\(\int \frac {(-4+e^{e^{x^2}} (4 x+(4 x+8 e^{x^2} x^3) \log (x))) \log ^3(\frac {1}{4} (25 \log (x)-25 e^{e^{x^2}} x \log (x)))}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx\) [2776]

Optimal result

Integrand size = 77, antiderivative size = 21 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log ^4\left (\frac {5}{4} \left (5-5 e^{e^{x^2}} x\right ) \log (x)\right ) \] Output:

ln(5/4*(5-5*exp(exp(x^2))*x)*ln(x))^4
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log ^4\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right ) \] Input:

Integrate[((-4 + E^E^x^2*(4*x + (4*x + 8*E^x^2*x^3)*Log[x]))*Log[(25*Log[x 
] - 25*E^E^x^2*x*Log[x])/4]^3)/(-(x*Log[x]) + E^E^x^2*x^2*Log[x]),x]
 

Output:

Log[(-25*(-1 + E^E^x^2*x)*Log[x])/4]^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.

\(\displaystyle \int \frac {\left (e^{e^{x^2}} \left (\left (8 e^{x^2} x^3+4 x\right ) \log (x)+4 x\right )-4\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{e^{e^{x^2}} x^2 \log (x)-x \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (4-e^{e^{x^2}} \left (\left (8 e^{x^2} x^3+4 x\right ) \log (x)+4 x\right )\right ) \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \left (1-e^{e^{x^2}} x\right ) \log (x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 \left (e^{e^{x^2}} x+e^{e^{x^2}} x \log (x)-1\right ) \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \left (e^{e^{x^2}} x-1\right ) \log (x)}+\frac {8 e^{x^2+e^{x^2}} x^2 \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{e^{e^{x^2}} x-1}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \int \frac {\log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x}dx+4 \int \frac {\log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \left (e^{e^{x^2}} x-1\right )}dx+8 \int \frac {e^{x^2+e^{x^2}} x^2 \log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{e^{e^{x^2}} x-1}dx+4 \int \frac {\log ^3\left (-\frac {25}{4} \left (e^{e^{x^2}} x-1\right ) \log (x)\right )}{x \log (x)}dx\)

Input:

Int[((-4 + E^E^x^2*(4*x + (4*x + 8*E^x^2*x^3)*Log[x]))*Log[(25*Log[x] - 25 
*E^E^x^2*x*Log[x])/4]^3)/(-(x*Log[x]) + E^E^x^2*x^2*Log[x]),x]
 

Output:

$Aborted
 

Maple [F]

\[\int \frac {\left (\left (\left (8 x^{3} {\mathrm e}^{x^{2}}+4 x \right ) \ln \left (x \right )+4 x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}-4\right ) \ln \left (-\frac {25 x \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}}{4}+\frac {25 \ln \left (x \right )}{4}\right )^{3}}{x^{2} \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}-x \ln \left (x \right )}d x\]

Input:

int((((8*x^3*exp(x^2)+4*x)*ln(x)+4*x)*exp(exp(x^2))-4)*ln(-25/4*x*ln(x)*ex 
p(exp(x^2))+25/4*ln(x))^3/(x^2*ln(x)*exp(exp(x^2))-x*ln(x)),x)
 

Output:

int((((8*x^3*exp(x^2)+4*x)*ln(x)+4*x)*exp(exp(x^2))-4)*ln(-25/4*x*ln(x)*ex 
p(exp(x^2))+25/4*ln(x))^3/(x^2*ln(x)*exp(exp(x^2))-x*ln(x)),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) + \frac {25}{4} \, \log \left (x\right )\right )^{4} \] Input:

integrate((((8*x^3*exp(x^2)+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x* 
log(x)*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x, 
 algorithm="fricas")
 

Output:

log(-25/4*x*e^(e^(x^2))*log(x) + 25/4*log(x))^4
 

Sympy [A] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\log {\left (- \frac {25 x e^{e^{x^{2}}} \log {\left (x \right )}}{4} + \frac {25 \log {\left (x \right )}}{4} \right )}^{4} \] Input:

integrate((((8*exp(x**2)*x**3+4*x)*ln(x)+4*x)*exp(exp(x**2))-4)*ln(-25/4*x 
*ln(x)*exp(exp(x**2))+25/4*ln(x))**3/(x**2*ln(x)*exp(exp(x**2))-x*ln(x)),x 
)
 

Output:

log(-25*x*exp(exp(x**2))*log(x)/4 + 25*log(x)/4)**4
 

Maxima [F]

\[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\int { \frac {4 \, {\left ({\left ({\left (2 \, x^{3} e^{\left (x^{2}\right )} + x\right )} \log \left (x\right ) + x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - 1\right )} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) + \frac {25}{4} \, \log \left (x\right )\right )^{3}}{x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - x \log \left (x\right )} \,d x } \] Input:

integrate((((8*x^3*exp(x^2)+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x* 
log(x)*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x, 
 algorithm="maxima")
 

Output:

4*integrate((((2*x^3*e^(x^2) + x)*log(x) + x)*e^(e^(x^2)) - 1)*log(-25/4*x 
*e^(e^(x^2))*log(x) + 25/4*log(x))^3/(x^2*e^(e^(x^2))*log(x) - x*log(x)), 
x)
 

Giac [F]

\[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\int { \frac {4 \, {\left ({\left ({\left (2 \, x^{3} e^{\left (x^{2}\right )} + x\right )} \log \left (x\right ) + x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - 1\right )} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) + \frac {25}{4} \, \log \left (x\right )\right )^{3}}{x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - x \log \left (x\right )} \,d x } \] Input:

integrate((((8*x^3*exp(x^2)+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x* 
log(x)*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x, 
 algorithm="giac")
 

Output:

integrate(4*(((2*x^3*e^(x^2) + x)*log(x) + x)*e^(e^(x^2)) - 1)*log(-25/4*x 
*e^(e^(x^2))*log(x) + 25/4*log(x))^3/(x^2*e^(e^(x^2))*log(x) - x*log(x)), 
x)
 

Mupad [B] (verification not implemented)

Time = 2.92 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx={\left (\ln \left (\ln \left (x\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \left (x\right )\right )+\ln \left (\frac {25}{4}\right )\right )}^4 \] Input:

int(-(log((25*log(x))/4 - (25*x*exp(exp(x^2))*log(x))/4)^3*(exp(exp(x^2))* 
(4*x + log(x)*(4*x + 8*x^3*exp(x^2))) - 4))/(x*log(x) - x^2*exp(exp(x^2))* 
log(x)),x)
 

Output:

(log(log(x) - x*exp(exp(x^2))*log(x)) + log(25/4))^4
 

Reduce [F]

\[ \int \frac {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} x^2 \log (x)} \, dx=\int \frac {\left (\left (\left (8 \,{\mathrm e}^{x^{2}} x^{3}+4 x \right ) \mathrm {log}\left (x \right )+4 x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}-4\right ) \mathrm {log}\left (-\frac {25 x \,\mathrm {log}\left (x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}}{4}+\frac {25 \,\mathrm {log}\left (x \right )}{4}\right )^{3}}{x^{2} \mathrm {log}\left (x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}-\mathrm {log}\left (x \right ) x}d x \] Input:

int((((8*exp(x^2)*x^3+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x*log(x) 
*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x)
 

Output:

int((((8*exp(x^2)*x^3+4*x)*log(x)+4*x)*exp(exp(x^2))-4)*log(-25/4*x*log(x) 
*exp(exp(x^2))+25/4*log(x))^3/(x^2*log(x)*exp(exp(x^2))-x*log(x)),x)