\(\int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} (8+5 e^3-x+(-8-5 e^3+2 x) \log (x) \log (\log (x)))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} (8 x^2+5 e^3 x^2-x^3+(-8 x^2-5 e^3 x^2+2 x^3) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x)))}{\log (x) \log ^2(\log (x))} \, dx\) [473]

Optimal result

Integrand size = 160, antiderivative size = 32 \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=3+\left (e^{-5+\frac {x \left (-4+\frac {1}{2} \left (-5 e^3+x\right )\right )}{\log (\log (x))}}+x^2\right )^2 \] Output:

(x^2+exp(x/ln(ln(x))*(1/2*x-5/2*exp(3)-4)-5))^2+3
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.94 \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=e^{-10-\frac {\left (8+5 e^3\right ) x}{\log (\log (x))}} \left (e^{\frac {x^2}{2 \log (\log (x))}}+e^{5+\frac {\left (8+5 e^3\right ) x}{2 \log (\log (x))}} x^2\right )^2 \] Input:

Integrate[(4*x^3*Log[x]*Log[Log[x]]^2 + E^((-8*x - 5*E^3*x + x^2 - 10*Log[ 
Log[x]])/Log[Log[x]])*(8 + 5*E^3 - x + (-8 - 5*E^3 + 2*x)*Log[x]*Log[Log[x 
]]) + E^((-8*x - 5*E^3*x + x^2 - 10*Log[Log[x]])/(2*Log[Log[x]]))*(8*x^2 + 
 5*E^3*x^2 - x^3 + (-8*x^2 - 5*E^3*x^2 + 2*x^3)*Log[x]*Log[Log[x]] + 4*x*L 
og[x]*Log[Log[x]]^2))/(Log[x]*Log[Log[x]]^2),x]
 

Output:

E^(-10 - ((8 + 5*E^3)*x)/Log[Log[x]])*(E^(x^2/(2*Log[Log[x]])) + E^(5 + (( 
8 + 5*E^3)*x)/(2*Log[Log[x]]))*x^2)^2
 

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.

\(\displaystyle \int \frac {\left (-x^3+5 e^3 x^2+8 x^2+\left (2 x^3-5 e^3 x^2-8 x^2\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right ) \exp \left (\frac {x^2-5 e^3 x-8 x-10 \log (\log (x))}{2 \log (\log (x))}\right )+4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {x^2-5 e^3 x-8 x-10 \log (\log (x))}{\log (\log (x))}} \left (-x+\left (2 x-5 e^3-8\right ) \log (x) \log (\log (x))+5 e^3+8\right )}{\log (x) \log ^2(\log (x))} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (4 x^3+\frac {x e^{\frac {x \left (x-5 e^3-8\right )}{2 \log (\log (x))}-5} \left (-x^2+2 x^2 \log (x) \log (\log (x))+8 \left (1+\frac {5 e^3}{8}\right ) x+4 \log (x) \log ^2(\log (x))-8 \left (1+\frac {5 e^3}{8}\right ) x \log (x) \log (\log (x))\right )}{\log (x) \log ^2(\log (x))}+\frac {e^{\frac {x \left (x-5 e^3-8\right )}{\log (\log (x))}-10} \left (-x+2 x \log (x) \log (\log (x))-8 \left (1+\frac {5 e^3}{8}\right ) \log (x) \log (\log (x))+8 \left (1+\frac {5 e^3}{8}\right )\right )}{\log (x) \log ^2(\log (x))}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \left (8+5 e^3\right ) \int \frac {e^{\frac {x \left (x-5 e^3-8\right )}{\log (\log (x))}-10}}{\log (x) \log ^2(\log (x))}dx-\int \frac {e^{\frac {x \left (x-5 e^3-8\right )}{\log (\log (x))}-10} x}{\log (x) \log ^2(\log (x))}dx-\left (8+5 e^3\right ) \int \frac {e^{\frac {x \left (x-5 e^3-8\right )}{\log (\log (x))}-10}}{\log (\log (x))}dx+2 \int \frac {e^{\frac {x \left (x-5 e^3-8\right )}{\log (\log (x))}-10} x}{\log (\log (x))}dx+\frac {2 x \left (-x^2+2 x^2 \log (x) \log (\log (x))+\left (8+5 e^3\right ) x-\left (8+5 e^3\right ) x \log (x) \log (\log (x))\right ) \exp \left (-\frac {\left (-x+5 e^3+8\right ) x}{2 \log (\log (x))}-5\right )}{\log (x) \left (\frac {-x+5 e^3+8}{\log (x) \log ^2(\log (x))}-\frac {-x+5 e^3+8}{\log (\log (x))}+\frac {x}{\log (\log (x))}\right ) \log ^2(\log (x))}+x^4\)

Input:

Int[(4*x^3*Log[x]*Log[Log[x]]^2 + E^((-8*x - 5*E^3*x + x^2 - 10*Log[Log[x] 
])/Log[Log[x]])*(8 + 5*E^3 - x + (-8 - 5*E^3 + 2*x)*Log[x]*Log[Log[x]]) + 
E^((-8*x - 5*E^3*x + x^2 - 10*Log[Log[x]])/(2*Log[Log[x]]))*(8*x^2 + 5*E^3 
*x^2 - x^3 + (-8*x^2 - 5*E^3*x^2 + 2*x^3)*Log[x]*Log[Log[x]] + 4*x*Log[x]* 
Log[Log[x]]^2))/(Log[x]*Log[Log[x]]^2),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(63\) vs. \(2(27)=54\).

Time = 33.74 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00

method	result	size
risch	\(x^{4}+2 x^{2} {\mathrm e}^{-\frac {5 x \,{\mathrm e}^{3}-x^{2}+10 \ln \left (\ln \left (x \right )\right )+8 x}{2 \ln \left (\ln \left (x \right )\right )}}+{\mathrm e}^{-\frac {5 x \,{\mathrm e}^{3}-x^{2}+10 \ln \left (\ln \left (x \right )\right )+8 x}{\ln \left (\ln \left (x \right )\right )}}\)	\(64\)
parallelrisch	\(x^{4}+2 x^{2} {\mathrm e}^{-\frac {5 x \,{\mathrm e}^{3}-x^{2}+10 \ln \left (\ln \left (x \right )\right )+8 x}{2 \ln \left (\ln \left (x \right )\right )}}+{\mathrm e}^{-\frac {5 x \,{\mathrm e}^{3}-x^{2}+10 \ln \left (\ln \left (x \right )\right )+8 x}{\ln \left (\ln \left (x \right )\right )}}\)	\(66\)

Input:

int((((-5*exp(3)+2*x-8)*ln(x)*ln(ln(x))+5*exp(3)+8-x)*exp(1/2*(-10*ln(ln(x 
))-5*x*exp(3)+x^2-8*x)/ln(ln(x)))^2+(4*x*ln(x)*ln(ln(x))^2+(-5*x^2*exp(3)+ 
2*x^3-8*x^2)*ln(x)*ln(ln(x))+5*x^2*exp(3)-x^3+8*x^2)*exp(1/2*(-10*ln(ln(x) 
)-5*x*exp(3)+x^2-8*x)/ln(ln(x)))+4*x^3*ln(x)*ln(ln(x))^2)/ln(x)/ln(ln(x))^ 
2,x,method=_RETURNVERBOSE)
 

Output:

x^4+2*x^2*exp(-1/2*(5*x*exp(3)-x^2+10*ln(ln(x))+8*x)/ln(ln(x)))+exp(-(5*x* 
exp(3)-x^2+10*ln(ln(x))+8*x)/ln(ln(x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=x^{4} + 2 \, x^{2} e^{\left (\frac {x^{2} - 5 \, x e^{3} - 8 \, x - 10 \, \log \left (\log \left (x\right )\right )}{2 \, \log \left (\log \left (x\right )\right )}\right )} + e^{\left (\frac {x^{2} - 5 \, x e^{3} - 8 \, x - 10 \, \log \left (\log \left (x\right )\right )}{\log \left (\log \left (x\right )\right )}\right )} \] Input:

integrate((((-5*exp(3)+2*x-8)*log(x)*log(log(x))+5*exp(3)+8-x)*exp(1/2*(-1 
0*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))^2+(4*x*log(x)*log(log(x))^2 
+(-5*x^2*exp(3)+2*x^3-8*x^2)*log(x)*log(log(x))+5*x^2*exp(3)-x^3+8*x^2)*ex 
p(1/2*(-10*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))+4*x^3*log(x)*log(l 
og(x))^2)/log(x)/log(log(x))^2,x, algorithm="fricas")
 

Output:

x^4 + 2*x^2*e^(1/2*(x^2 - 5*x*e^3 - 8*x - 10*log(log(x)))/log(log(x))) + e 
^((x^2 - 5*x*e^3 - 8*x - 10*log(log(x)))/log(log(x)))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((-5*exp(3)+2*x-8)*ln(x)*ln(ln(x))+5*exp(3)+8-x)*exp(1/2*(-10*l 
n(ln(x))-5*x*exp(3)+x**2-8*x)/ln(ln(x)))**2+(4*x*ln(x)*ln(ln(x))**2+(-5*x* 
*2*exp(3)+2*x**3-8*x**2)*ln(x)*ln(ln(x))+5*x**2*exp(3)-x**3+8*x**2)*exp(1/ 
2*(-10*ln(ln(x))-5*x*exp(3)+x**2-8*x)/ln(ln(x)))+4*x**3*ln(x)*ln(ln(x))**2 
)/ln(x)/ln(ln(x))**2,x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((((-5*exp(3)+2*x-8)*log(x)*log(log(x))+5*exp(3)+8-x)*exp(1/2*(-1 
0*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))^2+(4*x*log(x)*log(log(x))^2 
+(-5*x^2*exp(3)+2*x^3-8*x^2)*log(x)*log(log(x))+5*x^2*exp(3)-x^3+8*x^2)*ex 
p(1/2*(-10*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))+4*x^3*log(x)*log(l 
og(x))^2)/log(x)/log(log(x))^2,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [F]

\[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=\int { \frac {4 \, x^{3} \log \left (x\right ) \log \left (\log \left (x\right )\right )^{2} + {\left ({\left (2 \, x - 5 \, e^{3} - 8\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right ) - x + 5 \, e^{3} + 8\right )} e^{\left (\frac {x^{2} - 5 \, x e^{3} - 8 \, x - 10 \, \log \left (\log \left (x\right )\right )}{\log \left (\log \left (x\right )\right )}\right )} + {\left (4 \, x \log \left (x\right ) \log \left (\log \left (x\right )\right )^{2} - x^{3} + 5 \, x^{2} e^{3} + {\left (2 \, x^{3} - 5 \, x^{2} e^{3} - 8 \, x^{2}\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right ) + 8 \, x^{2}\right )} e^{\left (\frac {x^{2} - 5 \, x e^{3} - 8 \, x - 10 \, \log \left (\log \left (x\right )\right )}{2 \, \log \left (\log \left (x\right )\right )}\right )}}{\log \left (x\right ) \log \left (\log \left (x\right )\right )^{2}} \,d x } \] Input:

integrate((((-5*exp(3)+2*x-8)*log(x)*log(log(x))+5*exp(3)+8-x)*exp(1/2*(-1 
0*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))^2+(4*x*log(x)*log(log(x))^2 
+(-5*x^2*exp(3)+2*x^3-8*x^2)*log(x)*log(log(x))+5*x^2*exp(3)-x^3+8*x^2)*ex 
p(1/2*(-10*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))+4*x^3*log(x)*log(l 
og(x))^2)/log(x)/log(log(x))^2,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.34 \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=x^4+{\mathrm {e}}^{\frac {x^2}{\ln \left (\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^3}{\ln \left (\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-10}\,{\mathrm {e}}^{-\frac {8\,x}{\ln \left (\ln \left (x\right )\right )}}+2\,x^2\,{\mathrm {e}}^{\frac {x^2}{2\,\ln \left (\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^3}{2\,\ln \left (\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-\frac {4\,x}{\ln \left (\ln \left (x\right )\right )}} \] Input:

int((exp(-(4*x + 5*log(log(x)) + (5*x*exp(3))/2 - x^2/2)/log(log(x)))*(5*x 
^2*exp(3) + 8*x^2 - x^3 + 4*x*log(log(x))^2*log(x) - log(log(x))*log(x)*(5 
*x^2*exp(3) + 8*x^2 - 2*x^3)) - exp(-(2*(4*x + 5*log(log(x)) + (5*x*exp(3) 
)/2 - x^2/2))/log(log(x)))*(x - 5*exp(3) + log(log(x))*log(x)*(5*exp(3) - 
2*x + 8) - 8) + 4*x^3*log(log(x))^2*log(x))/(log(log(x))^2*log(x)),x)
 

Output:

x^4 + exp(x^2/log(log(x)))*exp(-(5*x*exp(3))/log(log(x)))*exp(-10)*exp(-(8 
*x)/log(log(x))) + 2*x^2*exp(x^2/(2*log(log(x))))*exp(-(5*x*exp(3))/(2*log 
(log(x))))*exp(-5)*exp(-(4*x)/log(log(x)))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.84 \[ \int \frac {4 x^3 \log (x) \log ^2(\log (x))+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}} \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+e^{\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}} \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx=\frac {e^{\frac {x^{2}}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}}+2 e^{\frac {5 e^{3} x +x^{2}+8 x}{2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}} e^{5} x^{2}+e^{\frac {5 e^{3} x +8 x}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}} e^{10} x^{4}}{e^{\frac {5 e^{3} x +8 x}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}} e^{10}} \] Input:

int((((-5*exp(3)+2*x-8)*log(x)*log(log(x))+5*exp(3)+8-x)*exp(1/2*(-10*log( 
log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))^2+(4*x*log(x)*log(log(x))^2+(-5*x 
^2*exp(3)+2*x^3-8*x^2)*log(x)*log(log(x))+5*x^2*exp(3)-x^3+8*x^2)*exp(1/2* 
(-10*log(log(x))-5*x*exp(3)+x^2-8*x)/log(log(x)))+4*x^3*log(x)*log(log(x)) 
^2)/log(x)/log(log(x))^2,x)
 

Output:

(e**(x**2/log(log(x))) + 2*e**((5*e**3*x + x**2 + 8*x)/(2*log(log(x))))*e* 
*5*x**2 + e**((5*e**3*x + 8*x)/log(log(x)))*e**10*x**4)/(e**((5*e**3*x + 8 
*x)/log(log(x)))*e**10)