\(\int \frac {e^{-\frac {x^2}{3}} ((12 e^x-24 x) \log (x)+(-12 e^x+24 x+(48 x+16 x^3+e^x (-12-12 x-8 x^2)) \log (x)) \log (2 x))}{(3 e^{2 x} x^2-12 e^x x^3+12 x^4) \log ^2(x)} \, dx\) [1590]

Optimal result

Integrand size = 93, antiderivative size = 31 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {4 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {4 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \] Input:

Integrate[((12*E^x - 24*x)*Log[x] + (-12*E^x + 24*x + (48*x + 16*x^3 + E^x 
*(-12 - 12*x - 8*x^2))*Log[x])*Log[2*x])/(E^(x^2/3)*(3*E^(2*x)*x^2 - 12*E^ 
x*x^3 + 12*x^4)*Log[x]^2),x]
 

Output:

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

\(\displaystyle \int \frac {e^{-\frac {x^2}{3}} \left (\left (\left (16 x^3+e^x \left (-8 x^2-12 x-12\right )+48 x\right ) \log (x)+24 x-12 e^x\right ) \log (2 x)+\left (12 e^x-24 x\right ) \log (x)\right )}{\left (12 x^4-12 e^x x^3+3 e^{2 x} x^2\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-\frac {x^2}{3}} \left (\left (\left (16 x^3+e^x \left (-8 x^2-12 x-12\right )+48 x\right ) \log (x)+24 x-12 e^x\right ) \log (2 x)+\left (12 e^x-24 x\right ) \log (x)\right )}{3 \left (e^x-2 x\right )^2 x^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {4 e^{-\frac {x^2}{3}} \left (3 \left (e^x-2 x\right ) \log (x)-\left (-6 x+3 e^x-\left (4 x^3+12 x-e^x \left (2 x^2+3 x+3\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (e^x-2 x\right )^2 x^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4}{3} \int \frac {e^{-\frac {x^2}{3}} \left (3 \left (e^x-2 x\right ) \log (x)-\left (-6 x+3 e^x-\left (4 x^3+12 x-e^x \left (2 x^2+3 x+3\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (e^x-2 x\right )^2 x^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[((12*E^x - 24*x)*Log[x] + (-12*E^x + 24*x + (48*x + 16*x^3 + E^x*(-12 
- 12*x - 8*x^2))*Log[x])*Log[2*x])/(E^(x^2/3)*(3*E^(2*x)*x^2 - 12*E^x*x^3 
+ 12*x^4)*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [F]

\[\int \frac {\left (\left (\left (\left (-8 x^{2}-12 x -12\right ) {\mathrm e}^{x}+16 x^{3}+48 x \right ) \ln \left (x \right )-12 \,{\mathrm e}^{x}+24 x \right ) \ln \left (2 x \right )+\left (12 \,{\mathrm e}^{x}-24 x \right ) \ln \left (x \right )\right ) {\mathrm e}^{-\frac {x^{2}}{3}}}{\left (3 \,{\mathrm e}^{2 x} x^{2}-12 \,{\mathrm e}^{x} x^{3}+12 x^{4}\right ) \ln \left (x \right )^{2}}d x\]

Input:

int(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*ln(x)-12*exp(x)+24*x)*ln(2*x)+ 
(12*exp(x)-24*x)*ln(x))/(3*exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2) 
/ln(x)^2,x)
 

Output:

int(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*ln(x)-12*exp(x)+24*x)*ln(2*x)+ 
(12*exp(x)-24*x)*ln(x))/(3*exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x^2) 
/ln(x)^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=-\frac {4 \, {\left (e^{\left (-\frac {1}{3} \, x^{2}\right )} \log \left (2\right ) + e^{\left (-\frac {1}{3} \, x^{2}\right )} \log \left (x\right )\right )}}{{\left (2 \, x^{2} - x e^{x}\right )} \log \left (x\right )} \] Input:

integrate(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*l 
og(2*x)+(12*exp(x)-24*x)*log(x))/(3*exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp 
(1/3*x^2)/log(x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {\left (- 4 \log {\left (x \right )} - 4 \log {\left (2 \right )}\right ) e^{- \frac {x^{2}}{3}}}{2 x^{2} \log {\left (x \right )} - x e^{x} \log {\left (x \right )}} \] Input:

integrate(((((-8*x**2-12*x-12)*exp(x)+16*x**3+48*x)*ln(x)-12*exp(x)+24*x)* 
ln(2*x)+(12*exp(x)-24*x)*ln(x))/(3*exp(x)**2*x**2-12*exp(x)*x**3+12*x**4)/ 
exp(1/3*x**2)/ln(x)**2,x)
 

Output:

(-4*log(x) - 4*log(2))*exp(-x**2/3)/(2*x**2*log(x) - x*exp(x)*log(x))
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=-\frac {4 \, {\left (\log \left (2\right ) + \log \left (x\right )\right )} e^{\left (-\frac {1}{3} \, x^{2}\right )}}{2 \, x^{2} \log \left (x\right ) - x e^{x} \log \left (x\right )} \] Input:

integrate(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*l 
og(2*x)+(12*exp(x)-24*x)*log(x))/(3*exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp 
(1/3*x^2)/log(x)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=-\frac {4 \, {\left (e^{\left (-\frac {1}{3} \, x^{2} + x\right )} \log \left (2\right ) + e^{\left (-\frac {1}{3} \, x^{2} + x\right )} \log \left (x\right )\right )}}{2 \, x^{2} e^{x} \log \left (x\right ) - x e^{\left (2 \, x\right )} \log \left (x\right )} \] Input:

integrate(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*l 
og(2*x)+(12*exp(x)-24*x)*log(x))/(3*exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp 
(1/3*x^2)/log(x)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{-\frac {x^2}{3}}\,\left (\ln \left (2\,x\right )\,\left (24\,x-12\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (48\,x-{\mathrm {e}}^x\,\left (8\,x^2+12\,x+12\right )+16\,x^3\right )\right )-\ln \left (x\right )\,\left (24\,x-12\,{\mathrm {e}}^x\right )\right )}{{\ln \left (x\right )}^2\,\left (3\,x^2\,{\mathrm {e}}^{2\,x}-12\,x^3\,{\mathrm {e}}^x+12\,x^4\right )} \,d x \] Input:

int((exp(-x^2/3)*(log(2*x)*(24*x - 12*exp(x) + log(x)*(48*x - exp(x)*(12*x 
 + 8*x^2 + 12) + 16*x^3)) - log(x)*(24*x - 12*exp(x))))/(log(x)^2*(3*x^2*e 
xp(2*x) - 12*x^3*exp(x) + 12*x^4)),x)
 

Output:

int((exp(-x^2/3)*(log(2*x)*(24*x - 12*exp(x) + log(x)*(48*x - exp(x)*(12*x 
 + 8*x^2 + 12) + 16*x^3)) - log(x)*(24*x - 12*exp(x))))/(log(x)^2*(3*x^2*e 
xp(2*x) - 12*x^3*exp(x) + 12*x^4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {4 \,\mathrm {log}\left (2 x \right )}{e^{\frac {x^{2}}{3}} \mathrm {log}\left (x \right ) x \left (e^{x}-2 x \right )} \] Input:

int(((((-8*x^2-12*x-12)*exp(x)+16*x^3+48*x)*log(x)-12*exp(x)+24*x)*log(2*x 
)+(12*exp(x)-24*x)*log(x))/(3*exp(x)^2*x^2-12*exp(x)*x^3+12*x^4)/exp(1/3*x 
^2)/log(x)^2,x)
 

Output:

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