+\int e^{-e^x x-\log ^2(-x+\log (x))} \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x^2 \, dx \\ & = 2 \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} \log (-x+\log (x))}{x-\log (x)} \, dx-2 \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \log (-x+\log (x))}{x-\log (x)} \, dx+\int e^{-e^x x-\log ^2(-x+\log (x))} \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x \, dx-\int e^{x-e^x x-\log ^2(-x+\log (x))} x^2 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=e^{-e^x x-\log ^2(-x+\log (x))} x \]

[In]

Integrate[(E^(-(E^x*x) - Log[-x + Log[x]]^2)*x*(-x + E^x*(x^2 + x^3) + (1 + E^x*(-x - x^2))*Log[x] + (-2 + 2*x
)*Log[-x + Log[x]]))/(-x^2 + x*Log[x]),x]

[Out]

E^(-(E^x*x) - Log[-x + Log[x]]^2)*x

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
risch	\(x \,{\mathrm e}^{-\ln \left (\ln \left (x \right )-x \right )^{2}-{\mathrm e}^{x} x}\)	\(21\)
parallelrisch	\({\mathrm e}^{-\ln \left (\ln \left (x \right )-x \right )^{2}+\ln \left (x \right )-{\mathrm e}^{x} x}\)	\(21\)

[In]

int(((-2+2*x)*ln(ln(x)-x)+((-x^2-x)*exp(x)+1)*ln(x)+(x^3+x^2)*exp(x)-x)*exp(-ln(ln(x)-x)^2+ln(x)-exp(x)*x)/(x*
ln(x)-x^2),x,method=_RETURNVERBOSE)

[Out]

x*exp(-ln(ln(x)-x)^2-exp(x)*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=e^{\left (-x e^{x} - \log \left (-x + \log \left (x\right )\right )^{2} + \log \left (x\right )\right )} \]

[In]

integrate(((-2+2*x)*log(log(x)-x)+((-x^2-x)*exp(x)+1)*log(x)+(x^3+x^2)*exp(x)-x)*exp(-log(log(x)-x)^2+log(x)-e
xp(x)*x)/(x*log(x)-x^2),x, algorithm="fricas")

[Out]

e^(-x*e^x - log(-x + log(x))^2 + log(x))

Sympy [A] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=x e^{- x e^{x} - \log {\left (- x + \log {\left (x \right )} \right )}^{2}} \]

[In]

integrate(((-2+2*x)*ln(ln(x)-x)+((-x**2-x)*exp(x)+1)*ln(x)+(x**3+x**2)*exp(x)-x)*exp(-ln(ln(x)-x)**2+ln(x)-exp
(x)*x)/(x*ln(x)-x**2),x)

[Out]

x*exp(-x*exp(x) - log(-x + log(x))**2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=x e^{\left (-x e^{x} - \log \left (-x + \log \left (x\right )\right )^{2}\right )} \]

[In]

integrate(((-2+2*x)*log(log(x)-x)+((-x^2-x)*exp(x)+1)*log(x)+(x^3+x^2)*exp(x)-x)*exp(-log(log(x)-x)^2+log(x)-e
xp(x)*x)/(x*log(x)-x^2),x, algorithm="maxima")

[Out]

x*e^(-x*e^x - log(-x + log(x))^2)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=e^{\left (-x e^{x} - \log \left (-x + \log \left (x\right )\right )^{2} + \log \left (x\right )\right )} \]

[In]

integrate(((-2+2*x)*log(log(x)-x)+((-x^2-x)*exp(x)+1)*log(x)+(x^3+x^2)*exp(x)-x)*exp(-log(log(x)-x)^2+log(x)-e
xp(x)*x)/(x*log(x)-x^2),x, algorithm="giac")

[Out]

e^(-x*e^x - log(-x + log(x))^2 + log(x))

Mupad [B] (verification not implemented)

Time = 12.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x x-\log ^2(-x+\log (x))} x \left (-x+e^x \left (x^2+x^3\right )+\left (1+e^x \left (-x-x^2\right )\right ) \log (x)+(-2+2 x) \log (-x+\log (x))\right )}{-x^2+x \log (x)} \, dx=x\,{\mathrm {e}}^{-x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\ln \left (\ln \left (x\right )-x\right )}^2} \]

[In]

int(-(exp(log(x) - x*exp(x) - log(log(x) - x)^2)*(x - log(log(x) - x)*(2*x - 2) - exp(x)*(x^2 + x^3) + log(x)*
(exp(x)*(x + x^2) - 1)))/(x*log(x) - x^2),x)

[Out]

x*exp(-x*exp(x))*exp(-log(log(x) - x)^2)