\(\int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x (-16-36 x-16 x^2)+(8 x-4 e^{3 x} x+e^{2 x} (16 x+16 x^2)+e^x (-20 x-28 x^2-16 x^3)) \log (x)+(2 x-e^{3 x} x+e^{2 x} (4 x+4 x^2)+e^x (-5 x-7 x^2-4 x^3)) \log ^2(x)}{(e^{4 x} x+x^3+e^{3 x} (-4 x-4 x^2)+e^x (-4 x^2-4 x^3)+e^{2 x} (4 x+10 x^2+4 x^3)) \log ^2(x)} \, dx\) [2708]

Optimal result

Integrand size = 207, antiderivative size = 27 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\frac {4+\log (x)}{\left (e^x+\frac {x}{-2+e^x-2 x}\right ) \log (x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\frac {\left (e^x-2 (1+x)\right ) (4+\log (x))}{\left (e^{2 x}+x-2 e^x (1+x)\right ) \log (x)} \] Input:

Integrate[(-4*E^(3*x) + 8*x + 8*x^2 + E^(2*x)*(16 + 16*x) + E^x*(-16 - 36* 
x - 16*x^2) + (8*x - 4*E^(3*x)*x + E^(2*x)*(16*x + 16*x^2) + E^x*(-20*x - 
28*x^2 - 16*x^3))*Log[x] + (2*x - E^(3*x)*x + E^(2*x)*(4*x + 4*x^2) + E^x* 
(-5*x - 7*x^2 - 4*x^3))*Log[x]^2)/((E^(4*x)*x + x^3 + E^(3*x)*(-4*x - 4*x^ 
2) + E^x*(-4*x^2 - 4*x^3) + E^(2*x)*(4*x + 10*x^2 + 4*x^3))*Log[x]^2),x]
 

Output:

((E^x - 2*(1 + x))*(4 + Log[x]))/((E^(2*x) + x - 2*E^x*(1 + 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.

\(\displaystyle \int \frac {8 x^2+e^x \left (-16 x^2-36 x-16\right )+\left (e^{2 x} \left (4 x^2+4 x\right )+e^x \left (-4 x^3-7 x^2-5 x\right )-e^{3 x} x+2 x\right ) \log ^2(x)+\left (e^{2 x} \left (16 x^2+16 x\right )+e^x \left (-16 x^3-28 x^2-20 x\right )-4 e^{3 x} x+8 x\right ) \log (x)+8 x-4 e^{3 x}+e^{2 x} (16 x+16)}{\left (x^3+e^{3 x} \left (-4 x^2-4 x\right )+e^x \left (-4 x^3-4 x^2\right )+e^{2 x} \left (4 x^3+10 x^2+4 x\right )+e^{4 x} x\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {8 x^2+e^x \left (-16 x^2-36 x-16\right )+\left (e^{2 x} \left (4 x^2+4 x\right )+e^x \left (-4 x^3-7 x^2-5 x\right )-e^{3 x} x+2 x\right ) \log ^2(x)+\left (e^{2 x} \left (16 x^2+16 x\right )+e^x \left (-16 x^3-28 x^2-20 x\right )-4 e^{3 x} x+8 x\right ) \log (x)+8 x-4 e^{3 x}+e^{2 x} (16 x+16)}{x \left (2 e^x x-x+2 e^x-e^{2 x}\right )^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-4*E^(3*x) + 8*x + 8*x^2 + E^(2*x)*(16 + 16*x) + E^x*(-16 - 36*x - 16 
*x^2) + (8*x - 4*E^(3*x)*x + E^(2*x)*(16*x + 16*x^2) + E^x*(-20*x - 28*x^2 
 - 16*x^3))*Log[x] + (2*x - E^(3*x)*x + E^(2*x)*(4*x + 4*x^2) + E^x*(-5*x 
- 7*x^2 - 4*x^3))*Log[x]^2)/((E^(4*x)*x + x^3 + E^(3*x)*(-4*x - 4*x^2) + E 
^x*(-4*x^2 - 4*x^3) + E^(2*x)*(4*x + 10*x^2 + 4*x^3))*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(68\) vs. \(2(25)=50\).

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56

\[\frac {-{\mathrm e}^{x}+2 x +2}{2 \,{\mathrm e}^{x} x -{\mathrm e}^{2 x}-x +2 \,{\mathrm e}^{x}}+\frac {-4 \,{\mathrm e}^{x}+8 x +8}{\left (2 \,{\mathrm e}^{x} x -{\mathrm e}^{2 x}-x +2 \,{\mathrm e}^{x}\right ) \ln \left (x \right )}\]

Input:

int(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x)*ln(x 
)^2+(-4*x*exp(x)^3+(16*x^2+16*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp(x)+8*x 
)*ln(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8*x^2+8*x)/ 
(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4*x^3-4*x^ 
2)*exp(x)+x^3)/ln(x)^2,x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\frac {{\left (2 \, x - e^{x} + 2\right )} \log \left (x\right ) + 8 \, x - 4 \, e^{x} + 8}{{\left (2 \, {\left (x + 1\right )} e^{x} - x - e^{\left (2 \, x\right )}\right )} \log \left (x\right )} \] Input:

integrate(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x 
)*log(x)^2+(-4*x*exp(x)^3+(16*x^2+16*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp 
(x)+8*x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8*x 
^2+8*x)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4* 
x^3-4*x^2)*exp(x)+x^3)/log(x)^2,x, algorithm="fricas")
 

Output:

((2*x - e^x + 2)*log(x) + 8*x - 4*e^x + 8)/((2*(x + 1)*e^x - x - e^(2*x))* 
log(x))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (20) = 40\).

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\frac {- 2 x \log {\left (x \right )} - 8 x + \left (\log {\left (x \right )} + 4\right ) e^{x} - 2 \log {\left (x \right )} - 8}{x \log {\left (x \right )} + \left (- 2 x \log {\left (x \right )} - 2 \log {\left (x \right )}\right ) e^{x} + e^{2 x} \log {\left (x \right )}} \] Input:

integrate(((-x*exp(x)**3+(4*x**2+4*x)*exp(x)**2+(-4*x**3-7*x**2-5*x)*exp(x 
)+2*x)*ln(x)**2+(-4*x*exp(x)**3+(16*x**2+16*x)*exp(x)**2+(-16*x**3-28*x**2 
-20*x)*exp(x)+8*x)*ln(x)-4*exp(x)**3+(16*x+16)*exp(x)**2+(-16*x**2-36*x-16 
)*exp(x)+8*x**2+8*x)/(x*exp(x)**4+(-4*x**2-4*x)*exp(x)**3+(4*x**3+10*x**2+ 
4*x)*exp(x)**2+(-4*x**3-4*x**2)*exp(x)+x**3)/ln(x)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=-\frac {{\left (\log \left (x\right ) + 4\right )} e^{x} - 2 \, {\left (x + 1\right )} \log \left (x\right ) - 8 \, x - 8}{2 \, {\left (x + 1\right )} e^{x} \log \left (x\right ) - x \log \left (x\right ) - e^{\left (2 \, x\right )} \log \left (x\right )} \] Input:

integrate(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x 
)*log(x)^2+(-4*x*exp(x)^3+(16*x^2+16*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp 
(x)+8*x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8*x 
^2+8*x)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4* 
x^3-4*x^2)*exp(x)+x^3)/log(x)^2,x, algorithm="maxima")
 

Output:

-((log(x) + 4)*e^x - 2*(x + 1)*log(x) - 8*x - 8)/(2*(x + 1)*e^x*log(x) - x 
*log(x) - e^(2*x)*log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (30) = 60\).

Time = 0.23 (sec) , antiderivative size = 209, normalized size of antiderivative = 7.74 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (8 \, x^{4} \log \left (x\right )^{2} - 4 \, x^{3} e^{x} \log \left (x\right )^{2} + 16 \, x^{4} \log \left (x\right ) - 8 \, x^{3} e^{x} \log \left (x\right ) + 8 \, x^{3} \log \left (x\right )^{2} + 16 \, x^{3} \log \left (x\right ) - 16 \, x^{3} + 8 \, x^{2} e^{x} - 16 \, x^{2} \log \left (x\right ) + 8 \, x e^{x} \log \left (x\right ) - 2 \, x \log \left (x\right )^{2} + e^{x} \log \left (x\right )^{2} - 24 \, x^{2} + 8 \, x e^{x} - 12 \, x \log \left (x\right ) + 2 \, e^{x} \log \left (x\right ) - 2 \, \log \left (x\right )^{2} - 12 \, x - 8 \, \log \left (x\right ) - 4\right )}}{8 \, x^{4} e^{x} \log \left (x\right )^{2} - 4 \, x^{4} \log \left (x\right )^{2} - 4 \, x^{3} e^{\left (2 \, x\right )} \log \left (x\right )^{2} + 8 \, x^{3} e^{x} \log \left (x\right )^{2} - 2 \, x e^{x} \log \left (x\right )^{2} + x \log \left (x\right )^{2} + e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 2 \, e^{x} \log \left (x\right )^{2}} \] Input:

integrate(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x 
)*log(x)^2+(-4*x*exp(x)^3+(16*x^2+16*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp 
(x)+8*x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8*x 
^2+8*x)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4* 
x^3-4*x^2)*exp(x)+x^3)/log(x)^2,x, algorithm="giac")
 

Output:

2*(8*x^4*log(x)^2 - 4*x^3*e^x*log(x)^2 + 16*x^4*log(x) - 8*x^3*e^x*log(x) 
+ 8*x^3*log(x)^2 + 16*x^3*log(x) - 16*x^3 + 8*x^2*e^x - 16*x^2*log(x) + 8* 
x*e^x*log(x) - 2*x*log(x)^2 + e^x*log(x)^2 - 24*x^2 + 8*x*e^x - 12*x*log(x 
) + 2*e^x*log(x) - 2*log(x)^2 - 12*x - 8*log(x) - 4)/(8*x^4*e^x*log(x)^2 - 
 4*x^4*log(x)^2 - 4*x^3*e^(2*x)*log(x)^2 + 8*x^3*e^x*log(x)^2 - 2*x*e^x*lo 
g(x)^2 + x*log(x)^2 + e^(2*x)*log(x)^2 - 2*e^x*log(x)^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\int \frac {8\,x-4\,{\mathrm {e}}^{3\,x}+\ln \left (x\right )\,\left (8\,x+{\mathrm {e}}^{2\,x}\,\left (16\,x^2+16\,x\right )-4\,x\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^x\,\left (16\,x^3+28\,x^2+20\,x\right )\right )-{\mathrm {e}}^x\,\left (16\,x^2+36\,x+16\right )+{\mathrm {e}}^{2\,x}\,\left (16\,x+16\right )+8\,x^2+{\ln \left (x\right )}^2\,\left (2\,x+{\mathrm {e}}^{2\,x}\,\left (4\,x^2+4\,x\right )-x\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^x\,\left (4\,x^3+7\,x^2+5\,x\right )\right )}{{\ln \left (x\right )}^2\,\left (x\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^x\,\left (4\,x^3+4\,x^2\right )-{\mathrm {e}}^{3\,x}\,\left (4\,x^2+4\,x\right )+{\mathrm {e}}^{2\,x}\,\left (4\,x^3+10\,x^2+4\,x\right )+x^3\right )} \,d x \] Input:

int((8*x - 4*exp(3*x) + log(x)*(8*x + exp(2*x)*(16*x + 16*x^2) - 4*x*exp(3 
*x) - exp(x)*(20*x + 28*x^2 + 16*x^3)) - exp(x)*(36*x + 16*x^2 + 16) + exp 
(2*x)*(16*x + 16) + 8*x^2 + log(x)^2*(2*x + exp(2*x)*(4*x + 4*x^2) - x*exp 
(3*x) - exp(x)*(5*x + 7*x^2 + 4*x^3)))/(log(x)^2*(x*exp(4*x) - exp(x)*(4*x 
^2 + 4*x^3) - exp(3*x)*(4*x + 4*x^2) + exp(2*x)*(4*x + 10*x^2 + 4*x^3) + x 
^3)),x)
 

Output:

int((8*x - 4*exp(3*x) + log(x)*(8*x + exp(2*x)*(16*x + 16*x^2) - 4*x*exp(3 
*x) - exp(x)*(20*x + 28*x^2 + 16*x^3)) - exp(x)*(36*x + 16*x^2 + 16) + exp 
(2*x)*(16*x + 16) + 8*x^2 + log(x)^2*(2*x + exp(2*x)*(4*x + 4*x^2) - x*exp 
(3*x) - exp(x)*(5*x + 7*x^2 + 4*x^3)))/(log(x)^2*(x*exp(4*x) - exp(x)*(4*x 
^2 + 4*x^3) - exp(3*x)*(4*x + 4*x^2) + exp(2*x)*(4*x + 10*x^2 + 4*x^3) + x 
^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {-4 e^{3 x}+8 x+8 x^2+e^{2 x} (16+16 x)+e^x \left (-16-36 x-16 x^2\right )+\left (8 x-4 e^{3 x} x+e^{2 x} \left (16 x+16 x^2\right )+e^x \left (-20 x-28 x^2-16 x^3\right )\right ) \log (x)+\left (2 x-e^{3 x} x+e^{2 x} \left (4 x+4 x^2\right )+e^x \left (-5 x-7 x^2-4 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+x^3+e^{3 x} \left (-4 x-4 x^2\right )+e^x \left (-4 x^2-4 x^3\right )+e^{2 x} \left (4 x+10 x^2+4 x^3\right )\right ) \log ^2(x)} \, dx=\frac {e^{x} \mathrm {log}\left (x \right )+4 e^{x}-2 \,\mathrm {log}\left (x \right ) x -2 \,\mathrm {log}\left (x \right )-8 x -8}{\mathrm {log}\left (x \right ) \left (e^{2 x}-2 e^{x} x -2 e^{x}+x \right )} \] Input:

int(((-x*exp(x)^3+(4*x^2+4*x)*exp(x)^2+(-4*x^3-7*x^2-5*x)*exp(x)+2*x)*log( 
x)^2+(-4*x*exp(x)^3+(16*x^2+16*x)*exp(x)^2+(-16*x^3-28*x^2-20*x)*exp(x)+8* 
x)*log(x)-4*exp(x)^3+(16*x+16)*exp(x)^2+(-16*x^2-36*x-16)*exp(x)+8*x^2+8*x 
)/(x*exp(x)^4+(-4*x^2-4*x)*exp(x)^3+(4*x^3+10*x^2+4*x)*exp(x)^2+(-4*x^3-4* 
x^2)*exp(x)+x^3)/log(x)^2,x)
 

Output:

(e**x*log(x) + 4*e**x - 2*log(x)*x - 2*log(x) - 8*x - 8)/(log(x)*(e**(2*x) 
 - 2*e**x*x - 2*e**x + x))