\(\int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} (x+(-4 \log (x)+e^{25-5 e^x} (6 x^2-30 e^x x^3) \log (x)) \log (\log (x)))}{x^2 \log (x)} \, dx\) [1673]

Optimal result

Integrand size = 76, antiderivative size = 30 \[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=-x+e^{\frac {4}{x}+6 e^{5 \left (5-e^x\right )} x} \log (\log (x)) \] Output:

ln(ln(x))*exp(6*x*exp(-5*exp(x)+25)+4/x)-x
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=-x+e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \] Input:

Integrate[(-(x^2*Log[x]) + E^((4 + 6*E^(25 - 5*E^x)*x^2)/x)*(x + (-4*Log[x 
] + E^(25 - 5*E^x)*(6*x^2 - 30*E^x*x^3)*Log[x])*Log[Log[x]]))/(x^2*Log[x]) 
,x]
 

Output:

-x + E^(4/x + 6*E^(25 - 5*E^x)*x)*Log[Log[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.

\(\displaystyle \int \frac {e^{\frac {6 e^{25-5 e^x} x^2+4}{x}} \left (\left (e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)-4 \log (x)\right ) \log (\log (x))+x\right )-x^2 \log (x)}{x^2 \log (x)} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{6 e^{25-5 e^x} x-5 e^x+\frac {4}{x}} \left (-30 e^{x+25} x^3 \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))+e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))\right )}{x^2 \log (x)}-1\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \int \frac {e^{6 e^{25-5 e^x} x+\frac {4}{x}} \log (\log (x))}{x^2}dx+\int \frac {e^{6 e^{25-5 e^x} x+\frac {4}{x}}}{x \log (x)}dx+6 \int e^{6 e^{25-5 e^x} x-5 e^x+25+\frac {4}{x}} \log (\log (x))dx-30 \int e^{6 e^{25-5 e^x} x+x-5 e^x+25+\frac {4}{x}} x \log (\log (x))dx-x\)

Input:

Int[(-(x^2*Log[x]) + E^((4 + 6*E^(25 - 5*E^x)*x^2)/x)*(x + (-4*Log[x] + E^ 
(25 - 5*E^x)*(6*x^2 - 30*E^x*x^3)*Log[x])*Log[Log[x]]))/(x^2*Log[x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 63.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(\ln \left (\ln \left (x \right )\right ) {\mathrm e}^{\frac {6 x^{2} {\mathrm e}^{-5 \,{\mathrm e}^{x}+25}+4}{x}}-x\)	\(29\)
parallelrisch	\(\ln \left (\ln \left (x \right )\right ) {\mathrm e}^{\frac {6 x^{2} {\mathrm e}^{-5 \,{\mathrm e}^{x}+25}+4}{x}}-x\)	\(29\)

Input:

int(((((-30*exp(x)*x^3+6*x^2)*ln(x)*exp(-5*exp(x)+25)-4*ln(x))*ln(ln(x))+x 
)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)-x^2*ln(x))/x^2/ln(x),x,method=_RETURN 
VERBOSE)
 

Output:

ln(ln(x))*exp(2*(3*x^2*exp(-5*exp(x)+25)+2)/x)-x
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )} \log \left (\log \left (x\right )\right ) - x \] Input:

integrate(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log 
(log(x))+x)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)-x^2*log(x))/x^2/log(x),x, a 
lgorithm="fricas")
 

Output:

e^(2*(3*x^2*e^(-5*e^x + 25) + 2)/x)*log(log(x)) - x
 

Sympy [F(-1)]

Timed out. \[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\text {Timed out} \] Input:

integrate(((((-30*exp(x)*x**3+6*x**2)*ln(x)*exp(-5*exp(x)+25)-4*ln(x))*ln( 
ln(x))+x)*exp((6*x**2*exp(-5*exp(x)+25)+4)/x)-x**2*ln(x))/x**2/ln(x),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\int { -\frac {x^{2} \log \left (x\right ) + {\left (2 \, {\left (3 \, {\left (5 \, x^{3} e^{x} - x^{2}\right )} e^{\left (-5 \, e^{x} + 25\right )} \log \left (x\right ) + 2 \, \log \left (x\right )\right )} \log \left (\log \left (x\right )\right ) - x\right )} e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )}}{x^{2} \log \left (x\right )} \,d x } \] Input:

integrate(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log 
(log(x))+x)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)-x^2*log(x))/x^2/log(x),x, a 
lgorithm="maxima")
 

Output:

-x - integrate((6*(5*x^3*e^(x + 25)*log(x) - x^2*e^25*log(x))*e^(4/x)*log( 
log(x)) + (4*log(x)*log(log(x)) - x)*e^(4/x + 5*e^x))*e^(6*x*e^(-5*e^x + 2 
5) - 5*e^x)/(x^2*log(x)), x)
 

Giac [F]

\[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\int { -\frac {x^{2} \log \left (x\right ) + {\left (2 \, {\left (3 \, {\left (5 \, x^{3} e^{x} - x^{2}\right )} e^{\left (-5 \, e^{x} + 25\right )} \log \left (x\right ) + 2 \, \log \left (x\right )\right )} \log \left (\log \left (x\right )\right ) - x\right )} e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )}}{x^{2} \log \left (x\right )} \,d x } \] Input:

integrate(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log 
(log(x))+x)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)-x^2*log(x))/x^2/log(x),x, a 
lgorithm="giac")
 

Output:

integrate(-(x^2*log(x) + (2*(3*(5*x^3*e^x - x^2)*e^(-5*e^x + 25)*log(x) + 
2*log(x))*log(log(x)) - x)*e^(2*(3*x^2*e^(-5*e^x + 25) + 2)/x))/(x^2*log(x 
)), x)
 

Mupad [B] (verification not implemented)

Time = 3.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{6\,x\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{-5\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{4/x}-x \] Input:

int(-(x^2*log(x) - exp((6*x^2*exp(25 - 5*exp(x)) + 4)/x)*(x - log(log(x))* 
(4*log(x) + exp(25 - 5*exp(x))*log(x)*(30*x^3*exp(x) - 6*x^2))))/(x^2*log( 
x)),x)
 

Output:

log(log(x))*exp(6*x*exp(25)*exp(-5*exp(x)))*exp(4/x) - x
 

Reduce [F]

\[ \int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\int \frac {\left (\left (\left (-30 \,{\mathrm e}^{x} x^{3}+6 x^{2}\right ) \mathrm {log}\left (x \right ) {\mathrm e}^{-5 \,{\mathrm e}^{x}+25}-4 \,\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )+x \right ) {\mathrm e}^{\frac {6 x^{2} {\mathrm e}^{-5 \,{\mathrm e}^{x}+25}+4}{x}}-\mathrm {log}\left (x \right ) x^{2}}{x^{2} \mathrm {log}\left (x \right )}d x \] Input:

int(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log(log(x 
))+x)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)-x^2*log(x))/x^2/log(x),x)
 

Output:

int(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log(log(x 
))+x)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)-x^2*log(x))/x^2/log(x),x)