\(\int \frac {e^{-x} (9 x+3 x^2+(-9 x^2-3 x^3+(-24 x^2+3 x^4) \log (x)+(9 x+3 x^2+(18 x-3 x^2-3 x^3) \log (x)) \log (3+x)) \log (\log (x))+(9 x+3 x^2+(18 x-3 x^2-3 x^3) \log (x)) \log (\log (x)) \log (\log (\log (x))))}{(3+x) \log (\log (x))} \, dx\) [55]

Optimal result

Integrand size = 119, antiderivative size = 24 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=3 e^{-x} x^2 \log (x) (-x+\log (3+x)+\log (\log (\log (x)))) \] Output:

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

Mathematica [A] (verified)

Time = 5.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 e^{-x} x^2 \log (x) (x-\log (3+x)-\log (\log (\log (x)))) \] Input:

Integrate[(9*x + 3*x^2 + (-9*x^2 - 3*x^3 + (-24*x^2 + 3*x^4)*Log[x] + (9*x 
 + 3*x^2 + (18*x - 3*x^2 - 3*x^3)*Log[x])*Log[3 + x])*Log[Log[x]] + (9*x + 
 3*x^2 + (18*x - 3*x^2 - 3*x^3)*Log[x])*Log[Log[x]]*Log[Log[Log[x]]])/(E^x 
*(3 + x)*Log[Log[x]]),x]
 

Output:

(-3*x^2*Log[x]*(x - Log[3 + x] - Log[Log[Log[x]]]))/E^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.

\(\displaystyle \int \frac {e^{-x} \left (3 x^2+\left (3 x^2+\left (-3 x^3-3 x^2+18 x\right ) \log (x)+9 x\right ) \log (\log (x)) \log (\log (\log (x)))+\left (-3 x^3-9 x^2+\left (3 x^4-24 x^2\right ) \log (x)+\left (3 x^2+\left (-3 x^3-3 x^2+18 x\right ) \log (x)+9 x\right ) \log (x+3)\right ) \log (\log (x))+9 x\right )}{(x+3) \log (\log (x))} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {3 e^{-x} x \left (x^3 \log (x) \log (\log (x))-x^2 \log (x) \log (x+3) \log (\log (x))-x^2 \log (\log (x))+x-8 x \log (x) \log (\log (x))-x \log (x) \log (x+3) \log (\log (x))+x \log (x+3) \log (\log (x))-3 x \log (\log (x))+6 \log (x) \log (x+3) \log (\log (x))+3 \log (x+3) \log (\log (x))+3\right )}{(x+3) \log (\log (x))}-3 e^{-x} x (x \log (x)-2 \log (x)-1) \log (\log (\log (x)))\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 \int e^{-x} x^2 \log (x) \log (\log (\log (x)))dx+3 \int \frac {e^{-x} x}{\log (\log (x))}dx+3 \int e^{-x} x \log (\log (\log (x)))dx+6 \int e^{-x} x \log (x) \log (\log (\log (x)))dx-3 e^{-x} x^3 \log (x)+3 e^{-x} x^2 \log (x) \log (x+3)\)

Input:

Int[(9*x + 3*x^2 + (-9*x^2 - 3*x^3 + (-24*x^2 + 3*x^4)*Log[x] + (9*x + 3*x 
^2 + (18*x - 3*x^2 - 3*x^3)*Log[x])*Log[3 + x])*Log[Log[x]] + (9*x + 3*x^2 
 + (18*x - 3*x^2 - 3*x^3)*Log[x])*Log[Log[x]]*Log[Log[Log[x]]])/(E^x*(3 + 
x)*Log[Log[x]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 193.85 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50

method	result	size
risch	\(3 x^{2} {\mathrm e}^{-x} \ln \left (x \right ) \ln \left (\ln \left (\ln \left (x \right )\right )\right )-3 x^{2} \ln \left (x \right ) \left (x -\ln \left (3+x \right )\right ) {\mathrm e}^{-x}\)	\(36\)
parallelrisch	\(-\frac {\left (18 x^{3} \ln \left (x \right )-18 \ln \left (3+x \right ) \ln \left (x \right ) x^{2}-18 \ln \left (\ln \left (\ln \left (x \right )\right )\right ) x^{2} \ln \left (x \right )\right ) {\mathrm e}^{-x}}{6}\)	\(37\)

Input:

int((((-3*x^3-3*x^2+18*x)*ln(x)+3*x^2+9*x)*ln(ln(x))*ln(ln(ln(x)))+(((-3*x 
^3-3*x^2+18*x)*ln(x)+3*x^2+9*x)*ln(3+x)+(3*x^4-24*x^2)*ln(x)-3*x^3-9*x^2)* 
ln(ln(x))+3*x^2+9*x)/(3+x)/exp(x)/ln(ln(x)),x,method=_RETURNVERBOSE)
 

Output:

3*x^2*exp(-x)*ln(x)*ln(ln(ln(x)))-3*x^2*ln(x)*(x-ln(3+x))*exp(-x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 \, x^{3} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x + 3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:

integrate((((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(log(x))*log(log(log( 
x)))+(((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(3+x)+(3*x^4-24*x^2)*log(x 
)-3*x^3-9*x^2)*log(log(x))+3*x^2+9*x)/(3+x)/exp(x)/log(log(x)),x, algorith 
m="fricas")
 

Output:

-3*x^3*e^(-x)*log(x) + 3*x^2*e^(-x)*log(x + 3)*log(x) + 3*x^2*e^(-x)*log(x 
)*log(log(log(x)))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=\text {Timed out} \] Input:

integrate((((-3*x**3-3*x**2+18*x)*ln(x)+3*x**2+9*x)*ln(ln(x))*ln(ln(ln(x)) 
)+(((-3*x**3-3*x**2+18*x)*ln(x)+3*x**2+9*x)*ln(3+x)+(3*x**4-24*x**2)*ln(x) 
-3*x**3-9*x**2)*ln(ln(x))+3*x**2+9*x)/(3+x)/exp(x)/ln(ln(x)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 \, x^{3} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x + 3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:

integrate((((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(log(x))*log(log(log( 
x)))+(((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(3+x)+(3*x^4-24*x^2)*log(x 
)-3*x^3-9*x^2)*log(log(x))+3*x^2+9*x)/(3+x)/exp(x)/log(log(x)),x, algorith 
m="maxima")
 

Output:

-3*x^3*e^(-x)*log(x) + 3*x^2*e^(-x)*log(x + 3)*log(x) + 3*x^2*e^(-x)*log(x 
)*log(log(log(x)))
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 \, x^{3} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x + 3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:

integrate((((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(log(x))*log(log(log( 
x)))+(((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(3+x)+(3*x^4-24*x^2)*log(x 
)-3*x^3-9*x^2)*log(log(x))+3*x^2+9*x)/(3+x)/exp(x)/log(log(x)),x, algorith 
m="giac")
 

Output:

-3*x^3*e^(-x)*log(x) + 3*x^2*e^(-x)*log(x + 3)*log(x) + 3*x^2*e^(-x)*log(x 
)*log(log(log(x)))
 

Mupad [B] (verification not implemented)

Time = 3.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=3\,x^2\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left (\ln \left (x+3\right )-x+\ln \left (\ln \left (\ln \left (x\right )\right )\right )\right ) \] Input:

int((exp(-x)*(9*x - log(log(x))*(log(x)*(24*x^2 - 3*x^4) + 9*x^2 + 3*x^3 - 
 log(x + 3)*(9*x + 3*x^2 - log(x)*(3*x^2 - 18*x + 3*x^3))) + 3*x^2 + log(l 
og(x))*log(log(log(x)))*(9*x + 3*x^2 - log(x)*(3*x^2 - 18*x + 3*x^3))))/(l 
og(log(x))*(x + 3)),x)
 

Output:

3*x^2*exp(-x)*log(x)*(log(x + 3) - x + log(log(log(x))))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=\frac {3 \,\mathrm {log}\left (x \right ) x^{2} \left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )\right )+\mathrm {log}\left (x +3\right )-x \right )}{e^{x}} \] Input:

int((((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(log(x))*log(log(log(x)))+( 
((-3*x^3-3*x^2+18*x)*log(x)+3*x^2+9*x)*log(3+x)+(3*x^4-24*x^2)*log(x)-3*x^ 
3-9*x^2)*log(log(x))+3*x^2+9*x)/(3+x)/exp(x)/log(log(x)),x)
 

Output:

(3*log(x)*x**2*(log(log(log(x))) + log(x + 3) - x))/e**x