\(\int \frac {16 x^2+4 x^5+e^{1-x} (-12 x-4 x^2+2 x^5)+(8 x^2-4 x^5+e^{1-x} (-4 x+2 x^4)) \log (\frac {4 x-2 x^4+e^{1-x} (-2+x^3)}{x^3})}{(e^{1-x} (-2+x^3) \log ^2(2)+(4 x-2 x^4) \log ^2(2)) \log ^3(\frac {4 x-2 x^4+e^{1-x} (-2+x^3)}{x^3})} \, dx\) [2222]

Optimal result

Integrand size = 153, antiderivative size = 39 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^2}{\log ^2(2) \log ^2\left (\frac {\left (-e^{1-x}+2 x\right ) \left (\frac {2}{x}-x^2\right )}{x^2}\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^2}{\log ^2(2) \log ^2\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (-2+x^3\right )}{x^3}\right )} \] Input:

Integrate[(16*x^2 + 4*x^5 + E^(1 - x)*(-12*x - 4*x^2 + 2*x^5) + (8*x^2 - 4 
*x^5 + E^(1 - x)*(-4*x + 2*x^4))*Log[(4*x - 2*x^4 + E^(1 - x)*(-2 + x^3))/ 
x^3])/((E^(1 - x)*(-2 + x^3)*Log[2]^2 + (4*x - 2*x^4)*Log[2]^2)*Log[(4*x - 
 2*x^4 + E^(1 - x)*(-2 + x^3))/x^3]^3),x]
 

Output:

x^2/(Log[2]^2*Log[((E - 2*E^x*x)*(-2 + x^3))/(E^x*x^3)]^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.

\(\displaystyle \int \frac {4 x^5+16 x^2+e^{1-x} \left (2 x^5-4 x^2-12 x\right )+\left (-4 x^5+e^{1-x} \left (2 x^4-4 x\right )+8 x^2\right ) \log \left (\frac {-2 x^4+e^{1-x} \left (x^3-2\right )+4 x}{x^3}\right )}{\left (\left (4 x-2 x^4\right ) \log ^2(2)+e^{1-x} \left (x^3-2\right ) \log ^2(2)\right ) \log ^3\left (\frac {-2 x^4+e^{1-x} \left (x^3-2\right )+4 x}{x^3}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (-4 x^5-16 x^2-e^{1-x} \left (2 x^5-4 x^2-12 x\right )-\left (-4 x^5+e^{1-x} \left (2 x^4-4 x\right )+8 x^2\right ) \log \left (\frac {-2 x^4+e^{1-x} \left (x^3-2\right )+4 x}{x^3}\right )\right )}{\left (e-2 e^x x\right ) \left (2-x^3\right ) \log ^2(2) \log ^3\left (-\frac {e^{-x} \left (2 e^x x-e\right ) \left (x^3-2\right )}{x^3}\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int -\frac {2 e^x \left (2 x^5+8 x^2-e^{1-x} \left (-x^5+2 x^2+6 x\right )+\left (-2 x^5+4 x^2-e^{1-x} \left (2 x-x^4\right )\right ) \log \left (\frac {-2 x^4+4 x-e^{1-x} \left (2-x^3\right )}{x^3}\right )\right )}{\left (e-2 e^x x\right ) \left (2-x^3\right ) \log ^3\left (-\frac {e^{-x} \left (e-2 e^x x\right ) \left (2-x^3\right )}{x^3}\right )}dx}{\log ^2(2)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \int \frac {e^x \left (2 x^5+8 x^2-e^{1-x} \left (-x^5+2 x^2+6 x\right )+\left (-2 x^5+4 x^2-e^{1-x} \left (2 x-x^4\right )\right ) \log \left (\frac {-2 x^4+4 x-e^{1-x} \left (2-x^3\right )}{x^3}\right )\right )}{\left (e-2 e^x x\right ) \left (2-x^3\right ) \log ^3\left (-\frac {e^{-x} \left (e-2 e^x x\right ) \left (2-x^3\right )}{x^3}\right )}dx}{\log ^2(2)}\)

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {2 \int \left (\frac {2 e^x x^2 (x+1)}{\left (2 e^x x-e\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (x^3-2\right )}{x^3}\right )}-\frac {x \left (x^4+\log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (x^3-2\right )}{x^3}\right ) x^3-2 x-2 \log \left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (x^3-2\right )}{x^3}\right )-6\right )}{\left (x^3-2\right ) \log ^3\left (\frac {e^{-x} \left (e-2 e^x x\right ) \left (x^3-2\right )}{x^3}\right )}\right )dx}{\log ^2(2)}\)

\(\Big \downarrow \) 2009

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

Input:

Int[(16*x^2 + 4*x^5 + E^(1 - x)*(-12*x - 4*x^2 + 2*x^5) + (8*x^2 - 4*x^5 + 
 E^(1 - x)*(-4*x + 2*x^4))*Log[(4*x - 2*x^4 + E^(1 - x)*(-2 + x^3))/x^3])/ 
((E^(1 - x)*(-2 + x^3)*Log[2]^2 + (4*x - 2*x^4)*Log[2]^2)*Log[(4*x - 2*x^4 
 + E^(1 - x)*(-2 + x^3))/x^3]^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 9.98 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {x^{2}}{{\ln \left (\frac {\left (x^{3}-2\right ) {\mathrm e}^{1-x}-2 x^{4}+4 x}{x^{3}}\right )}^{2} \ln \left (2\right )^{2}}\)	\(37\)
risch	\(-\frac {4 x^{2}}{{\left (\pi \,\operatorname {csgn}\left (i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right ) \operatorname {csgn}\left (\frac {i}{x^{3}}\right )+\pi {\operatorname {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )}^{3}-2 \pi {\operatorname {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )}{x^{3}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{3}}\right )+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}+2 \pi -2 i \ln \left (2\right )+6 i \ln \left (x \right )-2 i \ln \left (x^{4}-\frac {{\mathrm e}^{1-x} x^{3}}{2}-2 x +{\mathrm e}^{1-x}\right )\right )}^{2} \ln \left (2\right )^{2}}\)	\(411\)

Input:

int((((2*x^4-4*x)*exp(1-x)-4*x^5+8*x^2)*ln(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^ 
3)+(2*x^5-4*x^2-12*x)*exp(1-x)+4*x^5+16*x^2)/((x^3-2)*ln(2)^2*exp(1-x)+(-2 
*x^4+4*x)*ln(2)^2)/ln(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^3,x,method=_RETURN 
VERBOSE)
 

Output:

x^2/ln(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^2/ln(2)^2
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^{2}}{\log \left (2\right )^{2} \log \left (-\frac {2 \, x^{4} - {\left (x^{3} - 2\right )} e^{\left (-x + 1\right )} - 4 \, x}{x^{3}}\right )^{2}} \] Input:

integrate((((2*x^4-4*x)*exp(1-x)-4*x^5+8*x^2)*log(((x^3-2)*exp(1-x)-2*x^4+ 
4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(1-x)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp( 
1-x)+(-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^3,x, alg 
orithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^{2}}{\log {\left (2 \right )}^{2} \log {\left (\frac {- 2 x^{4} + 4 x + \left (x^{3} - 2\right ) e^{1 - x}}{x^{3}} \right )}^{2}} \] Input:

integrate((((2*x**4-4*x)*exp(1-x)-4*x**5+8*x**2)*ln(((x**3-2)*exp(1-x)-2*x 
**4+4*x)/x**3)+(2*x**5-4*x**2-12*x)*exp(1-x)+4*x**5+16*x**2)/((x**3-2)*ln( 
2)**2*exp(1-x)+(-2*x**4+4*x)*ln(2)**2)/ln(((x**3-2)*exp(1-x)-2*x**4+4*x)/x 
**3)**3,x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (37) = 74\).

Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.18 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^{2}}{x^{2} \log \left (2\right )^{2} + \log \left (2\right )^{2} \log \left (x^{3} - 2\right )^{2} + \log \left (2\right )^{2} \log \left (-2 \, x e^{x} + e\right )^{2} + 6 \, x \log \left (2\right )^{2} \log \left (x\right ) + 9 \, \log \left (2\right )^{2} \log \left (x\right )^{2} - 2 \, {\left (x \log \left (2\right )^{2} + 3 \, \log \left (2\right )^{2} \log \left (x\right )\right )} \log \left (x^{3} - 2\right ) - 2 \, {\left (x \log \left (2\right )^{2} - \log \left (2\right )^{2} \log \left (x^{3} - 2\right ) + 3 \, \log \left (2\right )^{2} \log \left (x\right )\right )} \log \left (-2 \, x e^{x} + e\right )} \] Input:

integrate((((2*x^4-4*x)*exp(1-x)-4*x^5+8*x^2)*log(((x^3-2)*exp(1-x)-2*x^4+ 
4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(1-x)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp( 
1-x)+(-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^3,x, alg 
orithm="maxima")
 

Output:

x^2/(x^2*log(2)^2 + log(2)^2*log(x^3 - 2)^2 + log(2)^2*log(-2*x*e^x + e)^2 
 + 6*x*log(2)^2*log(x) + 9*log(2)^2*log(x)^2 - 2*(x*log(2)^2 + 3*log(2)^2* 
log(x))*log(x^3 - 2) - 2*(x*log(2)^2 - log(2)^2*log(x^3 - 2) + 3*log(2)^2* 
log(x))*log(-2*x*e^x + e))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (37) = 74\).

Time = 0.49 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.33 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^{2}}{\log \left (2\right )^{2} \log \left (-2 \, x^{4} + x^{3} e^{\left (-x + 1\right )} + 4 \, x - 2 \, e^{\left (-x + 1\right )}\right )^{2} - 2 \, \log \left (2\right )^{2} \log \left (-2 \, x^{4} + x^{3} e^{\left (-x + 1\right )} + 4 \, x - 2 \, e^{\left (-x + 1\right )}\right ) \log \left (x^{3}\right ) + \log \left (2\right )^{2} \log \left (x^{3}\right )^{2}} \] Input:

integrate((((2*x^4-4*x)*exp(1-x)-4*x^5+8*x^2)*log(((x^3-2)*exp(1-x)-2*x^4+ 
4*x)/x^3)+(2*x^5-4*x^2-12*x)*exp(1-x)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp( 
1-x)+(-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^3,x, alg 
orithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=-\int \frac {{\mathrm {e}}^{1-x}\,\left (-2\,x^5+4\,x^2+12\,x\right )-16\,x^2-4\,x^5+\ln \left (\frac {4\,x+{\mathrm {e}}^{1-x}\,\left (x^3-2\right )-2\,x^4}{x^3}\right )\,\left ({\mathrm {e}}^{1-x}\,\left (4\,x-2\,x^4\right )-8\,x^2+4\,x^5\right )}{{\ln \left (\frac {4\,x+{\mathrm {e}}^{1-x}\,\left (x^3-2\right )-2\,x^4}{x^3}\right )}^3\,\left ({\ln \left (2\right )}^2\,\left (4\,x-2\,x^4\right )+{\mathrm {e}}^{1-x}\,{\ln \left (2\right )}^2\,\left (x^3-2\right )\right )} \,d x \] Input:

int(-(exp(1 - x)*(12*x + 4*x^2 - 2*x^5) - 16*x^2 - 4*x^5 + log((4*x + exp( 
1 - x)*(x^3 - 2) - 2*x^4)/x^3)*(exp(1 - x)*(4*x - 2*x^4) - 8*x^2 + 4*x^5)) 
/(log((4*x + exp(1 - x)*(x^3 - 2) - 2*x^4)/x^3)^3*(log(2)^2*(4*x - 2*x^4) 
+ exp(1 - x)*log(2)^2*(x^3 - 2))),x)
 

Output:

-int((exp(1 - x)*(12*x + 4*x^2 - 2*x^5) - 16*x^2 - 4*x^5 + log((4*x + exp( 
1 - x)*(x^3 - 2) - 2*x^4)/x^3)*(exp(1 - x)*(4*x - 2*x^4) - 8*x^2 + 4*x^5)) 
/(log((4*x + exp(1 - x)*(x^3 - 2) - 2*x^4)/x^3)^3*(log(2)^2*(4*x - 2*x^4) 
+ exp(1 - x)*log(2)^2*(x^3 - 2))), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {16 x^2+4 x^5+e^{1-x} \left (-12 x-4 x^2+2 x^5\right )+\left (8 x^2-4 x^5+e^{1-x} \left (-4 x+2 x^4\right )\right ) \log \left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )}{\left (e^{1-x} \left (-2+x^3\right ) \log ^2(2)+\left (4 x-2 x^4\right ) \log ^2(2)\right ) \log ^3\left (\frac {4 x-2 x^4+e^{1-x} \left (-2+x^3\right )}{x^3}\right )} \, dx=\frac {x^{2}}{\mathrm {log}\left (\frac {-2 e^{x} x^{4}+4 e^{x} x +e \,x^{3}-2 e}{e^{x} x^{3}}\right )^{2} \mathrm {log}\left (2\right )^{2}} \] Input:

int((((2*x^4-4*x)*exp(1-x)-4*x^5+8*x^2)*log(((x^3-2)*exp(1-x)-2*x^4+4*x)/x 
^3)+(2*x^5-4*x^2-12*x)*exp(1-x)+4*x^5+16*x^2)/((x^3-2)*log(2)^2*exp(1-x)+( 
-2*x^4+4*x)*log(2)^2)/log(((x^3-2)*exp(1-x)-2*x^4+4*x)/x^3)^3,x)
 

Output:

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