\(\int \frac {2 x-x^2+e^x (-4 x+2 x^2-2 x^3)+(-1+e^x (2-2 x)) \log (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} (8-16 x+8 x^2)})}{x^4+e^x (-2 x^4+2 x^5)+(-2 x^2+e^x (4 x^2-4 x^3)) \log (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} (8-16 x+8 x^2)})+(1+e^x (-2+2 x)) \log ^2(\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} (8-16 x+8 x^2)})} \, dx\) [1354]

Optimal result

Integrand size = 205, antiderivative size = 29 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=\frac {x}{x^2-\log \left (\frac {1}{2 \left (2-e^{-x}-2 x\right )^2}\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=-\frac {x}{-x^2+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \] Input:

Integrate[(2*x - x^2 + E^x*(-4*x + 2*x^2 - 2*x^3) + (-1 + E^x*(2 - 2*x))*L 
og[E^(2*x)/(2 + E^x*(-8 + 8*x) + E^(2*x)*(8 - 16*x + 8*x^2))])/(x^4 + E^x* 
(-2*x^4 + 2*x^5) + (-2*x^2 + E^x*(4*x^2 - 4*x^3))*Log[E^(2*x)/(2 + E^x*(-8 
 + 8*x) + E^(2*x)*(8 - 16*x + 8*x^2))] + (1 + E^x*(-2 + 2*x))*Log[E^(2*x)/ 
(2 + E^x*(-8 + 8*x) + E^(2*x)*(8 - 16*x + 8*x^2))]^2),x]
 

Output:

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

\(\displaystyle \int \frac {-x^2+\left (e^x (2-2 x)-1\right ) \log \left (\frac {e^{2 x}}{e^{2 x} \left (8 x^2-16 x+8\right )+e^x (8 x-8)+2}\right )+e^x \left (-2 x^3+2 x^2-4 x\right )+2 x}{x^4+\left (e^x (2 x-2)+1\right ) \log ^2\left (\frac {e^{2 x}}{e^{2 x} \left (8 x^2-16 x+8\right )+e^x (8 x-8)+2}\right )+e^x \left (2 x^5-2 x^4\right )+\left (e^x \left (4 x^2-4 x^3\right )-2 x^2\right ) \log \left (\frac {e^{2 x}}{e^{2 x} \left (8 x^2-16 x+8\right )+e^x (8 x-8)+2}\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(2*x - x^2 + E^x*(-4*x + 2*x^2 - 2*x^3) + (-1 + E^x*(2 - 2*x))*Log[E^( 
2*x)/(2 + E^x*(-8 + 8*x) + E^(2*x)*(8 - 16*x + 8*x^2))])/(x^4 + E^x*(-2*x^ 
4 + 2*x^5) + (-2*x^2 + E^x*(4*x^2 - 4*x^3))*Log[E^(2*x)/(2 + E^x*(-8 + 8*x 
) + E^(2*x)*(8 - 16*x + 8*x^2))] + (1 + E^x*(-2 + 2*x))*Log[E^(2*x)/(2 + E 
^x*(-8 + 8*x) + E^(2*x)*(8 - 16*x + 8*x^2))]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83

method	result	size
parallelrisch	\(\frac {x}{x^{2}-\ln \left (\frac {{\mathrm e}^{2 x}}{8 \,{\mathrm e}^{2 x} x^{2}+8 \,{\mathrm e}^{x} x -16 x \,{\mathrm e}^{2 x}-8 \,{\mathrm e}^{x}+8 \,{\mathrm e}^{2 x}+2}\right )}\)	\(53\)
risch	\(\frac {2 x}{-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )}^{3}-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )+i \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{3}-i \pi \,\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+i \pi \,\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )+2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )}^{2}+2 x^{2}+6 \ln \left (2\right )+4 \ln \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )-4 \ln \left ({\mathrm e}^{x}\right )}\)	\(332\)

Input:

int((((2-2*x)*exp(x)-1)*ln(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x 
)+2))+(-2*x^3+2*x^2-4*x)*exp(x)-x^2+2*x)/(((-2+2*x)*exp(x)+1)*ln(exp(x)^2/ 
((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*exp(x)-2*x^2 
)*ln(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2*x^4)*ex 
p(x)+x^4),x,method=_RETURNVERBOSE)
 

Output:

x/(x^2-ln(1/2*exp(x)^2/(4*exp(x)^2*x^2-8*x*exp(x)^2+4*exp(x)^2+4*exp(x)*x- 
4*exp(x)+1)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=\frac {x}{x^{2} - \log \left (\frac {e^{\left (2 \, x\right )}}{2 \, {\left (4 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x - 1\right )} e^{x} + 1\right )}}\right )} \] Input:

integrate((((2-2*x)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8 
)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(ex 
p(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*exp(x 
)-2*x^2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2 
*x^4)*exp(x)+x^4),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=- \frac {x}{- x^{2} + \log {\left (\frac {e^{2 x}}{\left (8 x - 8\right ) e^{x} + \left (8 x^{2} - 16 x + 8\right ) e^{2 x} + 2} \right )}} \] Input:

integrate((((2-2*x)*exp(x)-1)*ln(exp(x)**2/((8*x**2-16*x+8)*exp(x)**2+(8*x 
-8)*exp(x)+2))+(-2*x**3+2*x**2-4*x)*exp(x)-x**2+2*x)/(((2*x-2)*exp(x)+1)*l 
n(exp(x)**2/((8*x**2-16*x+8)*exp(x)**2+(8*x-8)*exp(x)+2))**2+((-4*x**3+4*x 
**2)*exp(x)-2*x**2)*ln(exp(x)**2/((8*x**2-16*x+8)*exp(x)**2+(8*x-8)*exp(x) 
+2))+(2*x**5-2*x**4)*exp(x)+x**4),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=\frac {x}{x^{2} - 2 \, x + \log \left (2\right ) + 2 \, \log \left (2 \, {\left (x - 1\right )} e^{x} + 1\right )} \] Input:

integrate((((2-2*x)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8 
)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(ex 
p(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*exp(x 
)-2*x^2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2 
*x^4)*exp(x)+x^4),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=\frac {x}{x^{2} - 2 \, x + \log \left (8 \, x^{2} e^{\left (2 \, x\right )} - 16 \, x e^{\left (2 \, x\right )} + 8 \, x e^{x} + 8 \, e^{\left (2 \, x\right )} - 8 \, e^{x} + 2\right )} \] Input:

integrate((((2-2*x)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8 
)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(ex 
p(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*exp(x 
)-2*x^2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2 
*x^4)*exp(x)+x^4),x, algorithm="giac")
 

Output:

x/(x^2 - 2*x + log(8*x^2*e^(2*x) - 16*x*e^(2*x) + 8*x*e^x + 8*e^(2*x) - 8* 
e^x + 2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=\int -\frac {\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (2\,x-2\right )+1\right )-2\,x+x^2+{\mathrm {e}}^x\,\left (2\,x^3-2\,x^2+4\,x\right )}{\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (4\,x^2-4\,x^3\right )-2\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^4-2\,x^5\right )+x^4+{\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )}^2\,\left ({\mathrm {e}}^x\,\left (2\,x-2\right )+1\right )} \,d x \] Input:

int(-(log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*( 
exp(x)*(2*x - 2) + 1) - 2*x + x^2 + exp(x)*(4*x - 2*x^2 + 2*x^3))/(log(exp 
(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*(exp(x)*(4*x^2 
 - 4*x^3) - 2*x^2) - exp(x)*(2*x^4 - 2*x^5) + x^4 + log(exp(2*x)/(exp(2*x) 
*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))^2*(exp(x)*(2*x - 2) + 1)),x)
 

Output:

int(-(log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*( 
exp(x)*(2*x - 2) + 1) - 2*x + x^2 + exp(x)*(4*x - 2*x^2 + 2*x^3))/(log(exp 
(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*(exp(x)*(4*x^2 
 - 4*x^3) - 2*x^2) - exp(x)*(2*x^4 - 2*x^5) + x^4 + log(exp(2*x)/(exp(2*x) 
*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))^2*(exp(x)*(2*x - 2) + 1)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx=-\frac {x}{\mathrm {log}\left (\frac {e^{2 x}}{8 e^{2 x} x^{2}-16 e^{2 x} x +8 e^{2 x}+8 e^{x} x -8 e^{x}+2}\right )-x^{2}} \] Input:

int((((2-2*x)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp( 
x)+2))+(-2*x^3+2*x^2-4*x)*exp(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(exp(x)^2 
/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*exp(x)-2*x^ 
2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2*x^4)* 
exp(x)+x^4),x)
 

Output:

( - x)/(log(e**(2*x)/(8*e**(2*x)*x**2 - 16*e**(2*x)*x + 8*e**(2*x) + 8*e** 
x*x - 8*e**x + 2)) - x**2)