\(\int \frac {e^{-e^{e+x}} (4 x \log (x)+e^{e^e} (-x+e^{e+x} x^2) \log ^2(x)+(-4 x^2+4 e^{e+x} x^3) \log ^2(x)+(-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} (-1-e^{e+x} x \log (x))) \log (\frac {1}{4} (e^{e^e}+4 x)))}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx\) [2144]

Optimal result

Integrand size = 132, antiderivative size = 31 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=e^{-e^{e+x}} \left (-x+\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \] Output:

(ln(1/4*exp(exp(exp(1)))+x)/ln(x)-x)/exp(exp(x+exp(1)))
 

Mathematica [A] (verified)

Time = 5.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=e^{-e^{e+x}} \left (-x+\frac {\log \left (\frac {e^{e^e}}{4}+x\right )}{\log (x)}\right ) \] Input:

Integrate[(4*x*Log[x] + E^E^E*(-x + E^(E + x)*x^2)*Log[x]^2 + (-4*x^2 + 4* 
E^(E + x)*x^3)*Log[x]^2 + (-4*x - 4*E^(E + x)*x^2*Log[x] + E^E^E*(-1 - E^( 
E + x)*x*Log[x]))*Log[(E^E^E + 4*x)/4])/(E^E^(E + x)*(E^E^E*x*Log[x]^2 + 4 
*x^2*Log[x]^2)),x]
 

Output:

(-x + Log[E^E^E/4 + x]/Log[x])/E^E^(E + 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.

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \int e^{x-e^{x+e}+e} xdx-\int \frac {e^{-e^{x+e}} \log \left (x+\frac {e^{e^e}}{4}\right )}{x \log ^2(x)}dx+4 \int \frac {e^{-e^{x+e}}}{\left (4 x+e^{e^e}\right ) \log (x)}dx-\int \frac {e^{x-e^{x+e}+e} \log \left (x+\frac {e^{e^e}}{4}\right )}{\log (x)}dx-\operatorname {ExpIntegralEi}\left (-e^{x+e}\right )\)

Input:

Int[(4*x*Log[x] + E^E^E*(-x + E^(E + x)*x^2)*Log[x]^2 + (-4*x^2 + 4*E^(E + 
 x)*x^3)*Log[x]^2 + (-4*x - 4*E^(E + x)*x^2*Log[x] + E^E^E*(-1 - E^(E + x) 
*x*Log[x]))*Log[(E^E^E + 4*x)/4])/(E^E^(E + x)*(E^E^E*x*Log[x]^2 + 4*x^2*L 
og[x]^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00

\[-\frac {\left (x \ln \left (x \right )-\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}}}}{4}+x \right )\right ) {\mathrm e}^{-{\mathrm e}^{x +{\mathrm e}}}}{\ln \left (x \right )}\]

Input:

int((((-x*exp(x+exp(1))*ln(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*ln(x 
)-4*x)*ln(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*ln(x)^2*exp(exp(ex 
p(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*ln(x)^2+4*x*ln(x))/(x*ln(x)^2*exp(exp(e 
xp(1)))+4*x^2*ln(x)^2)/exp(exp(x+exp(1))),x)
 

Output:

-(x*ln(x)-ln(1/4*exp(exp(exp(1)))+x))/ln(x)*exp(-exp(x+exp(1)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=-\frac {x e^{\left (-e^{\left (x + e\right )}\right )} \log \left (x\right ) - e^{\left (-e^{\left (x + e\right )}\right )} \log \left (x + \frac {1}{4} \, e^{\left (e^{e}\right )}\right )}{\log \left (x\right )} \] Input:

integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1 
))*log(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2* 
exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x 
)^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm="fric 
as")
 

Output:

-(x*e^(-e^(x + e))*log(x) - e^(-e^(x + e))*log(x + 1/4*e^(e^e)))/log(x)
 

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\frac {\left (- x \log {\left (x \right )} + \log {\left (x + \frac {e^{e^{e}}}{4} \right )}\right ) e^{- e^{x + e}}}{\log {\left (x \right )}} \] Input:

integrate((((-x*exp(x+exp(1))*ln(x)-1)*exp(exp(exp(1)))-4*x**2*exp(x+exp(1 
))*ln(x)-4*x)*ln(1/4*exp(exp(exp(1)))+x)+(x**2*exp(x+exp(1))-x)*ln(x)**2*e 
xp(exp(exp(1)))+(4*x**3*exp(x+exp(1))-4*x**2)*ln(x)**2+4*x*ln(x))/(x*ln(x) 
**2*exp(exp(exp(1)))+4*x**2*ln(x)**2)/exp(exp(x+exp(1))),x)
 

Output:

(-x*log(x) + log(x + exp(exp(E))/4))*exp(-exp(x + E))/log(x)
 

Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=-\frac {{\left (x \log \left (x\right ) + 2 \, \log \left (2\right )\right )} e^{\left (-e^{\left (x + e\right )}\right )} - e^{\left (-e^{\left (x + e\right )}\right )} \log \left (4 \, x + e^{\left (e^{e}\right )}\right )}{\log \left (x\right )} \] Input:

integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1 
))*log(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2* 
exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x 
)^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm="maxi 
ma")
 

Output:

-((x*log(x) + 2*log(2))*e^(-e^(x + e)) - e^(-e^(x + e))*log(4*x + e^(e^e)) 
)/log(x)
 

Giac [F]

\[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\int { \frac {{\left ({\left (x^{2} e^{\left (x + e\right )} - x\right )} e^{\left (e^{e}\right )} \log \left (x\right )^{2} + 4 \, {\left (x^{3} e^{\left (x + e\right )} - x^{2}\right )} \log \left (x\right )^{2} - {\left (4 \, x^{2} e^{\left (x + e\right )} \log \left (x\right ) + {\left (x e^{\left (x + e\right )} \log \left (x\right ) + 1\right )} e^{\left (e^{e}\right )} + 4 \, x\right )} \log \left (x + \frac {1}{4} \, e^{\left (e^{e}\right )}\right ) + 4 \, x \log \left (x\right )\right )} e^{\left (-e^{\left (x + e\right )}\right )}}{4 \, x^{2} \log \left (x\right )^{2} + x e^{\left (e^{e}\right )} \log \left (x\right )^{2}} \,d x } \] Input:

integrate((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1 
))*log(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2* 
exp(exp(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x 
)^2*exp(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x, algorithm="giac 
")
 

Output:

integrate(((x^2*e^(x + e) - x)*e^(e^e)*log(x)^2 + 4*(x^3*e^(x + e) - x^2)* 
log(x)^2 - (4*x^2*e^(x + e)*log(x) + (x*e^(x + e)*log(x) + 1)*e^(e^e) + 4* 
x)*log(x + 1/4*e^(e^e)) + 4*x*log(x))*e^(-e^(x + e))/(4*x^2*log(x)^2 + x*e 
^(e^e)*log(x)^2), x)
 

Mupad [B] (verification not implemented)

Time = 4.72 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-{\mathrm {e}}^{\mathrm {e}}\,{\mathrm {e}}^x}\,\left (\ln \left (x+\frac {{\mathrm {e}}^{{\mathrm {e}}^{\mathrm {e}}}}{4}\right )-x\,\ln \left (x\right )\right )}{\ln \left (x\right )} \] Input:

int(-(exp(-exp(x + exp(1)))*(log(x)^2*(4*x^2 - 4*x^3*exp(x + exp(1))) + lo 
g(x + exp(exp(exp(1)))/4)*(4*x + exp(exp(exp(1)))*(x*exp(x + exp(1))*log(x 
) + 1) + 4*x^2*exp(x + exp(1))*log(x)) - 4*x*log(x) + exp(exp(exp(1)))*log 
(x)^2*(x - x^2*exp(x + exp(1)))))/(4*x^2*log(x)^2 + x*exp(exp(exp(1)))*log 
(x)^2),x)
 

Output:

(exp(-exp(exp(1))*exp(x))*(log(x + exp(exp(exp(1)))/4) - x*log(x)))/log(x)
 

Reduce [F]

\[ \int \frac {e^{-e^{e+x}} \left (4 x \log (x)+e^{e^e} \left (-x+e^{e+x} x^2\right ) \log ^2(x)+\left (-4 x^2+4 e^{e+x} x^3\right ) \log ^2(x)+\left (-4 x-4 e^{e+x} x^2 \log (x)+e^{e^e} \left (-1-e^{e+x} x \log (x)\right )\right ) \log \left (\frac {1}{4} \left (e^{e^e}+4 x\right )\right )\right )}{e^{e^e} x \log ^2(x)+4 x^2 \log ^2(x)} \, dx=\int \frac {\left (\left (-x \,{\mathrm e}^{x +{\mathrm e}} \mathrm {log}\left (x \right )-1\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}}}-4 x^{2} {\mathrm e}^{x +{\mathrm e}} \mathrm {log}\left (x \right )-4 x \right ) \mathrm {log}\left (\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}}}}{4}+x \right )+\left (x^{2} {\mathrm e}^{x +{\mathrm e}}-x \right ) \mathrm {log}\left (x \right )^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}}}+\left (4 x^{3} {\mathrm e}^{x +{\mathrm e}}-4 x^{2}\right ) \mathrm {log}\left (x \right )^{2}+4 \,\mathrm {log}\left (x \right ) x}{\left (x \mathrm {log}\left (x \right )^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}}}+4 \mathrm {log}\left (x \right )^{2} x^{2}\right ) {\mathrm e}^{{\mathrm e}^{x +{\mathrm e}}}}d x \] Input:

int((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*log 
(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2*exp(ex 
p(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x)^2*ex 
p(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x)
 

Output:

int((((-x*exp(x+exp(1))*log(x)-1)*exp(exp(exp(1)))-4*x^2*exp(x+exp(1))*log 
(x)-4*x)*log(1/4*exp(exp(exp(1)))+x)+(x^2*exp(x+exp(1))-x)*log(x)^2*exp(ex 
p(exp(1)))+(4*x^3*exp(x+exp(1))-4*x^2)*log(x)^2+4*x*log(x))/(x*log(x)^2*ex 
p(exp(exp(1)))+4*x^2*log(x)^2)/exp(exp(x+exp(1))),x)