\(\int \frac {((40-5 x-20 x^2) \log (x)+(-10+5 x^2) \log ^2(x)) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} (x^2-x^2 \log (x) \log (\log (x))+(-2+x^2) \log (x) \log ^2(\log (x)))}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+(-20 x^2 \log (x)+5 x^2 \log ^2(x)) \log ^2(\log (x))} \, dx\) [2019]

Optimal result

Integrand size = 117, antiderivative size = 27 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=\frac {2}{x}+x-\log \left (-4+\frac {1}{5} e^{\frac {x}{\log (\log (x))}}+\log (x)\right ) \] Output:

2/x-ln(ln(x)-4+1/5*exp(x/ln(ln(x))))+x
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=\frac {2}{x}+x-\log \left (20-e^{\frac {x}{\log (\log (x))}}-5 \log (x)\right ) \] Input:

Integrate[(((40 - 5*x - 20*x^2)*Log[x] + (-10 + 5*x^2)*Log[x]^2)*Log[Log[x 
]]^2 + E^(x/Log[Log[x]])*(x^2 - x^2*Log[x]*Log[Log[x]] + (-2 + x^2)*Log[x] 
*Log[Log[x]]^2))/(E^(x/Log[Log[x]])*x^2*Log[x]*Log[Log[x]]^2 + (-20*x^2*Lo 
g[x] + 5*x^2*Log[x]^2)*Log[Log[x]]^2),x]
 

Output:

2/x + x - Log[20 - E^(x/Log[Log[x]]) - 5*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.

\(\displaystyle \int \frac {\left (\left (5 x^2-10\right ) \log ^2(x)+\left (-20 x^2-5 x+40\right ) \log (x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2+\left (x^2-2\right ) \log (x) \log ^2(\log (x))+x^2 (-\log (x)) \log (\log (x))\right )}{x^2 e^{\frac {x}{\log (\log (x))}} \log (x) \log ^2(\log (x))+\left (5 x^2 \log ^2(x)-20 x^2 \log (x)\right ) \log ^2(\log (x))} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-\left (\left (\left (5 x^2-10\right ) \log ^2(x)+\left (-20 x^2-5 x+40\right ) \log (x)\right ) \log ^2(\log (x))\right )-e^{\frac {x}{\log (\log (x))}} \left (x^2+\left (x^2-2\right ) \log (x) \log ^2(\log (x))+x^2 (-\log (x)) \log (\log (x))\right )}{x^2 \left (-5 \log (x)-e^{\frac {x}{\log (\log (x))}}+20\right ) \log (x) \log ^2(\log (x))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^2+x^2 \log (x) \log ^2(\log (x))-x^2 \log (x) \log (\log (x))-2 \log (x) \log ^2(\log (x))}{x^2 \log (x) \log ^2(\log (x))}+\frac {5 \left (4 x+x \log (\log (x)) \log ^2(x)-\log ^2(\log (x)) \log (x)-x \log (x)-4 x \log (\log (x)) \log (x)\right )}{x \log (x) \left (5 \log (x)+e^{\frac {x}{\log (\log (x))}}-20\right ) \log ^2(\log (x))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {1}{\log (x) \log ^2(\log (x))}dx-5 \int \frac {1}{\left (5 \log (x)+e^{\frac {x}{\log (\log (x))}}-20\right ) \log ^2(\log (x))}dx+20 \int \frac {1}{\log (x) \left (5 \log (x)+e^{\frac {x}{\log (\log (x))}}-20\right ) \log ^2(\log (x))}dx-5 \int \frac {1}{x \left (5 \log (x)+e^{\frac {x}{\log (\log (x))}}-20\right )}dx-\int \frac {1}{\log (\log (x))}dx-20 \int \frac {1}{\left (5 \log (x)+e^{\frac {x}{\log (\log (x))}}-20\right ) \log (\log (x))}dx+5 \int \frac {\log (x)}{\left (5 \log (x)+e^{\frac {x}{\log (\log (x))}}-20\right ) \log (\log (x))}dx+x+\frac {2}{x}\)

Input:

Int[(((40 - 5*x - 20*x^2)*Log[x] + (-10 + 5*x^2)*Log[x]^2)*Log[Log[x]]^2 + 
 E^(x/Log[Log[x]])*(x^2 - x^2*Log[x]*Log[Log[x]] + (-2 + x^2)*Log[x]*Log[L 
og[x]]^2))/(E^(x/Log[Log[x]])*x^2*Log[x]*Log[Log[x]]^2 + (-20*x^2*Log[x] + 
 5*x^2*Log[x]^2)*Log[Log[x]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 10.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {x^{2}+2}{x}-\ln \left (5 \ln \left (x \right )+{\mathrm e}^{\frac {x}{\ln \left (\ln \left (x \right )\right )}}-20\right )\)	\(28\)
parallelrisch	\(-\frac {\ln \left (\ln \left (x \right )-4+\frac {{\mathrm e}^{\frac {x}{\ln \left (\ln \left (x \right )\right )}}}{5}\right ) x -x^{2}-2}{x}\)	\(30\)

Input:

int((((x^2-2)*ln(x)*ln(ln(x))^2-x^2*ln(x)*ln(ln(x))+x^2)*exp(x/ln(ln(x)))+ 
((5*x^2-10)*ln(x)^2+(-20*x^2-5*x+40)*ln(x))*ln(ln(x))^2)/(x^2*ln(x)*ln(ln( 
x))^2*exp(x/ln(ln(x)))+(5*x^2*ln(x)^2-20*x^2*ln(x))*ln(ln(x))^2),x,method= 
_RETURNVERBOSE)
 

Output:

(x^2+2)/x-ln(5*ln(x)+exp(x/ln(ln(x)))-20)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=\frac {x^{2} - x \log \left (e^{\frac {x}{\log \left (\log \left (x\right )\right )}} + 5 \, \log \left (x\right ) - 20\right ) + 2}{x} \] Input:

integrate((((x^2-2)*log(x)*log(log(x))^2-x^2*log(x)*log(log(x))+x^2)*exp(x 
/log(log(x)))+((5*x^2-10)*log(x)^2+(-20*x^2-5*x+40)*log(x))*log(log(x))^2) 
/(x^2*log(x)*log(log(x))^2*exp(x/log(log(x)))+(5*x^2*log(x)^2-20*x^2*log(x 
))*log(log(x))^2),x, algorithm="fricas")
 

Output:

(x^2 - x*log(e^(x/log(log(x))) + 5*log(x) - 20) + 2)/x
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=x - \log {\left (e^{\frac {x}{\log {\left (\log {\left (x \right )} \right )}}} + 5 \log {\left (x \right )} - 20 \right )} + \frac {2}{x} \] Input:

integrate((((x**2-2)*ln(x)*ln(ln(x))**2-x**2*ln(x)*ln(ln(x))+x**2)*exp(x/l 
n(ln(x)))+((5*x**2-10)*ln(x)**2+(-20*x**2-5*x+40)*ln(x))*ln(ln(x))**2)/(x* 
*2*ln(x)*ln(ln(x))**2*exp(x/ln(ln(x)))+(5*x**2*ln(x)**2-20*x**2*ln(x))*ln( 
ln(x))**2),x)
 

Output:

x - log(exp(x/log(log(x))) + 5*log(x) - 20) + 2/x
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=\frac {x^{2} + 2}{x} - \log \left (e^{\frac {x}{\log \left (\log \left (x\right )\right )}} + 5 \, \log \left (x\right ) - 20\right ) \] Input:

integrate((((x^2-2)*log(x)*log(log(x))^2-x^2*log(x)*log(log(x))+x^2)*exp(x 
/log(log(x)))+((5*x^2-10)*log(x)^2+(-20*x^2-5*x+40)*log(x))*log(log(x))^2) 
/(x^2*log(x)*log(log(x))^2*exp(x/log(log(x)))+(5*x^2*log(x)^2-20*x^2*log(x 
))*log(log(x))^2),x, algorithm="maxima")
 

Output:

(x^2 + 2)/x - log(e^(x/log(log(x))) + 5*log(x) - 20)
 

Giac [F]

\[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=\int { \frac {5 \, {\left ({\left (x^{2} - 2\right )} \log \left (x\right )^{2} - {\left (4 \, x^{2} + x - 8\right )} \log \left (x\right )\right )} \log \left (\log \left (x\right )\right )^{2} - {\left (x^{2} \log \left (x\right ) \log \left (\log \left (x\right )\right ) - {\left (x^{2} - 2\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right )^{2} - x^{2}\right )} e^{\frac {x}{\log \left (\log \left (x\right )\right )}}}{x^{2} e^{\frac {x}{\log \left (\log \left (x\right )\right )}} \log \left (x\right ) \log \left (\log \left (x\right )\right )^{2} + 5 \, {\left (x^{2} \log \left (x\right )^{2} - 4 \, x^{2} \log \left (x\right )\right )} \log \left (\log \left (x\right )\right )^{2}} \,d x } \] Input:

integrate((((x^2-2)*log(x)*log(log(x))^2-x^2*log(x)*log(log(x))+x^2)*exp(x 
/log(log(x)))+((5*x^2-10)*log(x)^2+(-20*x^2-5*x+40)*log(x))*log(log(x))^2) 
/(x^2*log(x)*log(log(x))^2*exp(x/log(log(x)))+(5*x^2*log(x)^2-20*x^2*log(x 
))*log(log(x))^2),x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 3.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=x-\ln \left (5\,\ln \left (x\right )+{\mathrm {e}}^{\frac {x}{\ln \left (\ln \left (x\right )\right )}}-20\right )+\frac {2}{x} \] Input:

int(-(exp(x/log(log(x)))*(x^2 + log(log(x))^2*log(x)*(x^2 - 2) - x^2*log(l 
og(x))*log(x)) + log(log(x))^2*(log(x)^2*(5*x^2 - 10) - log(x)*(5*x + 20*x 
^2 - 40)))/(log(log(x))^2*(20*x^2*log(x) - 5*x^2*log(x)^2) - x^2*log(log(x 
))^2*exp(x/log(log(x)))*log(x)),x)
 

Output:

x - log(5*log(x) + exp(x/log(log(x))) - 20) + 2/x
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {\left (\left (40-5 x-20 x^2\right ) \log (x)+\left (-10+5 x^2\right ) \log ^2(x)\right ) \log ^2(\log (x))+e^{\frac {x}{\log (\log (x))}} \left (x^2-x^2 \log (x) \log (\log (x))+\left (-2+x^2\right ) \log (x) \log ^2(\log (x))\right )}{e^{\frac {x}{\log (\log (x))}} x^2 \log (x) \log ^2(\log (x))+\left (-20 x^2 \log (x)+5 x^2 \log ^2(x)\right ) \log ^2(\log (x))} \, dx=\frac {-\mathrm {log}\left (e^{\frac {x}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}}+5 \,\mathrm {log}\left (x \right )-20\right ) x +x^{2}+2}{x} \] Input:

int((((x^2-2)*log(x)*log(log(x))^2-x^2*log(x)*log(log(x))+x^2)*exp(x/log(l 
og(x)))+((5*x^2-10)*log(x)^2+(-20*x^2-5*x+40)*log(x))*log(log(x))^2)/(x^2* 
log(x)*log(log(x))^2*exp(x/log(log(x)))+(5*x^2*log(x)^2-20*x^2*log(x))*log 
(log(x))^2),x)
 

Output:

( - log(e**(x/log(log(x))) + 5*log(x) - 20)*x + x**2 + 2)/x