\(\int \frac {-5 x^2+e^x (5 x-x^2+x^3)+(5 x^2-5 e^x x^2) \log (\frac {x}{3})+(-12 x+4 x^2+(60-20 x) \log (\frac {x}{3})) \log (3 x^2)+(-15+11 x-x^2-15 \log (\frac {x}{3})) \log ^2(3 x^2)}{e^{2 x} x^2-2 e^x x^3+x^4+(6 x^2-2 x^3+e^x (-6 x+2 x^2)) \log ^2(3 x^2)+(9-6 x+x^2) \log ^4(3 x^2)} \, dx\) [2040]

Optimal result

Integrand size = 172, antiderivative size = 40 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=\frac {-x+5 \log \left (\frac {x}{3}\right )}{e^x-x-\frac {(3-x) \log ^2\left (3 x^2\right )}{x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=\frac {x \left (x-5 \log \left (\frac {x}{3}\right )\right )}{x \left (-e^x+x\right )-(-3+x) \log ^2\left (3 x^2\right )} \] Input:

Integrate[(-5*x^2 + E^x*(5*x - x^2 + x^3) + (5*x^2 - 5*E^x*x^2)*Log[x/3] + 
 (-12*x + 4*x^2 + (60 - 20*x)*Log[x/3])*Log[3*x^2] + (-15 + 11*x - x^2 - 1 
5*Log[x/3])*Log[3*x^2]^2)/(E^(2*x)*x^2 - 2*E^x*x^3 + x^4 + (6*x^2 - 2*x^3 
+ E^x*(-6*x + 2*x^2))*Log[3*x^2]^2 + (9 - 6*x + x^2)*Log[3*x^2]^4),x]
 

Output:

(x*(x - 5*Log[x/3]))/(x*(-E^x + x) - (-3 + x)*Log[3*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.

\(\displaystyle \int \frac {-5 x^2+\left (-x^2+11 x-15 \log \left (\frac {x}{3}\right )-15\right ) \log ^2\left (3 x^2\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (4 x^2-12 x+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+e^x \left (x^3-x^2+5 x\right )}{x^4-2 e^x x^3+e^{2 x} x^2+\left (x^2-6 x+9\right ) \log ^4\left (3 x^2\right )+\left (-2 x^3+6 x^2+e^x \left (2 x^2-6 x\right )\right ) \log ^2\left (3 x^2\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-5 x^2+\left (-x^2+11 x-15 \log \left (\frac {x}{3}\right )-15\right ) \log ^2\left (3 x^2\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (4 x^2-12 x+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+e^x \left (x^3-x^2+5 x\right )}{\left (-x^2+x \log ^2\left (3 x^2\right )-3 \log ^2\left (3 x^2\right )+e^x x\right )^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7299

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

Input:

Int[(-5*x^2 + E^x*(5*x - x^2 + x^3) + (5*x^2 - 5*E^x*x^2)*Log[x/3] + (-12* 
x + 4*x^2 + (60 - 20*x)*Log[x/3])*Log[3*x^2] + (-15 + 11*x - x^2 - 15*Log[ 
x/3])*Log[3*x^2]^2)/(E^(2*x)*x^2 - 2*E^x*x^3 + x^4 + (6*x^2 - 2*x^3 + E^x* 
(-6*x + 2*x^2))*Log[3*x^2]^2 + (9 - 6*x + x^2)*Log[3*x^2]^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 87.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(\frac {-120 x \ln \left (\frac {x}{3}\right )+24 x^{2}}{-24 x \ln \left (3 x^{2}\right )^{2}+24 x^{2}-24 \,{\mathrm e}^{x} x +72 \ln \left (3 x^{2}\right )^{2}}\)	\(48\)
risch	\(\frac {-20 x \ln \left (x \right )+20 x \ln \left (3\right )+4 x^{2}}{-3 \pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}-4 x \ln \left (3\right )^{2}+48 \ln \left (3\right ) \ln \left (x \right )-4 \,{\mathrm e}^{x} x -16 x \ln \left (3\right ) \ln \left (x \right )-16 x \ln \left (x \right )^{2}+12 \ln \left (3\right )^{2}+48 \ln \left (x \right )^{2}+4 x^{2}+48 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+4 i x \ln \left (3\right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-12 i \ln \left (3\right ) \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+24 i \ln \left (3\right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+8 i x \ln \left (x \right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-24 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-4 x \,\pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}+x \,\pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}-4 x \,\pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}+6 x \,\pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}-24 i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-12 i \ln \left (3\right ) \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i x \ln \left (3\right ) \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-8 i x \ln \left (3\right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+8 i x \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-3 \pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}+12 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}+x \,\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}-18 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}+12 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-16 i x \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}}\)	\(492\)

Input:

int(((-15*ln(1/3*x)-x^2+11*x-15)*ln(3*x^2)^2+((-20*x+60)*ln(1/3*x)+4*x^2-1 
2*x)*ln(3*x^2)+(-5*exp(x)*x^2+5*x^2)*ln(1/3*x)+(x^3-x^2+5*x)*exp(x)-5*x^2) 
/((x^2-6*x+9)*ln(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*ln(3*x^2)^2+exp 
(x)^2*x^2-2*exp(x)*x^3+x^4),x,method=_RETURNVERBOSE)
 

Output:

1/24*(-120*x*ln(1/3*x)+24*x^2)/(-x*ln(3*x^2)^2+x^2-exp(x)*x+3*ln(3*x^2)^2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.40 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=-\frac {x^{2} - 5 \, x \log \left (\frac {1}{3} \, x\right )}{9 \, {\left (x - 3\right )} \log \left (3\right )^{2} + 12 \, {\left (x - 3\right )} \log \left (3\right ) \log \left (\frac {1}{3} \, x\right ) + 4 \, {\left (x - 3\right )} \log \left (\frac {1}{3} \, x\right )^{2} - x^{2} + x e^{x}} \] Input:

integrate(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x 
)+4*x^2-12*x)*log(3*x^2)+(-5*exp(x)*x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*ex 
p(x)-5*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*log 
(3*x^2)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x, algorithm="fricas")
 

Output:

-(x^2 - 5*x*log(1/3*x))/(9*(x - 3)*log(3)^2 + 12*(x - 3)*log(3)*log(1/3*x) 
 + 4*(x - 3)*log(1/3*x)^2 - x^2 + x*e^x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (26) = 52\).

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.88 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=\frac {- x^{2} + 5 x \log {\left (\frac {x}{3} \right )}}{- x^{2} + x e^{x} + 4 x \log {\left (\frac {x}{3} \right )}^{2} + 12 x \log {\left (3 \right )} \log {\left (\frac {x}{3} \right )} + 9 x \log {\left (3 \right )}^{2} - 12 \log {\left (\frac {x}{3} \right )}^{2} - 36 \log {\left (3 \right )} \log {\left (\frac {x}{3} \right )} - 27 \log {\left (3 \right )}^{2}} \] Input:

integrate(((-15*ln(1/3*x)-x**2+11*x-15)*ln(3*x**2)**2+((-20*x+60)*ln(1/3*x 
)+4*x**2-12*x)*ln(3*x**2)+(-5*exp(x)*x**2+5*x**2)*ln(1/3*x)+(x**3-x**2+5*x 
)*exp(x)-5*x**2)/((x**2-6*x+9)*ln(3*x**2)**4+((2*x**2-6*x)*exp(x)-2*x**3+6 
*x**2)*ln(3*x**2)**2+exp(x)**2*x**2-2*exp(x)*x**3+x**4),x)
 

Output:

(-x**2 + 5*x*log(x/3))/(-x**2 + x*exp(x) + 4*x*log(x/3)**2 + 12*x*log(3)*l 
og(x/3) + 9*x*log(3)**2 - 12*log(x/3)**2 - 36*log(3)*log(x/3) - 27*log(3)* 
*2)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.55 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=-\frac {x^{2} + 5 \, x \log \left (3\right ) - 5 \, x \log \left (x\right )}{x \log \left (3\right )^{2} + 4 \, {\left (x - 3\right )} \log \left (x\right )^{2} - x^{2} + x e^{x} - 3 \, \log \left (3\right )^{2} + 4 \, {\left (x \log \left (3\right ) - 3 \, \log \left (3\right )\right )} \log \left (x\right )} \] Input:

integrate(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x 
)+4*x^2-12*x)*log(3*x^2)+(-5*exp(x)*x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*ex 
p(x)-5*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*log 
(3*x^2)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1267 vs. \(2 (33) = 66\).

Time = 147.04 (sec) , antiderivative size = 1267, normalized size of antiderivative = 31.68 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x 
)+4*x^2-12*x)*log(3*x^2)+(-5*exp(x)*x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*ex 
p(x)-5*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*log 
(3*x^2)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x, algorithm="giac")
 

Output:

-(2*x^7*e^(2*x) - 3*x^7*e^x + 10*x^6*e^(2*x)*log(3) - 19*x^6*e^x*log(3) - 
30*x^5*e^x*log(3)^2 - 10*x^6*e^(2*x)*log(x) + 7*x^6*e^x*log(x) - 60*x^5*e^ 
x*log(3)*log(x) + x^7 - 13*x^6*e^(2*x) + 38*x^6*e^x + 5*x^6*log(3) - 60*x^ 
5*e^(2*x)*log(3) + 175*x^5*e^x*log(3) + 270*x^4*e^x*log(3)^2 - 5*x^6*log(x 
) + 60*x^5*e^(2*x)*log(x) - 55*x^5*e^x*log(x) + 540*x^4*e^x*log(3)*log(x) 
- 28*x^6 + 10*x^5*e^(2*x) - 106*x^5*e^x - 140*x^5*log(3) + 45*x^4*e^(2*x)* 
log(3) - 324*x^4*e^x*log(3) - 810*x^3*e^x*log(3)^2 + 140*x^5*log(x) - 45*x 
^4*e^(2*x)*log(x) - 168*x^4*e^x*log(x) - 1620*x^3*e^x*log(3)*log(x) + 180* 
x^5 + 51*x^4*e^(2*x) - 27*x^4*e^x + 900*x^4*log(3) + 135*x^3*e^(2*x)*log(3 
) - 504*x^3*e^x*log(3) + 810*x^2*e^x*log(3)^2 - 900*x^4*log(x) - 135*x^3*e 
^(2*x)*log(x) + 1692*x^3*e^x*log(x) + 1620*x^2*e^x*log(3)*log(x) - 432*x^4 
 - 60*x^3*e^(2*x) + 342*x^3*e^x - 2160*x^3*log(3) + 45*x^2*e^(2*x)*log(3) 
+ 1512*x^2*e^x*log(3) + 2160*x^3*log(x) - 45*x^2*e^(2*x)*log(x) - 3456*x^2 
*e^x*log(x) + 432*x^3 + 45*x^2*e^(2*x) - 432*x^2*e^x + 2160*x^2*log(3) - 1 
620*x*e^x*log(3) - 2160*x^2*log(x) + 3240*x*e^x*log(x))/(x^6*e^(2*x)*log(3 
)^2 - 2*x^6*e^x*log(3)^2 + 4*x^6*e^(2*x)*log(3)*log(x) - 8*x^6*e^x*log(3)* 
log(x) + 4*x^6*e^(2*x)*log(x)^2 - 8*x^6*e^x*log(x)^2 - x^7*e^(2*x) + 2*x^7 
*e^x + x^6*log(3)^2 - 9*x^5*e^(2*x)*log(3)^2 + 24*x^5*e^x*log(3)^2 + 4*x^6 
*log(3)*log(x) - 36*x^5*e^(2*x)*log(3)*log(x) + 96*x^5*e^x*log(3)*log(x) + 
 4*x^6*log(x)^2 - 36*x^5*e^(2*x)*log(x)^2 + 96*x^5*e^x*log(x)^2 - x^7 +...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=\int -\frac {{\ln \left (3\,x^2\right )}^2\,\left (15\,\ln \left (\frac {x}{3}\right )-11\,x+x^2+15\right )+\ln \left (3\,x^2\right )\,\left (12\,x-4\,x^2+\ln \left (\frac {x}{3}\right )\,\left (20\,x-60\right )\right )+\ln \left (\frac {x}{3}\right )\,\left (5\,x^2\,{\mathrm {e}}^x-5\,x^2\right )-{\mathrm {e}}^x\,\left (x^3-x^2+5\,x\right )+5\,x^2}{{\ln \left (3\,x^2\right )}^4\,\left (x^2-6\,x+9\right )-2\,x^3\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^{2\,x}+x^4-{\ln \left (3\,x^2\right )}^2\,\left ({\mathrm {e}}^x\,\left (6\,x-2\,x^2\right )-6\,x^2+2\,x^3\right )} \,d x \] Input:

int(-(log(3*x^2)^2*(15*log(x/3) - 11*x + x^2 + 15) + log(3*x^2)*(12*x - 4* 
x^2 + log(x/3)*(20*x - 60)) + log(x/3)*(5*x^2*exp(x) - 5*x^2) - exp(x)*(5* 
x - x^2 + x^3) + 5*x^2)/(log(3*x^2)^4*(x^2 - 6*x + 9) - 2*x^3*exp(x) + x^2 
*exp(2*x) + x^4 - log(3*x^2)^2*(exp(x)*(6*x - 2*x^2) - 6*x^2 + 2*x^3)),x)
 

Output:

int(-(log(3*x^2)^2*(15*log(x/3) - 11*x + x^2 + 15) + log(3*x^2)*(12*x - 4* 
x^2 + log(x/3)*(20*x - 60)) + log(x/3)*(5*x^2*exp(x) - 5*x^2) - exp(x)*(5* 
x - x^2 + x^3) + 5*x^2)/(log(3*x^2)^4*(x^2 - 6*x + 9) - 2*x^3*exp(x) + x^2 
*exp(2*x) + x^4 - log(3*x^2)^2*(exp(x)*(6*x - 2*x^2) - 6*x^2 + 2*x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.12 \[ \int \frac {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx=\frac {x \left (5 \,\mathrm {log}\left (\frac {x}{3}\right )-x \right )}{e^{x} x +\mathrm {log}\left (3 x^{2}\right )^{2} x -3 \mathrm {log}\left (3 x^{2}\right )^{2}-x^{2}} \] Input:

int(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x)+4*x^ 
2-12*x)*log(3*x^2)+(-5*exp(x)*x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*exp(x)-5 
*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*log(3*x^2 
)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x)
 

Output:

(x*(5*log(x/3) - x))/(e**x*x + log(3*x**2)**2*x - 3*log(3*x**2)**2 - x**2)