\(\int \frac {-x+\log (\log (5))+e^{e^{e^{3+e^x}+\log ^2(\frac {-x+\log (\log (5))}{\log (\log (5))})}} (x-\log (\log (5))+e^{e^{3+e^x}+\log ^2(\frac {-x+\log (\log (5))}{\log (\log (5))})} (e^{3+e^x} (e^x x^2-e^x x \log (\log (5)))+2 x \log (\frac {-x+\log (\log (5))}{\log (\log (5))})))}{-x+\log (\log (5))} \, dx\) [79]

Optimal result

Integrand size = 123, antiderivative size = 30 \[ \int \frac {-x+\log (\log (5))+e^{e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )}} \left (x-\log (\log (5))+e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )} \left (e^{3+e^x} \left (e^x x^2-e^x x \log (\log (5))\right )+2 x \log \left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )\right )\right )}{-x+\log (\log (5))} \, dx=x-e^{e^{e^{3+e^x}+\log ^2\left (1-\frac {x}{\log (\log (5))}\right )}} x \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-x+\log (\log (5))+e^{e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )}} \left (x-\log (\log (5))+e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )} \left (e^{3+e^x} \left (e^x x^2-e^x x \log (\log (5))\right )+2 x \log \left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )\right )\right )}{-x+\log (\log (5))} \, dx=x-e^{e^{e^{3+e^x}+\log ^2\left (1-\frac {x}{\log (\log (5))}\right )}} x \] Input:

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

Output:

x - E^E^(E^(3 + E^x) + Log[1 - x/Log[Log[5]]]^2)*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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 110.92 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00

method	result	size
risch	\(-{\mathrm e}^{{\mathrm e}^{\ln \left (\frac {\ln \left (\ln \left (5\right )\right )-x}{\ln \left (\ln \left (5\right )\right )}\right )^{2}+{\mathrm e}^{3+{\mathrm e}^{x}}}} x +x\)	\(30\)
parallelrisch	\(-{\mathrm e}^{{\mathrm e}^{\ln \left (\frac {\ln \left (\ln \left (5\right )\right )-x}{\ln \left (\ln \left (5\right )\right )}\right )^{2}+{\mathrm e}^{3+{\mathrm e}^{x}}}} x +2 \ln \left (\ln \left (5\right )\right )+x\)	\(35\)

Input:

int((((2*x*ln((ln(ln(5))-x)/ln(ln(5)))+(-x*exp(x)*ln(ln(5))+exp(x)*x^2)*ex 
p(3+exp(x)))*exp(ln((ln(ln(5))-x)/ln(ln(5)))^2+exp(3+exp(x)))-ln(ln(5))+x) 
*exp(exp(ln((ln(ln(5))-x)/ln(ln(5)))^2+exp(3+exp(x))))+ln(ln(5))-x)/(ln(ln 
(5))-x),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-x+\log (\log (5))+e^{e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )}} \left (x-\log (\log (5))+e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )} \left (e^{3+e^x} \left (e^x x^2-e^x x \log (\log (5))\right )+2 x \log \left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )\right )\right )}{-x+\log (\log (5))} \, dx=-x e^{\left (e^{\left (\log \left (-\frac {x - \log \left (\log \left (5\right )\right )}{\log \left (\log \left (5\right )\right )}\right )^{2} + e^{\left (e^{x} + 3\right )}\right )}\right )} + x \] Input:

integrate((((2*x*log((log(log(5))-x)/log(log(5)))+(-x*exp(x)*log(log(5))+e 
xp(x)*x^2)*exp(3+exp(x)))*exp(log((log(log(5))-x)/log(log(5)))^2+exp(3+exp 
(x)))-log(log(5))+x)*exp(exp(log((log(log(5))-x)/log(log(5)))^2+exp(3+exp( 
x))))+log(log(5))-x)/(log(log(5))-x),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((((2*x*log((log(log(5))-x)/log(log(5)))+(-x*exp(x)*log(log(5))+e 
xp(x)*x^2)*exp(3+exp(x)))*exp(log((log(log(5))-x)/log(log(5)))^2+exp(3+exp 
(x)))-log(log(5))+x)*exp(exp(log((log(log(5))-x)/log(log(5)))^2+exp(3+exp( 
x))))+log(log(5))-x)/(log(log(5))-x),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((((2*x*log((log(log(5))-x)/log(log(5)))+(-x*exp(x)*log(log(5))+e 
xp(x)*x^2)*exp(3+exp(x)))*exp(log((log(log(5))-x)/log(log(5)))^2+exp(3+exp 
(x)))-log(log(5))+x)*exp(exp(log((log(log(5))-x)/log(log(5)))^2+exp(3+exp( 
x))))+log(log(5))-x)/(log(log(5))-x),x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {-x+\log (\log (5))+e^{e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )}} \left (x-\log (\log (5))+e^{e^{3+e^x}+\log ^2\left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )} \left (e^{3+e^x} \left (e^x x^2-e^x x \log (\log (5))\right )+2 x \log \left (\frac {-x+\log (\log (5))}{\log (\log (5))}\right )\right )\right )}{-x+\log (\log (5))} \, dx=-x\,\left ({\mathrm {e}}^{\frac {{\mathrm {e}}^{{\ln \left (\ln \left (\ln \left (5\right )\right )\right )}^2}\,{\mathrm {e}}^{{\ln \left (\ln \left (\ln \left (5\right )\right )-x\right )}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^3}}{{\left (\ln \left (\ln \left (5\right )\right )-x\right )}^{2\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )}}}-1\right ) \] Input:

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

Output:

-x*(exp((exp(log(log(log(5)))^2)*exp(log(log(log(5)) - x)^2)*exp(exp(exp(x 
))*exp(3)))/(log(log(5)) - x)^(2*log(log(log(5))))) - 1)
 

Reduce [F]

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

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

Output:

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