\(\int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} (e^x+(-2 x^2+x^3) \log (x))+(-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)) \log (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}) \log (\log (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}))}{(e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)) \log (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)})} \, dx\) [505]

Optimal result

Integrand size = 212, antiderivative size = 33 \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=-2-x \log \left (\log \left (\frac {1}{5} \left (x+\frac {e^{e^{-x} x^2}-x}{\log (x)}\right )\right )\right ) \] Output:

-ln(ln(1/5*x+1/5*(exp(x^2/exp(x))-x)/ln(x)))*x-2
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=-x \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right ) \] Input:

Integrate[(-(E^x*x) + E^x*x*Log[x] - E^x*x*Log[x]^2 + E^(x^2/E^x)*(E^x + ( 
-2*x^2 + x^3)*Log[x]) + (-(E^(x + x^2/E^x)*Log[x]) + E^x*x*Log[x] - E^x*x* 
Log[x]^2)*Log[(E^(x^2/E^x) - x + x*Log[x])/(5*Log[x])]*Log[Log[(E^(x^2/E^x 
) - x + x*Log[x])/(5*Log[x])]])/((E^(x + x^2/E^x)*Log[x] - E^x*x*Log[x] + 
E^x*x*Log[x]^2)*Log[(E^(x^2/E^x) - x + x*Log[x])/(5*Log[x])]),x]
 

Output:

-(x*Log[Log[(E^(x^2/E^x) - x + x*Log[x])/(5*Log[x])]])
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (-e^{e^{-x} x^2+x} \log (x)-e^x x \log ^2(x)+e^x x \log (x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )+e^{e^{-x} x^2} \left (\left (x^3-2 x^2\right ) \log (x)+e^x\right )-e^x x-e^x x \log ^2(x)+e^x x \log (x)}{\left (e^{e^{-x} x^2+x} \log (x)+e^x x \log ^2(x)-e^x x \log (x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-x} \left (\left (-e^{e^{-x} x^2+x} \log (x)-e^x x \log ^2(x)+e^x x \log (x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )+e^{e^{-x} x^2} \left (\left (x^3-2 x^2\right ) \log (x)+e^x\right )-e^x x-e^x x \log ^2(x)+e^x x \log (x)\right )}{\log (x) \left (e^{e^{-x} x^2}-x+x \log (x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {e^{-x} x \left (-x^3+x^3 \log (x)+2 x^2-2 x^2 \log (x)+e^x \log (x)\right )}{\left (e^{e^{-x} x^2}-x+x \log (x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )}-\frac {e^{-x} \left (x^3 (-\log (x))+2 x^2 \log (x)+e^x \log (x) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )-e^x\right )}{\log (x) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \int \frac {e^{-x} x^2}{\log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx+\int \frac {1}{\log (x) \log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx-\int \frac {x \log (x)}{\left (\log (x) x-x+e^{e^{-x} x^2}\right ) \log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx-\int \log \left (\log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )\right )dx+\int \frac {e^{-x} x^4}{\left (\log (x) x-x+e^{e^{-x} x^2}\right ) \log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx-\int \frac {e^{-x} x^4 \log (x)}{\left (\log (x) x-x+e^{e^{-x} x^2}\right ) \log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx+\int \frac {e^{-x} x^3}{\log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx-2 \int \frac {e^{-x} x^3}{\left (\log (x) x-x+e^{e^{-x} x^2}\right ) \log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx+2 \int \frac {e^{-x} x^3 \log (x)}{\left (\log (x) x-x+e^{e^{-x} x^2}\right ) \log \left (\frac {\log (x) x-x+e^{e^{-x} x^2}}{5 \log (x)}\right )}dx\)

Input:

Int[(-(E^x*x) + E^x*x*Log[x] - E^x*x*Log[x]^2 + E^(x^2/E^x)*(E^x + (-2*x^2 
 + x^3)*Log[x]) + (-(E^(x + x^2/E^x)*Log[x]) + E^x*x*Log[x] - E^x*x*Log[x] 
^2)*Log[(E^(x^2/E^x) - x + x*Log[x])/(5*Log[x])]*Log[Log[(E^(x^2/E^x) - x 
+ x*Log[x])/(5*Log[x])]])/((E^(x + x^2/E^x)*Log[x] - E^x*x*Log[x] + E^x*x* 
Log[x]^2)*Log[(E^(x^2/E^x) - x + x*Log[x])/(5*Log[x])]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 4.30

\[-x \ln \left (-\ln \left (5\right )-\ln \left (\ln \left (x \right )\right )+\ln \left (\left (\ln \left (x \right )-1\right ) x +{\mathrm e}^{{\mathrm e}^{-x} x^{2}}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \right )-1\right ) x +{\mathrm e}^{{\mathrm e}^{-x} x^{2}}\right )}{\ln \left (x \right )}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \right )-1\right ) x +{\mathrm e}^{{\mathrm e}^{-x} x^{2}}\right )}{\ln \left (x \right )}\right )+\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \right )-1\right ) x +{\mathrm e}^{{\mathrm e}^{-x} x^{2}}\right )}{\ln \left (x \right )}\right )+\operatorname {csgn}\left (i \left (\left (\ln \left (x \right )-1\right ) x +{\mathrm e}^{{\mathrm e}^{-x} x^{2}}\right )\right )\right )}{2}\right )\]

Input:

int(((-exp(x)*ln(x)*exp(x^2/exp(x))-x*exp(x)*ln(x)^2+x*exp(x)*ln(x))*ln(1/ 
5*(exp(x^2/exp(x))+x*ln(x)-x)/ln(x))*ln(ln(1/5*(exp(x^2/exp(x))+x*ln(x)-x) 
/ln(x)))+((x^3-2*x^2)*ln(x)+exp(x))*exp(x^2/exp(x))-x*exp(x)*ln(x)^2+x*exp 
(x)*ln(x)-exp(x)*x)/(exp(x)*ln(x)*exp(x^2/exp(x))+x*exp(x)*ln(x)^2-x*exp(x 
)*ln(x))/ln(1/5*(exp(x^2/exp(x))+x*ln(x)-x)/ln(x)),x)
 

Output:

-x*ln(-ln(5)-ln(ln(x))+ln((ln(x)-1)*x+exp(exp(-x)*x^2))-1/2*I*Pi*csgn(I/ln 
(x)*((ln(x)-1)*x+exp(exp(-x)*x^2)))*(-csgn(I/ln(x)*((ln(x)-1)*x+exp(exp(-x 
)*x^2)))+csgn(I/ln(x)))*(-csgn(I/ln(x)*((ln(x)-1)*x+exp(exp(-x)*x^2)))+csg 
n(I*((ln(x)-1)*x+exp(exp(-x)*x^2)))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=-x \log \left (\log \left (\frac {{\left (x e^{x} \log \left (x\right ) - x e^{x} + e^{\left ({\left (x^{2} + x e^{x}\right )} e^{\left (-x\right )}\right )}\right )} e^{\left (-x\right )}}{5 \, \log \left (x\right )}\right )\right ) \] Input:

integrate(((-exp(x)*log(x)*exp(x^2/exp(x))-x*exp(x)*log(x)^2+x*exp(x)*log( 
x))*log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x))*log(log(1/5*(exp(x^2/exp( 
x))+x*log(x)-x)/log(x)))+((x^3-2*x^2)*log(x)+exp(x))*exp(x^2/exp(x))-x*exp 
(x)*log(x)^2+x*exp(x)*log(x)-exp(x)*x)/(exp(x)*log(x)*exp(x^2/exp(x))+x*ex 
p(x)*log(x)^2-x*exp(x)*log(x))/log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x) 
),x, algorithm="fricas")
 

Output:

-x*log(log(1/5*(x*e^x*log(x) - x*e^x + e^((x^2 + x*e^x)*e^(-x)))*e^(-x)/lo 
g(x)))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=\text {Timed out} \] Input:

integrate(((-exp(x)*ln(x)*exp(x**2/exp(x))-x*exp(x)*ln(x)**2+x*exp(x)*ln(x 
))*ln(1/5*(exp(x**2/exp(x))+x*ln(x)-x)/ln(x))*ln(ln(1/5*(exp(x**2/exp(x))+ 
x*ln(x)-x)/ln(x)))+((x**3-2*x**2)*ln(x)+exp(x))*exp(x**2/exp(x))-x*exp(x)* 
ln(x)**2+x*exp(x)*ln(x)-exp(x)*x)/(exp(x)*ln(x)*exp(x**2/exp(x))+x*exp(x)* 
ln(x)**2-x*exp(x)*ln(x))/ln(1/5*(exp(x**2/exp(x))+x*ln(x)-x)/ln(x)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=-x \log \left (-\log \left (5\right ) + \log \left (x \log \left (x\right ) - x + e^{\left (x^{2} e^{\left (-x\right )}\right )}\right ) - \log \left (\log \left (x\right )\right )\right ) \] Input:

integrate(((-exp(x)*log(x)*exp(x^2/exp(x))-x*exp(x)*log(x)^2+x*exp(x)*log( 
x))*log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x))*log(log(1/5*(exp(x^2/exp( 
x))+x*log(x)-x)/log(x)))+((x^3-2*x^2)*log(x)+exp(x))*exp(x^2/exp(x))-x*exp 
(x)*log(x)^2+x*exp(x)*log(x)-exp(x)*x)/(exp(x)*log(x)*exp(x^2/exp(x))+x*ex 
p(x)*log(x)^2-x*exp(x)*log(x))/log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x) 
),x, algorithm="maxima")
 

Output:

-x*log(-log(5) + log(x*log(x) - x + e^(x^2*e^(-x))) - log(log(x)))
 

Giac [F]

\[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=\int { -\frac {x e^{x} \log \left (x\right )^{2} - x e^{x} \log \left (x\right ) + {\left (x e^{x} \log \left (x\right )^{2} - x e^{x} \log \left (x\right ) + e^{\left (x^{2} e^{\left (-x\right )} + x\right )} \log \left (x\right )\right )} \log \left (\frac {x \log \left (x\right ) - x + e^{\left (x^{2} e^{\left (-x\right )}\right )}}{5 \, \log \left (x\right )}\right ) \log \left (\log \left (\frac {x \log \left (x\right ) - x + e^{\left (x^{2} e^{\left (-x\right )}\right )}}{5 \, \log \left (x\right )}\right )\right ) - {\left ({\left (x^{3} - 2 \, x^{2}\right )} \log \left (x\right ) + e^{x}\right )} e^{\left (x^{2} e^{\left (-x\right )}\right )} + x e^{x}}{{\left (x e^{x} \log \left (x\right )^{2} - x e^{x} \log \left (x\right ) + e^{\left (x^{2} e^{\left (-x\right )} + x\right )} \log \left (x\right )\right )} \log \left (\frac {x \log \left (x\right ) - x + e^{\left (x^{2} e^{\left (-x\right )}\right )}}{5 \, \log \left (x\right )}\right )} \,d x } \] Input:

integrate(((-exp(x)*log(x)*exp(x^2/exp(x))-x*exp(x)*log(x)^2+x*exp(x)*log( 
x))*log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x))*log(log(1/5*(exp(x^2/exp( 
x))+x*log(x)-x)/log(x)))+((x^3-2*x^2)*log(x)+exp(x))*exp(x^2/exp(x))-x*exp 
(x)*log(x)^2+x*exp(x)*log(x)-exp(x)*x)/(exp(x)*log(x)*exp(x^2/exp(x))+x*ex 
p(x)*log(x)^2-x*exp(x)*log(x))/log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x) 
),x, algorithm="giac")
 

Output:

integrate(-(x*e^x*log(x)^2 - x*e^x*log(x) + (x*e^x*log(x)^2 - x*e^x*log(x) 
 + e^(x^2*e^(-x) + x)*log(x))*log(1/5*(x*log(x) - x + e^(x^2*e^(-x)))/log( 
x))*log(log(1/5*(x*log(x) - x + e^(x^2*e^(-x)))/log(x))) - ((x^3 - 2*x^2)* 
log(x) + e^x)*e^(x^2*e^(-x)) + x*e^x)/((x*e^x*log(x)^2 - x*e^x*log(x) + e^ 
(x^2*e^(-x) + x)*log(x))*log(1/5*(x*log(x) - x + e^(x^2*e^(-x)))/log(x))), 
 x)
 

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=-x\,\ln \left (\ln \left (\frac {{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-x}}-x+x\,\ln \left (x\right )}{5\,\ln \left (x\right )}\right )\right ) \] Input:

int(-(x*exp(x) - exp(x^2*exp(-x))*(exp(x) - log(x)*(2*x^2 - x^3)) + log((e 
xp(x^2*exp(-x))/5 - x/5 + (x*log(x))/5)/log(x))*log(log((exp(x^2*exp(-x))/ 
5 - x/5 + (x*log(x))/5)/log(x)))*(x*exp(x)*log(x)^2 - x*exp(x)*log(x) + ex 
p(x^2*exp(-x))*exp(x)*log(x)) - x*exp(x)*log(x) + x*exp(x)*log(x)^2)/(log( 
(exp(x^2*exp(-x))/5 - x/5 + (x*log(x))/5)/log(x))*(x*exp(x)*log(x)^2 - x*e 
xp(x)*log(x) + exp(x^2*exp(-x))*exp(x)*log(x))),x)
 

Output:

-x*log(log((exp(x^2*exp(-x)) - x + x*log(x))/(5*log(x))))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {-e^x x+e^x x \log (x)-e^x x \log ^2(x)+e^{e^{-x} x^2} \left (e^x+\left (-2 x^2+x^3\right ) \log (x)\right )+\left (-e^{x+e^{-x} x^2} \log (x)+e^x x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right ) \log \left (\log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )\right )}{\left (e^{x+e^{-x} x^2} \log (x)-e^x x \log (x)+e^x x \log ^2(x)\right ) \log \left (\frac {e^{e^{-x} x^2}-x+x \log (x)}{5 \log (x)}\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (\frac {e^{\frac {x^{2}}{e^{x}}}+\mathrm {log}\left (x \right ) x -x}{5 \,\mathrm {log}\left (x \right )}\right )\right ) x \] Input:

int(((-exp(x)*log(x)*exp(x^2/exp(x))-x*exp(x)*log(x)^2+x*exp(x)*log(x))*lo 
g(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x))*log(log(1/5*(exp(x^2/exp(x))+x* 
log(x)-x)/log(x)))+((x^3-2*x^2)*log(x)+exp(x))*exp(x^2/exp(x))-x*exp(x)*lo 
g(x)^2+x*exp(x)*log(x)-exp(x)*x)/(exp(x)*log(x)*exp(x^2/exp(x))+x*exp(x)*l 
og(x)^2-x*exp(x)*log(x))/log(1/5*(exp(x^2/exp(x))+x*log(x)-x)/log(x)),x)
 

Output:

 - log(log((e**(x**2/e**x) + log(x)*x - x)/(5*log(x))))*x