\(\int \frac {e^{2 x} (4 x+16 x^2)+e^x (-8 x-32 x^2) \log (3)+(4 x+16 x^2) \log ^2(3)+(e^{2 x} (-8 x-16 x^2)-8 x^3 \log (3)+(-8 x-16 x^2) \log ^2(3)+e^x (8 x^3+(16 x+32 x^2) \log (3))) \log (x)-4 e^x x^4 \log ^2(x)}{(e^{2 x} (1+8 x+16 x^2)+e^x (-2-16 x-32 x^2) \log (3)+(1+8 x+16 x^2) \log ^2(3)) \log ^2(x)+(e^x (2 x^2+8 x^3)+(-2 x^2-8 x^3) \log (3)) \log ^3(x)+x^4 \log ^4(x)} \, dx\) [250]

Optimal result

Integrand size = 219, antiderivative size = 34 \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=\frac {4 x}{\log (x) \left (-\frac {1+4 x}{x}+\frac {x \log (x)}{-e^x+\log (3)}\right )} \] Output:

4*x/ln(x)/(x*ln(x)/(ln(3)-exp(x))-(1+4*x)/x)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=\int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx \] Input:

Integrate[(E^(2*x)*(4*x + 16*x^2) + E^x*(-8*x - 32*x^2)*Log[3] + (4*x + 16 
*x^2)*Log[3]^2 + (E^(2*x)*(-8*x - 16*x^2) - 8*x^3*Log[3] + (-8*x - 16*x^2) 
*Log[3]^2 + E^x*(8*x^3 + (16*x + 32*x^2)*Log[3]))*Log[x] - 4*E^x*x^4*Log[x 
]^2)/((E^(2*x)*(1 + 8*x + 16*x^2) + E^x*(-2 - 16*x - 32*x^2)*Log[3] + (1 + 
 8*x + 16*x^2)*Log[3]^2)*Log[x]^2 + (E^x*(2*x^2 + 8*x^3) + (-2*x^2 - 8*x^3 
)*Log[3])*Log[x]^3 + x^4*Log[x]^4),x]
 

Output:

Integrate[(E^(2*x)*(4*x + 16*x^2) + E^x*(-8*x - 32*x^2)*Log[3] + (4*x + 16 
*x^2)*Log[3]^2 + (E^(2*x)*(-8*x - 16*x^2) - 8*x^3*Log[3] + (-8*x - 16*x^2) 
*Log[3]^2 + E^x*(8*x^3 + (16*x + 32*x^2)*Log[3]))*Log[x] - 4*E^x*x^4*Log[x 
]^2)/((E^(2*x)*(1 + 8*x + 16*x^2) + E^x*(-2 - 16*x - 32*x^2)*Log[3] + (1 + 
 8*x + 16*x^2)*Log[3]^2)*Log[x]^2 + (E^x*(2*x^2 + 8*x^3) + (-2*x^2 - 8*x^3 
)*Log[3])*Log[x]^3 + x^4*Log[x]^4), 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.

\(\displaystyle \int \frac {-4 e^x x^4 \log ^2(x)+e^{2 x} \left (16 x^2+4 x\right )+\left (16 x^2+4 x\right ) \log ^2(3)+e^x \left (-32 x^2-8 x\right ) \log (3)+\left (-8 x^3 \log (3)+e^{2 x} \left (-16 x^2-8 x\right )+\left (-16 x^2-8 x\right ) \log ^2(3)+e^x \left (8 x^3+\left (32 x^2+16 x\right ) \log (3)\right )\right ) \log (x)}{x^4 \log ^4(x)+\left (e^{2 x} \left (16 x^2+8 x+1\right )+\left (16 x^2+8 x+1\right ) \log ^2(3)+e^x \left (-32 x^2-16 x-2\right ) \log (3)\right ) \log ^2(x)+\left (e^x \left (8 x^3+2 x^2\right )+\left (-8 x^3-2 x^2\right ) \log (3)\right ) \log ^3(x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {4 x \left (-e^x x^3 \log ^2(x)-\left (x^2 \log (9)-2 e^x \left (x^2+x \log (81)+\log (9)\right )+e^{2 x} (4 x+2)+4 x \log ^2(3)+2 \log ^2(3)\right ) \log (x)+e^{2 x} (4 x+1)+(4 x+1) \log ^2(3)-e^x (8 x \log (3)+\log (9))\right )}{\log ^2(x) \left (x^2 \log (x)+(4 x+1) \left (e^x-\log (3)\right )\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int \frac {x \left (-e^x \log ^2(x) x^3+e^{2 x} (4 x+1)-e^x (8 \log (3) x+\log (9))-\left (\log (9) x^2+4 \log ^2(3) x+2 e^{2 x} (2 x+1)-2 e^x \left (x^2+\log (81) x+\log (9)\right )+2 \log ^2(3)\right ) \log (x)+(4 x+1) \log ^2(3)\right )}{\log ^2(x) \left (\log (x) x^2+(4 x+1) \left (e^x-\log (3)\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 4 \int \left (-\frac {\left (4 x^2-7 x-4\right ) x^3}{(4 x+1)^2 \left (\log (x) x^2+4 e^x x-4 \log (3) x+e^x-\log (3)\right )}-\frac {(4 \log (x) x-4 x+2 \log (x)-1) x}{(4 x+1)^2 \log ^2(x)}+\frac {\left (4 \log ^2(x) x^6-3 \log ^2(x) x^5-4 (1+\log (81)) \log (x) x^5-2 \log ^2(x) x^4-(1+\log (6561)) \log (x) x^4-\log (3) \log (x) x^3-40 \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right ) x-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )\right ) x}{(4 x+1)^2 \log (x) \left (\log (x) x^2+4 e^x x-4 \log (3) x+e^x-\log (3)\right )^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle 4 \int \left (-\frac {\left (4 x^2-7 x-4\right ) x^3}{(4 x+1)^2 \left (\log (x) x^2+4 e^x x-4 \log (3) x+e^x-\log (3)\right )}-\frac {(4 \log (x) x-4 x+2 \log (x)-1) x}{(4 x+1)^2 \log ^2(x)}+\frac {\left (4 \log ^2(x) x^6-3 \log ^2(x) x^5-4 (1+\log (81)) \log (x) x^5-2 \log ^2(x) x^4-(1+\log (6561)) \log (x) x^4-\log (3) \log (x) x^3-40 \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right ) x-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )\right ) x}{(4 x+1)^2 \log (x) \left (\log (x) x^2+4 e^x x-4 \log (3) x+e^x-\log (3)\right )^2}\right )dx\)

Input:

Int[(E^(2*x)*(4*x + 16*x^2) + E^x*(-8*x - 32*x^2)*Log[3] + (4*x + 16*x^2)* 
Log[3]^2 + (E^(2*x)*(-8*x - 16*x^2) - 8*x^3*Log[3] + (-8*x - 16*x^2)*Log[3 
]^2 + E^x*(8*x^3 + (16*x + 32*x^2)*Log[3]))*Log[x] - 4*E^x*x^4*Log[x]^2)/( 
(E^(2*x)*(1 + 8*x + 16*x^2) + E^x*(-2 - 16*x - 32*x^2)*Log[3] + (1 + 8*x + 
 16*x^2)*Log[3]^2)*Log[x]^2 + (E^x*(2*x^2 + 8*x^3) + (-2*x^2 - 8*x^3)*Log[ 
3])*Log[x]^3 + x^4*Log[x]^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26

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

Input:

int((-4*x^4*exp(x)*ln(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*ln(3)+8* 
x^3)*exp(x)+(-16*x^2-8*x)*ln(3)^2-8*x^3*ln(3))*ln(x)+(16*x^2+4*x)*exp(x)^2 
+(-32*x^2-8*x)*ln(3)*exp(x)+(16*x^2+4*x)*ln(3)^2)/(x^4*ln(x)^4+((8*x^3+2*x 
^2)*exp(x)+(-8*x^3-2*x^2)*ln(3))*ln(x)^3+((16*x^2+8*x+1)*exp(x)^2+(-32*x^2 
-16*x-2)*ln(3)*exp(x)+(16*x^2+8*x+1)*ln(3)^2)*ln(x)^2),x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=-\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \left (3\right )\right )}}{x^{2} \log \left (x\right )^{2} + {\left ({\left (4 \, x + 1\right )} e^{x} - {\left (4 \, x + 1\right )} \log \left (3\right )\right )} \log \left (x\right )} \] Input:

integrate((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*l 
og(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*log(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4* 
x)*exp(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(x) 
^4+((8*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*e 
xp(x)^2+(-32*x^2-16*x-2)*log(3)*exp(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2), 
x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (24) = 48\).

Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.91 \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=\frac {4 x^{4}}{4 x^{3} \log {\left (x \right )} + x^{2} \log {\left (x \right )} - 16 x^{2} \log {\left (3 \right )} - 8 x \log {\left (3 \right )} + \left (16 x^{2} + 8 x + 1\right ) e^{x} - \log {\left (3 \right )}} - \frac {4 x^{2}}{\left (4 x + 1\right ) \log {\left (x \right )}} \] Input:

integrate((-4*x**4*exp(x)*ln(x)**2+((-16*x**2-8*x)*exp(x)**2+((32*x**2+16* 
x)*ln(3)+8*x**3)*exp(x)+(-16*x**2-8*x)*ln(3)**2-8*x**3*ln(3))*ln(x)+(16*x* 
*2+4*x)*exp(x)**2+(-32*x**2-8*x)*ln(3)*exp(x)+(16*x**2+4*x)*ln(3)**2)/(x** 
4*ln(x)**4+((8*x**3+2*x**2)*exp(x)+(-8*x**3-2*x**2)*ln(3))*ln(x)**3+((16*x 
**2+8*x+1)*exp(x)**2+(-32*x**2-16*x-2)*ln(3)*exp(x)+(16*x**2+8*x+1)*ln(3)* 
*2)*ln(x)**2),x)
 

Output:

4*x**4/(4*x**3*log(x) + x**2*log(x) - 16*x**2*log(3) - 8*x*log(3) + (16*x* 
*2 + 8*x + 1)*exp(x) - log(3)) - 4*x**2/((4*x + 1)*log(x))
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=-\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \left (3\right )\right )}}{x^{2} \log \left (x\right )^{2} + {\left (4 \, x + 1\right )} e^{x} \log \left (x\right ) - {\left (4 \, x \log \left (3\right ) + \log \left (3\right )\right )} \log \left (x\right )} \] Input:

integrate((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*l 
og(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*log(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4* 
x)*exp(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(x) 
^4+((8*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*e 
xp(x)^2+(-32*x^2-16*x-2)*log(3)*exp(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2), 
x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=-\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \left (3\right )\right )}}{x^{2} \log \left (x\right )^{2} + 4 \, x e^{x} \log \left (x\right ) - 4 \, x \log \left (3\right ) \log \left (x\right ) + e^{x} \log \left (x\right ) - \log \left (3\right ) \log \left (x\right )} \] Input:

integrate((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*l 
og(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*log(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4* 
x)*exp(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(x) 
^4+((8*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*e 
xp(x)^2+(-32*x^2-16*x-2)*log(3)*exp(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2), 
x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=\int -\frac {\ln \left (x\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+8\,x\right )-{\mathrm {e}}^x\,\left (\ln \left (3\right )\,\left (32\,x^2+16\,x\right )+8\,x^3\right )+{\ln \left (3\right )}^2\,\left (16\,x^2+8\,x\right )+8\,x^3\,\ln \left (3\right )\right )-{\ln \left (3\right )}^2\,\left (16\,x^2+4\,x\right )-{\mathrm {e}}^{2\,x}\,\left (16\,x^2+4\,x\right )+{\mathrm {e}}^x\,\ln \left (3\right )\,\left (32\,x^2+8\,x\right )+4\,x^4\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2}{x^4\,{\ln \left (x\right )}^4+{\ln \left (x\right )}^3\,\left ({\mathrm {e}}^x\,\left (8\,x^3+2\,x^2\right )-\ln \left (3\right )\,\left (8\,x^3+2\,x^2\right )\right )+{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+8\,x+1\right )+{\ln \left (3\right )}^2\,\left (16\,x^2+8\,x+1\right )-{\mathrm {e}}^x\,\ln \left (3\right )\,\left (32\,x^2+16\,x+2\right )\right )} \,d x \] Input:

int(-(log(x)*(exp(2*x)*(8*x + 16*x^2) - exp(x)*(log(3)*(16*x + 32*x^2) + 8 
*x^3) + log(3)^2*(8*x + 16*x^2) + 8*x^3*log(3)) - log(3)^2*(4*x + 16*x^2) 
- exp(2*x)*(4*x + 16*x^2) + exp(x)*log(3)*(8*x + 32*x^2) + 4*x^4*exp(x)*lo 
g(x)^2)/(x^4*log(x)^4 + log(x)^3*(exp(x)*(2*x^2 + 8*x^3) - log(3)*(2*x^2 + 
 8*x^3)) + log(x)^2*(exp(2*x)*(8*x + 16*x^2 + 1) + log(3)^2*(8*x + 16*x^2 
+ 1) - exp(x)*log(3)*(16*x + 32*x^2 + 2))),x)
 

Output:

int(-(log(x)*(exp(2*x)*(8*x + 16*x^2) - exp(x)*(log(3)*(16*x + 32*x^2) + 8 
*x^3) + log(3)^2*(8*x + 16*x^2) + 8*x^3*log(3)) - log(3)^2*(4*x + 16*x^2) 
- exp(2*x)*(4*x + 16*x^2) + exp(x)*log(3)*(8*x + 32*x^2) + 4*x^4*exp(x)*lo 
g(x)^2)/(x^4*log(x)^4 + log(x)^3*(exp(x)*(2*x^2 + 8*x^3) - log(3)*(2*x^2 + 
 8*x^3)) + log(x)^2*(exp(2*x)*(8*x + 16*x^2 + 1) + log(3)^2*(8*x + 16*x^2 
+ 1) - exp(x)*log(3)*(16*x + 32*x^2 + 2))), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29 \[ \int \frac {e^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx=\frac {4 x^{2} \left (-e^{x}+\mathrm {log}\left (3\right )\right )}{\mathrm {log}\left (x \right ) \left (4 e^{x} x +e^{x}+\mathrm {log}\left (x \right ) x^{2}-4 \,\mathrm {log}\left (3\right ) x -\mathrm {log}\left (3\right )\right )} \] Input:

int((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*log(3)+ 
8*x^3)*exp(x)+(-16*x^2-8*x)*log(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4*x)*exp 
(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(x)^4+((8 
*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*exp(x)^ 
2+(-32*x^2-16*x-2)*log(3)*exp(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2),x)
 

Output:

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