\(\int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} (-59 x+20 x^2)+(-10+e^{-2-6 x+x^2} (58 x-20 x^2)) \log (\frac {3}{x})+x \log ^2(\frac {3}{x})+(2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} (x-12 x^2+4 x^3)+(-2 x+e^{-2-6 x+x^2} (-2 x+12 x^2-4 x^3)) \log (\frac {3}{x})+x \log ^2(\frac {3}{x})) \log (x)}{x} \, dx\) [1006]

Optimal result

Integrand size = 166, antiderivative size = 31 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=x \left (e^{-2-6 x+x^2}-\log \left (\frac {3}{x}\right )\right )^2 \left (\frac {5}{x}+\log (x)\right ) \] Output:

x*(ln(x)+5/x)*(exp(x^2-6*x-2)-ln(3/x))^2
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=e^{-4-12 x} \left (e^{x^2}-e^{2+6 x} \log \left (\frac {3}{x}\right )\right )^2 (5+x \log (x)) \] Input:

Integrate[(10*E^(-2 - 6*x + x^2) + E^(-4 - 12*x + 2*x^2)*(-59*x + 20*x^2) 
+ (-10 + E^(-2 - 6*x + x^2)*(58*x - 20*x^2))*Log[3/x] + x*Log[3/x]^2 + (2* 
E^(-2 - 6*x + x^2)*x + E^(-4 - 12*x + 2*x^2)*(x - 12*x^2 + 4*x^3) + (-2*x 
+ E^(-2 - 6*x + x^2)*(-2*x + 12*x^2 - 4*x^3))*Log[3/x] + x*Log[3/x]^2)*Log 
[x])/x,x]
 

Output:

E^(-4 - 12*x)*(E^x^2 - E^(2 + 6*x)*Log[3/x])^2*(5 + 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.

\(\displaystyle \int \frac {10 e^{x^2-6 x-2}+e^{2 x^2-12 x-4} \left (20 x^2-59 x\right )+\left (e^{x^2-6 x-2} \left (58 x-20 x^2\right )-10\right ) \log \left (\frac {3}{x}\right )+\left (2 e^{x^2-6 x-2} x+e^{2 x^2-12 x-4} \left (4 x^3-12 x^2+x\right )+\left (e^{x^2-6 x-2} \left (-4 x^3+12 x^2-2 x\right )-2 x\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)+x \log ^2\left (\frac {3}{x}\right )}{x} \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (e^{2 x^2-12 x-4} \left (4 x^2 \log (x)+20 x-12 x \log (x)+\log (x)-59\right )-\frac {2 e^{x^2-6 x-2} \left (2 x^3 \log \left (\frac {3}{x}\right ) \log (x)+10 x^2 \log \left (\frac {3}{x}\right )-6 x^2 \log \left (\frac {3}{x}\right ) \log (x)-29 x \log \left (\frac {3}{x}\right )+x \log \left (\frac {3}{x}\right ) \log (x)-x \log (x)-5\right )}{x}+\frac {\log \left (\frac {3}{x}\right ) \left (x \log \left (\frac {3}{x}\right )+x \log (x) \log \left (\frac {3}{x}\right )-2 x \log (x)-10\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 \int \frac {e^{2 x^2-12 x-4}}{x}dx+12 \int \frac {\int \frac {e^{x^2-6 x-2}}{x}dx}{x}dx-\frac {12 \int \frac {\int \frac {e^{x^2-6 x+9}}{x}dx}{x}dx}{e^{11}}+6 \log \left (\frac {3}{x}\right ) \int \frac {e^{x^2-6 x-2}}{x}dx-\frac {6 \log \left (\frac {3}{x}\right ) \int \frac {e^{x^2-6 x+9}}{x}dx}{e^{11}}-6 \log (x) \int \frac {e^{x^2-6 x-2}}{x}dx+\frac {6 \log (x) \int \frac {e^{x^2-6 x+9}}{x}dx}{e^{11}}+\frac {3 \int \frac {e^{2 (x-3)^2}}{x}dx}{e^{22}}+5 e^{2 x^2-12 x-4}-10 e^{x^2-6 x-2} \log \left (\frac {3}{x}\right )-2 e^{x^2-6 x-2} x \log (x) \log \left (\frac {3}{x}\right )+e^{2 x^2-12 x-4} x \log (x)+x \log (x) \log ^2\left (\frac {3}{x}\right )+5 \log ^2\left (\frac {3}{x}\right )\)

Input:

Int[(10*E^(-2 - 6*x + x^2) + E^(-4 - 12*x + 2*x^2)*(-59*x + 20*x^2) + (-10 
 + E^(-2 - 6*x + x^2)*(58*x - 20*x^2))*Log[3/x] + x*Log[3/x]^2 + (2*E^(-2 
- 6*x + x^2)*x + E^(-4 - 12*x + 2*x^2)*(x - 12*x^2 + 4*x^3) + (-2*x + E^(- 
2 - 6*x + x^2)*(-2*x + 12*x^2 - 4*x^3))*Log[3/x] + x*Log[3/x]^2)*Log[x])/x 
,x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(111\) vs. \(2(30)=60\).

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 3.61

\[x \ln \left (x \right )^{3}+\left (5-2 x \ln \left (3\right )+2 x \,{\mathrm e}^{x^{2}-6 x -2}\right ) \ln \left (x \right )^{2}+\left (x \ln \left (3\right )^{2}+{\mathrm e}^{2 x^{2}-12 x -4} x -2 \,{\mathrm e}^{x^{2}-6 x -2} \ln \left (3\right ) x +10 \,{\mathrm e}^{x^{2}-6 x -2}\right ) \ln \left (x \right )-10 \ln \left (3\right ) \ln \left (x \right )+5 \,{\mathrm e}^{2 x^{2}-12 x -4}-10 \ln \left (3\right ) {\mathrm e}^{x^{2}-6 x -2}\]

Input:

int(((x*ln(3/x)^2+((-4*x^3+12*x^2-2*x)*exp(x^2-6*x-2)-2*x)*ln(3/x)+(4*x^3- 
12*x^2+x)*exp(x^2-6*x-2)^2+2*x*exp(x^2-6*x-2))*ln(x)+x*ln(3/x)^2+((-20*x^2 
+58*x)*exp(x^2-6*x-2)-10)*ln(3/x)+(20*x^2-59*x)*exp(x^2-6*x-2)^2+10*exp(x^ 
2-6*x-2))/x,x)
 

Output:

x*ln(x)^3+(5-2*x*ln(3)+2*x*exp(x^2-6*x-2))*ln(x)^2+(x*ln(3)^2+exp(2*x^2-12 
*x-4)*x-2*exp(x^2-6*x-2)*ln(3)*x+10*exp(x^2-6*x-2))*ln(x)-10*ln(3)*ln(x)+5 
*exp(2*x^2-12*x-4)-10*ln(3)*exp(x^2-6*x-2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (30) = 60\).

Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.10 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=-x \log \left (\frac {3}{x}\right )^{3} + {\left (2 \, x e^{\left (x^{2} - 6 \, x - 2\right )} + x \log \left (3\right ) + 5\right )} \log \left (\frac {3}{x}\right )^{2} + {\left (x \log \left (3\right ) + 5\right )} e^{\left (2 \, x^{2} - 12 \, x - 4\right )} - {\left (x e^{\left (2 \, x^{2} - 12 \, x - 4\right )} + 2 \, {\left (x \log \left (3\right ) + 5\right )} e^{\left (x^{2} - 6 \, x - 2\right )}\right )} \log \left (\frac {3}{x}\right ) \] Input:

integrate(((x*log(3/x)^2+((-4*x^3+12*x^2-2*x)*exp(x^2-6*x-2)-2*x)*log(3/x) 
+(4*x^3-12*x^2+x)*exp(x^2-6*x-2)^2+2*x*exp(x^2-6*x-2))*log(x)+x*log(3/x)^2 
+((-20*x^2+58*x)*exp(x^2-6*x-2)-10)*log(3/x)+(20*x^2-59*x)*exp(x^2-6*x-2)^ 
2+10*exp(x^2-6*x-2))/x,x, algorithm="fricas")
 

Output:

-x*log(3/x)^3 + (2*x*e^(x^2 - 6*x - 2) + x*log(3) + 5)*log(3/x)^2 + (x*log 
(3) + 5)*e^(2*x^2 - 12*x - 4) - (x*e^(2*x^2 - 12*x - 4) + 2*(x*log(3) + 5) 
*e^(x^2 - 6*x - 2))*log(3/x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (24) = 48\).

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.06 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=x \log {\left (x \right )}^{3} + x \log {\left (3 \right )}^{2} \log {\left (x \right )} + \left (- 2 x \log {\left (3 \right )} + 5\right ) \log {\left (x \right )}^{2} + \left (x \log {\left (x \right )} + 5\right ) e^{2 x^{2} - 12 x - 4} + \left (2 x \log {\left (x \right )}^{2} - 2 x \log {\left (3 \right )} \log {\left (x \right )} + 10 \log {\left (x \right )} - 10 \log {\left (3 \right )}\right ) e^{x^{2} - 6 x - 2} - 10 \log {\left (3 \right )} \log {\left (x \right )} \] Input:

integrate(((x*ln(3/x)**2+((-4*x**3+12*x**2-2*x)*exp(x**2-6*x-2)-2*x)*ln(3/ 
x)+(4*x**3-12*x**2+x)*exp(x**2-6*x-2)**2+2*x*exp(x**2-6*x-2))*ln(x)+x*ln(3 
/x)**2+((-20*x**2+58*x)*exp(x**2-6*x-2)-10)*ln(3/x)+(20*x**2-59*x)*exp(x** 
2-6*x-2)**2+10*exp(x**2-6*x-2))/x,x)
 

Output:

x*log(x)**3 + x*log(3)**2*log(x) + (-2*x*log(3) + 5)*log(x)**2 + (x*log(x) 
 + 5)*exp(2*x**2 - 12*x - 4) + (2*x*log(x)**2 - 2*x*log(3)*log(x) + 10*log 
(x) - 10*log(3))*exp(x**2 - 6*x - 2) - 10*log(3)*log(x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (30) = 60\).

Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 5.00 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=x \log \left (\frac {3}{x}\right )^{2} + {\left ({\left (x \log \left (x\right ) + 5\right )} e^{\left (2 \, x^{2}\right )} + 2 \, {\left (x e^{2} \log \left (x\right )^{2} - 5 \, e^{2} \log \left (3\right ) - {\left (x e^{2} \log \left (3\right ) - 5 \, e^{2}\right )} \log \left (x\right )\right )} e^{\left (x^{2} + 6 \, x\right )} - {\left (x {\left (2 \, \log \left (3\right ) + 1\right )} e^{4} \log \left (x\right )^{2} - x e^{4} \log \left (x\right )^{3} - {\left (\log \left (3\right )^{2} + 2 \, \log \left (3\right ) + 2\right )} x e^{4} \log \left (x\right ) + {\left (\log \left (3\right )^{2} + 2 \, \log \left (3\right ) + 2\right )} x e^{4}\right )} e^{\left (12 \, x\right )}\right )} e^{\left (-12 \, x - 4\right )} + 2 \, x \log \left (\frac {3}{x}\right ) + 5 \, \log \left (\frac {3}{x}\right )^{2} + 2 \, x \] Input:

integrate(((x*log(3/x)^2+((-4*x^3+12*x^2-2*x)*exp(x^2-6*x-2)-2*x)*log(3/x) 
+(4*x^3-12*x^2+x)*exp(x^2-6*x-2)^2+2*x*exp(x^2-6*x-2))*log(x)+x*log(3/x)^2 
+((-20*x^2+58*x)*exp(x^2-6*x-2)-10)*log(3/x)+(20*x^2-59*x)*exp(x^2-6*x-2)^ 
2+10*exp(x^2-6*x-2))/x,x, algorithm="maxima")
 

Output:

x*log(3/x)^2 + ((x*log(x) + 5)*e^(2*x^2) + 2*(x*e^2*log(x)^2 - 5*e^2*log(3 
) - (x*e^2*log(3) - 5*e^2)*log(x))*e^(x^2 + 6*x) - (x*(2*log(3) + 1)*e^4*l 
og(x)^2 - x*e^4*log(x)^3 - (log(3)^2 + 2*log(3) + 2)*x*e^4*log(x) + (log(3 
)^2 + 2*log(3) + 2)*x*e^4)*e^(12*x))*e^(-12*x - 4) + 2*x*log(3/x) + 5*log( 
3/x)^2 + 2*x
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (30) = 60\).

Time = 0.13 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.29 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx={\left (x e^{4} \log \left (3\right )^{2} \log \left (x\right ) - 2 \, x e^{4} \log \left (3\right ) \log \left (x\right )^{2} + x e^{4} \log \left (x\right )^{3} - 2 \, x e^{\left (x^{2} - 6 \, x + 2\right )} \log \left (3\right ) \log \left (x\right ) + 2 \, x e^{\left (x^{2} - 6 \, x + 2\right )} \log \left (x\right )^{2} + x e^{\left (2 \, x^{2} - 12 \, x\right )} \log \left (x\right ) - 10 \, e^{4} \log \left (3\right ) \log \left (x\right ) + 5 \, e^{4} \log \left (x\right )^{2} - 10 \, e^{\left (x^{2} - 6 \, x + 2\right )} \log \left (3\right ) + 10 \, e^{\left (x^{2} - 6 \, x + 2\right )} \log \left (x\right ) + 5 \, e^{\left (2 \, x^{2} - 12 \, x\right )}\right )} e^{\left (-4\right )} \] Input:

integrate(((x*log(3/x)^2+((-4*x^3+12*x^2-2*x)*exp(x^2-6*x-2)-2*x)*log(3/x) 
+(4*x^3-12*x^2+x)*exp(x^2-6*x-2)^2+2*x*exp(x^2-6*x-2))*log(x)+x*log(3/x)^2 
+((-20*x^2+58*x)*exp(x^2-6*x-2)-10)*log(3/x)+(20*x^2-59*x)*exp(x^2-6*x-2)^ 
2+10*exp(x^2-6*x-2))/x,x, algorithm="giac")
 

Output:

(x*e^4*log(3)^2*log(x) - 2*x*e^4*log(3)*log(x)^2 + x*e^4*log(x)^3 - 2*x*e^ 
(x^2 - 6*x + 2)*log(3)*log(x) + 2*x*e^(x^2 - 6*x + 2)*log(x)^2 + x*e^(2*x^ 
2 - 12*x)*log(x) - 10*e^4*log(3)*log(x) + 5*e^4*log(x)^2 - 10*e^(x^2 - 6*x 
 + 2)*log(3) + 10*e^(x^2 - 6*x + 2)*log(x) + 5*e^(2*x^2 - 12*x))*e^(-4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=\int \frac {10\,{\mathrm {e}}^{x^2-6\,x-2}-{\mathrm {e}}^{2\,x^2-12\,x-4}\,\left (59\,x-20\,x^2\right )+\ln \left (\frac {3}{x}\right )\,\left ({\mathrm {e}}^{x^2-6\,x-2}\,\left (58\,x-20\,x^2\right )-10\right )+x\,{\ln \left (\frac {3}{x}\right )}^2+\ln \left (x\right )\,\left (x\,{\ln \left (\frac {3}{x}\right )}^2+\left (-2\,x-{\mathrm {e}}^{x^2-6\,x-2}\,\left (4\,x^3-12\,x^2+2\,x\right )\right )\,\ln \left (\frac {3}{x}\right )+2\,x\,{\mathrm {e}}^{x^2-6\,x-2}+{\mathrm {e}}^{2\,x^2-12\,x-4}\,\left (4\,x^3-12\,x^2+x\right )\right )}{x} \,d x \] Input:

int((10*exp(x^2 - 6*x - 2) - exp(2*x^2 - 12*x - 4)*(59*x - 20*x^2) + log(3 
/x)*(exp(x^2 - 6*x - 2)*(58*x - 20*x^2) - 10) + x*log(3/x)^2 + log(x)*(2*x 
*exp(x^2 - 6*x - 2) + exp(2*x^2 - 12*x - 4)*(x - 12*x^2 + 4*x^3) - log(3/x 
)*(2*x + exp(x^2 - 6*x - 2)*(2*x - 12*x^2 + 4*x^3)) + x*log(3/x)^2))/x,x)
 

Output:

int((10*exp(x^2 - 6*x - 2) - exp(2*x^2 - 12*x - 4)*(59*x - 20*x^2) + log(3 
/x)*(exp(x^2 - 6*x - 2)*(58*x - 20*x^2) - 10) + x*log(3/x)^2 + log(x)*(2*x 
*exp(x^2 - 6*x - 2) + exp(2*x^2 - 12*x - 4)*(x - 12*x^2 + 4*x^3) - log(3/x 
)*(2*x + exp(x^2 - 6*x - 2)*(2*x - 12*x^2 + 4*x^3)) + x*log(3/x)^2))/x, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 3.65 \[ \int \frac {10 e^{-2-6 x+x^2}+e^{-4-12 x+2 x^2} \left (-59 x+20 x^2\right )+\left (-10+e^{-2-6 x+x^2} \left (58 x-20 x^2\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )+\left (2 e^{-2-6 x+x^2} x+e^{-4-12 x+2 x^2} \left (x-12 x^2+4 x^3\right )+\left (-2 x+e^{-2-6 x+x^2} \left (-2 x+12 x^2-4 x^3\right )\right ) \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )\right ) \log (x)}{x} \, dx=\frac {e^{2 x^{2}} \mathrm {log}\left (x \right ) x +5 e^{2 x^{2}}-2 e^{x^{2}+6 x} \mathrm {log}\left (\frac {3}{x}\right ) \mathrm {log}\left (x \right ) e^{2} x -10 e^{x^{2}+6 x} \mathrm {log}\left (\frac {3}{x}\right ) e^{2}+e^{12 x} \mathrm {log}\left (\frac {3}{x}\right )^{2} \mathrm {log}\left (x \right ) e^{4} x +5 e^{12 x} \mathrm {log}\left (\frac {3}{x}\right )^{2} e^{4}}{e^{12 x} e^{4}} \] Input:

int(((x*log(3/x)^2+((-4*x^3+12*x^2-2*x)*exp(x^2-6*x-2)-2*x)*log(3/x)+(4*x^ 
3-12*x^2+x)*exp(x^2-6*x-2)^2+2*x*exp(x^2-6*x-2))*log(x)+x*log(3/x)^2+((-20 
*x^2+58*x)*exp(x^2-6*x-2)-10)*log(3/x)+(20*x^2-59*x)*exp(x^2-6*x-2)^2+10*e 
xp(x^2-6*x-2))/x,x)
 

Output:

(e**(2*x**2)*log(x)*x + 5*e**(2*x**2) - 2*e**(x**2 + 6*x)*log(3/x)*log(x)* 
e**2*x - 10*e**(x**2 + 6*x)*log(3/x)*e**2 + e**(12*x)*log(3/x)**2*log(x)*e 
**4*x + 5*e**(12*x)*log(3/x)**2*e**4)/(e**(12*x)*e**4)