\(\int \frac {e^{\frac {36+12 \log (\frac {4}{x})}{x \log (e^{-e^4 x} x)}} (-36+36 e^4 x+(-12+12 e^4 x) \log (\frac {4}{x})+(-48-12 \log (\frac {4}{x})) \log (e^{-e^4 x} x))}{x^2 \log ^2(e^{-e^4 x} x)} \, dx\) [1715]

Optimal result

Integrand size = 91, antiderivative size = 28 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \] Output:

exp(12*(ln(4/x)+3)/x/ln(x/exp(x*exp(4))))
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \] Input:

Integrate[(E^((36 + 12*Log[4/x])/(x*Log[x/E^(E^4*x)]))*(-36 + 36*E^4*x + ( 
-12 + 12*E^4*x)*Log[4/x] + (-48 - 12*Log[4/x])*Log[x/E^(E^4*x)]))/(x^2*Log 
[x/E^(E^4*x)]^2),x]
 

Output:

E^((12*(3 + Log[4/x]))/(x*Log[x/E^(E^4*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.

\(\displaystyle \int \frac {e^{\frac {12 \log \left (\frac {4}{x}\right )+36}{x \log \left (e^{-e^4 x} x\right )}} \left (36 e^4 x+\left (12 e^4 x-12\right ) \log \left (\frac {4}{x}\right )+\left (-12 \log \left (\frac {4}{x}\right )-48\right ) \log \left (e^{-e^4 x} x\right )-36\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (36 e^4 x+\left (12 e^4 x-12\right ) \log \left (\frac {4}{x}\right )+\left (-12 \log \left (\frac {4}{x}\right )-48\right ) \log \left (e^{-e^4 x} x\right )-36\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {12 \left (e^4 x-1\right ) e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (\log \left (\frac {4}{x}\right )+3\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )}-\frac {12 e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (\log \left (\frac {4}{x}\right )+4\right )}{x^2 \log \left (e^{-e^4 x} x\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -36 \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log ^2\left (e^{-e^4 x} x\right )}dx-12 \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )}dx-48 \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log \left (e^{-e^4 x} x\right )}dx-12 \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log \left (e^{-e^4 x} x\right )}dx+36 \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}+4}}{x \log ^2\left (e^{-e^4 x} x\right )}dx+12 \int \frac {e^{\frac {12 \left (\log \left (\frac {4}{x}\right )+3\right )}{x \log \left (e^{-e^4 x} x\right )}+4} \log \left (\frac {4}{x}\right )}{x \log ^2\left (e^{-e^4 x} x\right )}dx\)

Input:

Int[(E^((36 + 12*Log[4/x])/(x*Log[x/E^(E^4*x)]))*(-36 + 36*E^4*x + (-12 + 
12*E^4*x)*Log[4/x] + (-48 - 12*Log[4/x])*Log[x/E^(E^4*x)]))/(x^2*Log[x/E^( 
E^4*x)]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 88.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\({\mathrm e}^{\frac {12 \ln \left (\frac {4}{x}\right )+36}{x \ln \left (x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )}}\)	\(27\)
risch	\({\mathrm e}^{\frac {48 \ln \left (2\right )-24 \ln \left (x \right )+72}{x \left (i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x \,{\mathrm e}^{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x \,{\mathrm e}^{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )^{3}+i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{4}}\right )\right )}}\)	\(128\)

Input:

int(((-12*ln(4/x)-48)*ln(x/exp(x*exp(4)))+(12*x*exp(4)-12)*ln(4/x)+36*x*ex 
p(4)-36)*exp((12*ln(4/x)+36)/x/ln(x/exp(x*exp(4))))/x^2/ln(x/exp(x*exp(4)) 
)^2,x,method=_RETURNVERBOSE)
 

Output:

exp(12*(ln(4/x)+3)/x/ln(x/exp(x*exp(4))))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=e^{\left (-\frac {12 \, {\left (\log \left (\frac {4}{x}\right ) + 3\right )}}{x^{2} e^{4} - 2 \, x \log \left (2\right ) + x \log \left (\frac {4}{x}\right )}\right )} \] Input:

integrate(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x 
)+36*x*exp(4)-36)*exp((12*log(4/x)+36)/x/log(x/exp(x*exp(4))))/x^2/log(x/e 
xp(x*exp(4)))^2,x, algorithm="fricas")
 

Output:

e^(-12*(log(4/x) + 3)/(x^2*e^4 - 2*x*log(2) + x*log(4/x)))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=\text {Timed out} \] Input:

integrate(((-12*ln(4/x)-48)*ln(x/exp(x*exp(4)))+(12*x*exp(4)-12)*ln(4/x)+3 
6*x*exp(4)-36)*exp((12*ln(4/x)+36)/x/ln(x/exp(x*exp(4))))/x**2/ln(x/exp(x* 
exp(4)))**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (25) = 50\).

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.93 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=e^{\left (-\frac {24 \, e^{4} \log \left (2\right )}{x e^{4} \log \left (x\right ) - \log \left (x\right )^{2}} - \frac {36 \, e^{4}}{x e^{4} \log \left (x\right ) - \log \left (x\right )^{2}} + \frac {12 \, e^{4}}{x e^{4} - \log \left (x\right )} - \frac {12}{x} + \frac {24 \, \log \left (2\right )}{x \log \left (x\right )} + \frac {36}{x \log \left (x\right )}\right )} \] Input:

integrate(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x 
)+36*x*exp(4)-36)*exp((12*log(4/x)+36)/x/log(x/exp(x*exp(4))))/x^2/log(x/e 
xp(x*exp(4)))^2,x, algorithm="maxima")
 

Output:

e^(-24*e^4*log(2)/(x*e^4*log(x) - log(x)^2) - 36*e^4/(x*e^4*log(x) - log(x 
)^2) + 12*e^4/(x*e^4 - log(x)) - 12/x + 24*log(2)/(x*log(x)) + 36/(x*log(x 
)))
 

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=e^{\left (-\frac {12 \, {\left (2 \, \log \left (2\right ) - \log \left (x\right ) + 3\right )}}{x^{2} e^{4} - x \log \left (x\right )}\right )} \] Input:

integrate(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x 
)+36*x*exp(4)-36)*exp((12*log(4/x)+36)/x/log(x/exp(x*exp(4))))/x^2/log(x/e 
xp(x*exp(4)))^2,x, algorithm="giac")
 

Output:

e^(-12*(2*log(2) - log(x) + 3)/(x^2*e^4 - x*log(x)))
 

Mupad [B] (verification not implemented)

Time = 2.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=\frac {{\mathrm {e}}^{-\frac {36}{x^2\,{\mathrm {e}}^4-x\,\ln \left (x\right )}}}{2^{\frac {24}{x^2\,{\mathrm {e}}^4-x\,\ln \left (x\right )}}\,{\left (\frac {1}{x}\right )}^{\frac {12}{x^2\,{\mathrm {e}}^4-x\,\ln \left (x\right )}}} \] Input:

int((exp((12*log(4/x) + 36)/(x*log(x*exp(-x*exp(4)))))*(36*x*exp(4) + log( 
4/x)*(12*x*exp(4) - 12) - log(x*exp(-x*exp(4)))*(12*log(4/x) + 48) - 36))/ 
(x^2*log(x*exp(-x*exp(4)))^2),x)
 

Output:

exp(-36/(x^2*exp(4) - x*log(x)))/(2^(24/(x^2*exp(4) - x*log(x)))*(1/x)^(12 
/(x^2*exp(4) - x*log(x))))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx=e^{\frac {12 \,\mathrm {log}\left (\frac {4}{x}\right )+36}{\mathrm {log}\left (\frac {x}{e^{e^{4} x}}\right ) x}} \] Input:

int(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x)+36*x 
*exp(4)-36)*exp((12*log(4/x)+36)/x/log(x/exp(x*exp(4))))/x^2/log(x/exp(x*e 
xp(4)))^2,x)
 

Output:

e**((12*log(4/x) + 36)/(log(x/e**(e**4*x))*x))