Integrand size = 119, antiderivative size = 24 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=3 e^{-x} x^2 \log (x) (-x+\log (3+x)+\log (\log (\log (x)))) \]
Time = 5.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 e^{-x} x^2 \log (x) (x-\log (3+x)-\log (\log (\log (x)))) \]
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]
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 |
\(\displaystyle -3 \int e^{-x} x^2 \log (x) \log (\log (\log (x)))dx+3 \int \frac {e^{-x} x}{\log (\log (x))}dx+3 \int e^{-x} x \log (\log (\log (x)))dx+6 \int e^{-x} x \log (x) \log (\log (\log (x)))dx-3 e^{-x} x^3 \log (x)+3 e^{-x} x^2 \log (x) \log (x+3)\) |
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]
3.1.55.3.1 Defintions of rubi rules used
Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50
\[3 x^{2} {\mathrm e}^{-x} \ln \left (x \right ) \ln \left (\ln \left (\ln \left (x \right )\right )\right )-3 x^{2} \ln \left (x \right ) \left (-\ln \left (3+x \right )+x \right ) {\mathrm e}^{-x}\]
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)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 \, x^{3} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x + 3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )\right ) \]
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=\
-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)))
Timed out. \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=\text {Timed out} \]
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)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 \, x^{3} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x + 3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )\right ) \]
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=\
-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)))
Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=-3 \, x^{3} e^{\left (-x\right )} \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x + 3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (\log \left (x\right )\right )\right ) \]
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=\
-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)))
Time = 12.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (9 x+3 x^2+\left (-9 x^2-3 x^3+\left (-24 x^2+3 x^4\right ) \log (x)+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (3+x)\right ) \log (\log (x))+\left (9 x+3 x^2+\left (18 x-3 x^2-3 x^3\right ) \log (x)\right ) \log (\log (x)) \log (\log (\log (x)))\right )}{(3+x) \log (\log (x))} \, dx=3\,x^2\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left (\ln \left (x+3\right )-x+\ln \left (\ln \left (\ln \left (x\right )\right )\right )\right ) \]