\(\int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} (e^x (16 x-16 x^2)+4 e^x x \log (-x^4)) \log (4-4 x+\log (-x^4))}{4 x-4 x^2+x \log (-x^4)} \, dx\) [205]

Optimal result

Integrand size = 80, antiderivative size = 21 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \] Output:

4*exp(exp(x)-4)*ln(ln(-x^4)-4*x+4)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \] Input:

Integrate[(E^(-4 + E^x)*(16 - 16*x) + E^(-4 + E^x)*(E^x*(16*x - 16*x^2) + 
4*E^x*x*Log[-x^4])*Log[4 - 4*x + Log[-x^4]])/(4*x - 4*x^2 + x*Log[-x^4]),x 
]
 

Output:

4*E^(-4 + E^x)*Log[4 - 4*x + Log[-x^4]]
 

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.

\(\displaystyle \int \frac {e^{e^x-4} \left (4 e^x x \log \left (-x^4\right )+e^x \left (16 x-16 x^2\right )\right ) \log \left (\log \left (-x^4\right )-4 x+4\right )+e^{e^x-4} (16-16 x)}{x \log \left (-x^4\right )-4 x^2+4 x} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {16 e^{e^x-4} (x-1)}{x \left (-\log \left (-x^4\right )+4 x-4\right )}+4 e^{x+e^x-4} \log \left (\log \left (-x^4\right )-4 x+4\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 16 \int \frac {e^{-4+e^x}}{4 x-\log \left (-x^4\right )-4}dx-16 \int \frac {e^{-4+e^x}}{x \left (4 x-\log \left (-x^4\right )-4\right )}dx+4 \int e^{x+e^x-4} \log \left (-4 x+\log \left (-x^4\right )+4\right )dx\)

Input:

Int[(E^(-4 + E^x)*(16 - 16*x) + E^(-4 + E^x)*(E^x*(16*x - 16*x^2) + 4*E^x* 
x*Log[-x^4])*Log[4 - 4*x + Log[-x^4]])/(4*x - 4*x^2 + x*Log[-x^4]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.61 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (\ln \left (-x^{4}\right )-4 x +4\right )\)	\(20\)
risch	\(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (i \pi +4 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{2}\right )\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{3}\right )\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}+i \pi \operatorname {csgn}\left (i x^{4}\right )^{2} \left (\operatorname {csgn}\left (i x^{4}\right )-1\right )-4 x +4\right )\)	\(158\)

Input:

int(((4*x*exp(x)*ln(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*ln(ln(-x^4) 
-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*ln(-x^4)-4*x^2+4*x),x,method=_RETURNV 
ERBOSE)
 

Output:

4*exp(exp(x)-4)*ln(ln(-x^4)-4*x+4)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \] Input:

integrate(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log( 
log(-x^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x, algo 
rithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(((4*x*exp(x)*ln(-x**4)+(-16*x**2+16*x)*exp(x))*exp(exp(x)-4)*ln( 
ln(-x**4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*ln(-x**4)-4*x**2+4*x),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (i \, \pi - 4 \, x + 4 \, \log \left (x\right ) + 4\right ) \] Input:

integrate(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log( 
log(-x^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x, algo 
rithm="maxima")
 

Output:

4*e^(e^x - 4)*log(I*pi - 4*x + 4*log(x) + 4)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \] Input:

integrate(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log( 
log(-x^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x, algo 
rithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.59 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=4\,\ln \left (\ln \left (-x^4\right )-4\,x+4\right )\,{\mathrm {e}}^{{\mathrm {e}}^x-4} \] Input:

int(-(exp(exp(x) - 4)*(16*x - 16) - log(log(-x^4) - 4*x + 4)*exp(exp(x) - 
4)*(exp(x)*(16*x - 16*x^2) + 4*x*exp(x)*log(-x^4)))/(4*x + x*log(-x^4) - 4 
*x^2),x)
 

Output:

4*log(log(-x^4) - 4*x + 4)*exp(exp(x) - 4)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} \left (e^x \left (16 x-16 x^2\right )+4 e^x x \log \left (-x^4\right )\right ) \log \left (4-4 x+\log \left (-x^4\right )\right )}{4 x-4 x^2+x \log \left (-x^4\right )} \, dx=\frac {4 e^{e^{x}} \mathrm {log}\left (\mathrm {log}\left (-x^{4}\right )-4 x +4\right )}{e^{4}} \] Input:

int(((4*x*exp(x)*log(-x^4)+(-16*x^2+16*x)*exp(x))*exp(exp(x)-4)*log(log(-x 
^4)-4*x+4)+(-16*x+16)*exp(exp(x)-4))/(x*log(-x^4)-4*x^2+4*x),x)
 

Output:

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