\(\int \frac {e^{\frac {2 (4 x-36 x^2+(3-108 x) \log (\log (\frac {e^x}{x})))}{x+3 \log (\log (\frac {e^x}{x}))}} (18 x^2-18 x^3+(2 x^3-72 x^4) \log (\frac {e^x}{x})+(30 x^2-432 x^3) \log (\frac {e^x}{x}) \log (\log (\frac {e^x}{x}))+(18 x-648 x^2) \log (\frac {e^x}{x}) \log ^2(\log (\frac {e^x}{x})))}{x^2 \log (\frac {e^x}{x})+6 x \log (\frac {e^x}{x}) \log (\log (\frac {e^x}{x}))+9 \log (\frac {e^x}{x}) \log ^2(\log (\frac {e^x}{x}))} \, dx\) [1407]

Optimal result

Integrand size = 189, antiderivative size = 30 \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=e^{2-2 x \left (36-\frac {3}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )} x^2 \] Output:

exp(1-(36-3/(3*ln(ln(exp(x)/x))+x))*x)^2*x^2
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=e^{\frac {8 (1-9 x) x+(6-216 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \] Input:

Integrate[(E^((2*(4*x - 36*x^2 + (3 - 108*x)*Log[Log[E^x/x]]))/(x + 3*Log[ 
Log[E^x/x]]))*(18*x^2 - 18*x^3 + (2*x^3 - 72*x^4)*Log[E^x/x] + (30*x^2 - 4 
32*x^3)*Log[E^x/x]*Log[Log[E^x/x]] + (18*x - 648*x^2)*Log[E^x/x]*Log[Log[E 
^x/x]]^2))/(x^2*Log[E^x/x] + 6*x*Log[E^x/x]*Log[Log[E^x/x]] + 9*Log[E^x/x] 
*Log[Log[E^x/x]]^2),x]
 

Output:

E^((8*(1 - 9*x)*x + (6 - 216*x)*Log[Log[E^x/x]])/(x + 3*Log[Log[E^x/x]]))* 
x^2
 

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.

\(\displaystyle \int \frac {\left (-18 x^3+18 x^2+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right ) \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )}{x^2 \log \left (\frac {e^x}{x}\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-18 x^3+18 x^2+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right ) \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )}{\log \left (\frac {e^x}{x}\right ) \left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {6 x^2 \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}-\frac {6 x^2 \left (3 x+x \log \left (\frac {e^x}{x}\right )-3\right ) \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )}{\log \left (\frac {e^x}{x}\right ) \left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}-2 (36 x-1) x \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) xdx-72 \int \exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) x^2dx+18 \int \frac {\exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) x^2}{\log \left (\frac {e^x}{x}\right ) \left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}dx+6 \int \frac {\exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) x^2}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}dx-6 \int \frac {\exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) x^3}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}dx-18 \int \frac {\exp \left (\frac {2 \left (-36 x^2+4 x+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) x^3}{\log \left (\frac {e^x}{x}\right ) \left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}dx\)

Input:

Int[(E^((2*(4*x - 36*x^2 + (3 - 108*x)*Log[Log[E^x/x]]))/(x + 3*Log[Log[E^ 
x/x]]))*(18*x^2 - 18*x^3 + (2*x^3 - 72*x^4)*Log[E^x/x] + (30*x^2 - 432*x^3 
)*Log[E^x/x]*Log[Log[E^x/x]] + (18*x - 648*x^2)*Log[E^x/x]*Log[Log[E^x/x]] 
^2))/(x^2*Log[E^x/x] + 6*x*Log[E^x/x]*Log[Log[E^x/x]] + 9*Log[E^x/x]*Log[L 
og[E^x/x]]^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 6.83

\[x^{2} {\mathrm e}^{-\frac {2 \left (108 \ln \left (-\ln \left (x \right )+\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}\right ) x +36 x^{2}-3 \ln \left (-\ln \left (x \right )+\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}\right )-4 x \right )}{3 \ln \left (-\ln \left (x \right )+\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}\right )+x}}\]

Input:

int(((-648*x^2+18*x)*ln(exp(x)/x)*ln(ln(exp(x)/x))^2+(-432*x^3+30*x^2)*ln( 
exp(x)/x)*ln(ln(exp(x)/x))+(-72*x^4+2*x^3)*ln(exp(x)/x)-18*x^3+18*x^2)*exp 
(((-108*x+3)*ln(ln(exp(x)/x))-36*x^2+4*x)/(3*ln(ln(exp(x)/x))+x))^2/(9*ln( 
exp(x)/x)*ln(ln(exp(x)/x))^2+6*x*ln(exp(x)/x)*ln(ln(exp(x)/x))+x^2*ln(exp( 
x)/x)),x)
 

Output:

x^2*exp(-2*(108*ln(-ln(x)+ln(exp(x))-1/2*I*Pi*csgn(I/x*exp(x))*(-csgn(I/x* 
exp(x))+csgn(I/x))*(-csgn(I/x*exp(x))+csgn(I*exp(x))))*x+36*x^2-3*ln(-ln(x 
)+ln(exp(x))-1/2*I*Pi*csgn(I/x*exp(x))*(-csgn(I/x*exp(x))+csgn(I/x))*(-csg 
n(I/x*exp(x))+csgn(I*exp(x))))-4*x)/(3*ln(-ln(x)+ln(exp(x))-1/2*I*Pi*csgn( 
I/x*exp(x))*(-csgn(I/x*exp(x))+csgn(I/x))*(-csgn(I/x*exp(x))+csgn(I*exp(x) 
)))+x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=x^{2} e^{\left (-\frac {2 \, {\left (36 \, x^{2} + 3 \, {\left (36 \, x - 1\right )} \log \left (\log \left (\frac {e^{x}}{x}\right )\right ) - 4 \, x\right )}}{x + 3 \, \log \left (\log \left (\frac {e^{x}}{x}\right )\right )}\right )} \] Input:

integrate(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30 
*x^2)*log(exp(x)/x)*log(log(exp(x)/x))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^ 
3+18*x^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(x) 
/x))+x))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log 
(exp(x)/x))+x^2*log(exp(x)/x)),x, algorithm="fricas")
 

Output:

x^2*e^(-2*(36*x^2 + 3*(36*x - 1)*log(log(e^x/x)) - 4*x)/(x + 3*log(log(e^x 
/x))))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=\text {Timed out} \] Input:

integrate(((-648*x**2+18*x)*ln(exp(x)/x)*ln(ln(exp(x)/x))**2+(-432*x**3+30 
*x**2)*ln(exp(x)/x)*ln(ln(exp(x)/x))+(-72*x**4+2*x**3)*ln(exp(x)/x)-18*x** 
3+18*x**2)*exp(((-108*x+3)*ln(ln(exp(x)/x))-36*x**2+4*x)/(3*ln(ln(exp(x)/x 
))+x))**2/(9*ln(exp(x)/x)*ln(ln(exp(x)/x))**2+6*x*ln(exp(x)/x)*ln(ln(exp(x 
)/x))+x**2*ln(exp(x)/x)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30 
*x^2)*log(exp(x)/x)*log(log(exp(x)/x))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^ 
3+18*x^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(x) 
/x))+x))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log 
(exp(x)/x))+x^2*log(exp(x)/x)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=x^{2} e^{\left (-\frac {2 \, {\left (36 \, x^{2} + 108 \, x \log \left (x - \log \left (x\right )\right ) - 4 \, x - 3 \, \log \left (x - \log \left (x\right )\right )\right )}}{x + 3 \, \log \left (x - \log \left (x\right )\right )}\right )} \] Input:

integrate(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30 
*x^2)*log(exp(x)/x)*log(log(exp(x)/x))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^ 
3+18*x^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(x) 
/x))+x))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log 
(exp(x)/x))+x^2*log(exp(x)/x)),x, algorithm="giac")
 

Output:

x^2*e^(-2*(36*x^2 + 108*x*log(x - log(x)) - 4*x - 3*log(x - log(x)))/(x + 
3*log(x - log(x))))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=\int \frac {{\mathrm {e}}^{-\frac {2\,\left (\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )\,\left (108\,x-3\right )-4\,x+36\,x^2\right )}{x+3\,\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )}}\,\left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,\left (2\,x^3-72\,x^4\right )+18\,x^2-18\,x^3+\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )\,\left (30\,x^2-432\,x^3\right )+\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,{\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )}^2\,\left (18\,x-648\,x^2\right )\right )}{9\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,{\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )}^2+x^2\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )+6\,x\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )} \,d x \] Input:

int((exp(-(2*(log(log(exp(x)/x))*(108*x - 3) - 4*x + 36*x^2))/(x + 3*log(l 
og(exp(x)/x))))*(log(exp(x)/x)*(2*x^3 - 72*x^4) + 18*x^2 - 18*x^3 + log(ex 
p(x)/x)*log(log(exp(x)/x))*(30*x^2 - 432*x^3) + log(exp(x)/x)*log(log(exp( 
x)/x))^2*(18*x - 648*x^2)))/(9*log(exp(x)/x)*log(log(exp(x)/x))^2 + x^2*lo 
g(exp(x)/x) + 6*x*log(exp(x)/x)*log(log(exp(x)/x))),x)
 

Output:

int((exp(-(2*(log(log(exp(x)/x))*(108*x - 3) - 4*x + 36*x^2))/(x + 3*log(l 
og(exp(x)/x))))*(log(exp(x)/x)*(2*x^3 - 72*x^4) + 18*x^2 - 18*x^3 + log(ex 
p(x)/x)*log(log(exp(x)/x))*(30*x^2 - 432*x^3) + log(exp(x)/x)*log(log(exp( 
x)/x))^2*(18*x - 648*x^2)))/(9*log(exp(x)/x)*log(log(exp(x)/x))^2 + x^2*lo 
g(exp(x)/x) + 6*x*log(exp(x)/x)*log(log(exp(x)/x))), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx=\frac {e^{\frac {6 x}{3 \,\mathrm {log}\left (\mathrm {log}\left (\frac {e^{x}}{x}\right )\right )+x}} e^{2} x^{2}}{e^{72 x}} \] Input:

int(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30*x^2)* 
log(exp(x)/x)*log(log(exp(x)/x))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^3+18*x 
^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(x)/x))+x 
))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log(exp(x 
)/x))+x^2*log(exp(x)/x)),x)
 

Output:

(e**((6*x)/(3*log(log(e**x/x)) + x))*e**2*x**2)/e**(72*x)