\(\int \frac {e^{\frac {e^9-x \log (\frac {e^{e^2} \log ^2(-4+x)}{x^2})}{\log (\frac {e^{e^2} \log ^2(-4+x)}{x^2})}} (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+(4 x-x^2) \log (-4+x) \log ^2(\frac {e^{e^2} \log ^2(-4+x)}{x^2}))}{(-4 x+x^2) \log (-4+x) \log ^2(\frac {e^{e^2} \log ^2(-4+x)}{x^2})} \, dx\) [1860]

Optimal result

Integrand size = 130, antiderivative size = 28 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{-x+\frac {e^9}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \] Output:

exp(exp(9)/ln(exp(exp(2))*ln(-4+x)^2/x^2)-x)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \] Input:

Integrate[(E^((E^9 - x*Log[(E^E^2*Log[-4 + x]^2)/x^2])/Log[(E^E^2*Log[-4 + 
 x]^2)/x^2])*(-2*E^9*x + E^9*(-8 + 2*x)*Log[-4 + x] + (4*x - x^2)*Log[-4 + 
 x]*Log[(E^E^2*Log[-4 + x]^2)/x^2]^2))/((-4*x + x^2)*Log[-4 + x]*Log[(E^E^ 
2*Log[-4 + x]^2)/x^2]^2),x]
 

Output:

E^(-x + E^9/(E^2 + Log[Log[-4 + x]^2/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.

\(\displaystyle \int \frac {\left (\left (4 x-x^2\right ) \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )-2 e^9 x+e^9 (2 x-8) \log (x-4)\right ) \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )}{\left (x^2-4 x\right ) \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (\left (4 x-x^2\right ) \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )-2 e^9 x+e^9 (2 x-8) \log (x-4)\right ) \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )}{(x-4) x \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 (-x+x \log (x-4)-4 \log (x-4)) \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}+9\right )}{(x-4) x \log (x-4) \left (\log \left (\frac {\log ^2(x-4)}{x^2}\right )+e^2\right )^2}-\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )dx+2 \int \frac {\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}+9\right )}{x \left (\log \left (\frac {\log ^2(x-4)}{x^2}\right )+e^2\right )^2}dx-2 \int \frac {\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}+9\right )}{(x-4) \log (x-4) \left (\log \left (\frac {\log ^2(x-4)}{x^2}\right )+e^2\right )^2}dx\)

Input:

Int[(E^((E^9 - x*Log[(E^E^2*Log[-4 + x]^2)/x^2])/Log[(E^E^2*Log[-4 + x]^2) 
/x^2])*(-2*E^9*x + E^9*(-8 + 2*x)*Log[-4 + x] + (4*x - x^2)*Log[-4 + x]*Lo 
g[(E^E^2*Log[-4 + x]^2)/x^2]^2))/((-4*x + x^2)*Log[-4 + x]*Log[(E^E^2*Log[ 
-4 + x]^2)/x^2]^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.27 (sec) , antiderivative size = 497, normalized size of antiderivative = 17.75

\[{\mathrm e}^{-\frac {-i x \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{3}+2 i x \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (x -4\right )\right )-i x \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x -4\right )\right )^{2}+i x \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2}-i x \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i x \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-i x \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{3}+i x \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+2 \,{\mathrm e}^{2} x -4 x \ln \left (x \right )+4 x \ln \left (\ln \left (x -4\right )\right )-2 \,{\mathrm e}^{9}}{-i \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (x -4\right )\right )-i \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x -4\right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-i \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+2 \,{\mathrm e}^{2}-4 \ln \left (x \right )+4 \ln \left (\ln \left (x -4\right )\right )}}\]

Input:

int(((-x^2+4*x)*ln(x-4)*ln(exp(exp(2))*ln(x-4)^2/x^2)^2+(2*x-8)*exp(9)*ln( 
x-4)-2*x*exp(9))*exp((-x*ln(exp(exp(2))*ln(x-4)^2/x^2)+exp(9))/ln(exp(exp( 
2))*ln(x-4)^2/x^2))/(x^2-4*x)/ln(x-4)/ln(exp(exp(2))*ln(x-4)^2/x^2)^2,x)
 

Output:

exp(-(-I*x*Pi*csgn(I*ln(x-4)^2)^3+2*I*x*Pi*csgn(I*ln(x-4)^2)^2*csgn(I*ln(x 
-4))-I*x*Pi*csgn(I*ln(x-4)^2)*csgn(I*ln(x-4))^2+I*x*Pi*csgn(I*ln(x-4)^2)*c 
sgn(I/x^2*ln(x-4)^2)^2-I*x*Pi*csgn(I*ln(x-4)^2)*csgn(I/x^2*ln(x-4)^2)*csgn 
(I/x^2)+I*x*Pi*csgn(I*x^2)^3-2*I*Pi*x*csgn(I*x)*csgn(I*x^2)^2+I*x*Pi*csgn( 
I*x^2)*csgn(I*x)^2-I*x*Pi*csgn(I/x^2*ln(x-4)^2)^3+I*x*Pi*csgn(I/x^2*ln(x-4 
)^2)^2*csgn(I/x^2)+2*exp(2)*x-4*x*ln(x)+4*x*ln(ln(x-4))-2*exp(9))/(-I*Pi*c 
sgn(I*ln(x-4)^2)^3+2*I*Pi*csgn(I*ln(x-4)^2)^2*csgn(I*ln(x-4))-I*Pi*csgn(I* 
ln(x-4)^2)*csgn(I*ln(x-4))^2+I*Pi*csgn(I*ln(x-4)^2)*csgn(I/x^2*ln(x-4)^2)^ 
2-I*Pi*csgn(I*ln(x-4)^2)*csgn(I/x^2*ln(x-4)^2)*csgn(I/x^2)+I*Pi*csgn(I*x^2 
)^3-2*I*Pi*csgn(I*x)*csgn(I*x^2)^2+I*Pi*csgn(I*x)^2*csgn(I*x^2)-I*Pi*csgn( 
I/x^2*ln(x-4)^2)^3+I*Pi*csgn(I/x^2*ln(x-4)^2)^2*csgn(I/x^2)+2*exp(2)-4*ln( 
x)+4*ln(ln(x-4))))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )} \] Input:

integrate(((-x^2+4*x)*log(-4+x)*log(exp(exp(2))*log(-4+x)^2/x^2)^2+(2*x-8) 
*exp(9)*log(-4+x)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(-4+x)^2/x^2)+exp 
(9))/log(exp(exp(2))*log(-4+x)^2/x^2))/(x^2-4*x)/log(-4+x)/log(exp(exp(2)) 
*log(-4+x)^2/x^2)^2,x, algorithm="fricas")
 

Output:

e^(-(x*log(e^(e^2)*log(x - 4)^2/x^2) - e^9)/log(e^(e^2)*log(x - 4)^2/x^2))
 

Sympy [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{\frac {- x \log {\left (\frac {e^{e^{2}} \log {\left (x - 4 \right )}^{2}}{x^{2}} \right )} + e^{9}}{\log {\left (\frac {e^{e^{2}} \log {\left (x - 4 \right )}^{2}}{x^{2}} \right )}}} \] Input:

integrate(((-x**2+4*x)*ln(-4+x)*ln(exp(exp(2))*ln(-4+x)**2/x**2)**2+(2*x-8 
)*exp(9)*ln(-4+x)-2*x*exp(9))*exp((-x*ln(exp(exp(2))*ln(-4+x)**2/x**2)+exp 
(9))/ln(exp(exp(2))*ln(-4+x)**2/x**2))/(x**2-4*x)/ln(-4+x)/ln(exp(exp(2))* 
ln(-4+x)**2/x**2)**2,x)
 

Output:

exp((-x*log(exp(exp(2))*log(x - 4)**2/x**2) + exp(9))/log(exp(exp(2))*log( 
x - 4)**2/x**2))
 

Maxima [F]

\[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\int { -\frac {{\left ({\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2} - 2 \, {\left (x - 4\right )} e^{9} \log \left (x - 4\right ) + 2 \, x e^{9}\right )} e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )}}{{\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2}} \,d x } \] Input:

integrate(((-x^2+4*x)*log(-4+x)*log(exp(exp(2))*log(-4+x)^2/x^2)^2+(2*x-8) 
*exp(9)*log(-4+x)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(-4+x)^2/x^2)+exp 
(9))/log(exp(exp(2))*log(-4+x)^2/x^2))/(x^2-4*x)/log(-4+x)/log(exp(exp(2)) 
*log(-4+x)^2/x^2)^2,x, algorithm="maxima")
 

Output:

-integrate(((x^2 - 4*x)*log(x - 4)*log(e^(e^2)*log(x - 4)^2/x^2)^2 - 2*(x 
- 4)*e^9*log(x - 4) + 2*x*e^9)*e^(-(x*log(e^(e^2)*log(x - 4)^2/x^2) - e^9) 
/log(e^(e^2)*log(x - 4)^2/x^2))/((x^2 - 4*x)*log(x - 4)*log(e^(e^2)*log(x 
- 4)^2/x^2)^2), x)
 

Giac [F]

\[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\int { -\frac {{\left ({\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2} - 2 \, {\left (x - 4\right )} e^{9} \log \left (x - 4\right ) + 2 \, x e^{9}\right )} e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )}}{{\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2}} \,d x } \] Input:

integrate(((-x^2+4*x)*log(-4+x)*log(exp(exp(2))*log(-4+x)^2/x^2)^2+(2*x-8) 
*exp(9)*log(-4+x)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(-4+x)^2/x^2)+exp 
(9))/log(exp(exp(2))*log(-4+x)^2/x^2))/(x^2-4*x)/log(-4+x)/log(exp(exp(2)) 
*log(-4+x)^2/x^2)^2,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 2.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.29 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\frac {{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^2}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^9}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}\,{\left (x^2\right )}^{\frac {x}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}}{{\left ({\ln \left (x-4\right )}^2\right )}^{\frac {x}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}} \] Input:

int(-(exp((exp(9) - x*log((log(x - 4)^2*exp(exp(2)))/x^2))/log((log(x - 4) 
^2*exp(exp(2)))/x^2))*(log(x - 4)*exp(9)*(2*x - 8) - 2*x*exp(9) + log((log 
(x - 4)^2*exp(exp(2)))/x^2)^2*log(x - 4)*(4*x - x^2)))/(log((log(x - 4)^2* 
exp(exp(2)))/x^2)^2*log(x - 4)*(4*x - x^2)),x)
 

Output:

(exp(-(x*exp(2))/(log(log(x - 4)^2) + log(1/x^2) + exp(2)))*exp(exp(9)/(lo 
g(log(x - 4)^2) + log(1/x^2) + exp(2)))*(x^2)^(x/(log(log(x - 4)^2) + log( 
1/x^2) + exp(2))))/(log(x - 4)^2)^(x/(log(log(x - 4)^2) + log(1/x^2) + exp 
(2)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\frac {e^{\frac {e^{9}}{\mathrm {log}\left (\frac {e^{e^{2}} \mathrm {log}\left (x -4\right )^{2}}{x^{2}}\right )}}}{e^{x}} \] Input:

int(((-x^2+4*x)*log(-4+x)*log(exp(exp(2))*log(-4+x)^2/x^2)^2+(2*x-8)*exp(9 
)*log(-4+x)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(-4+x)^2/x^2)+exp(9))/l 
og(exp(exp(2))*log(-4+x)^2/x^2))/(x^2-4*x)/log(-4+x)/log(exp(exp(2))*log(- 
4+x)^2/x^2)^2,x)
 

Output:

e**(e**9/log((e**(e**2)*log(x - 4)**2)/x**2))/e**x