Integrand size = 252, antiderivative size = 31 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4 x (1+x)}{\left (4-\frac {e^x}{x}\right ) \left (-e^{e^2}+x+\log (x)\right )} \] Output:
4*x*(1+x)/(x+ln(x)-exp(exp(2)))/(4-exp(x)/x)
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4 x^2 (1+x)}{\left (-e^x+4 x\right ) \left (-e^{e^2}+x+\log (x)\right )} \] Input:
Integrate[(-16*x^2 - 16*x^3 + 16*x^4 + E^x*(4*x - 4*x^3 + 4*x^4) + E^E^2*( -16*x^2 - 32*x^3 + E^x*(8*x + 8*x^2 - 4*x^3)) + (16*x^2 + 32*x^3 + E^x*(-8 *x - 8*x^2 + 4*x^3))*Log[x])/(E^(2*x)*x^2 - 8*E^x*x^3 + 16*x^4 + E^(2*E^2) *(E^(2*x) - 8*E^x*x + 16*x^2) + E^E^2*(-2*E^(2*x)*x + 16*E^x*x^2 - 32*x^3) + (2*E^(2*x)*x - 16*E^x*x^2 + 32*x^3 + E^E^2*(-2*E^(2*x) + 16*E^x*x - 32* x^2))*Log[x] + (E^(2*x) - 8*E^x*x + 16*x^2)*Log[x]^2),x]
Output:
(4*x^2*(1 + x))/((-E^x + 4*x)*(-E^E^2 + x + Log[x]))
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.
\(\displaystyle \int \frac {16 x^4-16 x^3-16 x^2+e^x \left (4 x^4-4 x^3+4 x\right )+e^{e^2} \left (-32 x^3-16 x^2+e^x \left (-4 x^3+8 x^2+8 x\right )\right )+\left (32 x^3+16 x^2+e^x \left (4 x^3-8 x^2-8 x\right )\right ) \log (x)}{16 x^4-8 e^x x^3+e^{2 x} x^2+e^{2 e^2} \left (16 x^2-8 e^x x+e^{2 x}\right )+\left (16 x^2-8 e^x x+e^{2 x}\right ) \log ^2(x)+e^{e^2} \left (-32 x^3+16 e^x x^2-2 e^{2 x} x\right )+\left (32 x^3-16 e^x x^2+e^{e^2} \left (-32 x^2+16 e^x x-2 e^{2 x}\right )+2 e^{2 x} x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 x \left (e^{x+e^2} \left (-x^2+2 x+2\right )+4 x \left (x^2-x-1\right )+\left (e^x \left (x^2-2 x-2\right )+4 x (2 x+1)\right ) \log (x)+e^x \left (x^3-x^2+1\right )-4 e^{e^2} x (2 x+1)\right )}{\left (e^x-4 x\right )^2 \left (-x-\log (x)+e^{e^2}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {x \left (4 e^{e^2} x (2 x+1)+4 x \left (-x^2+x+1\right )-e^{x+e^2} \left (-x^2+2 x+2\right )-e^x \left (x^3-x^2+1\right )-\left (4 x (2 x+1)-e^x \left (-x^2+2 x+2\right )\right ) \log (x)\right )}{\left (e^x-4 x\right )^2 \left (-x-\log (x)+e^{e^2}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {x \left (4 e^{e^2} x (2 x+1)+4 x \left (-x^2+x+1\right )-e^{x+e^2} \left (-x^2+2 x+2\right )-e^x \left (x^3-x^2+1\right )-\left (4 x (2 x+1)-e^x \left (-x^2+2 x+2\right )\right ) \log (x)\right )}{\left (e^x-4 x\right )^2 \left (-x-\log (x)+e^{e^2}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (\frac {x \left (-x^3-\log (x) x^2+\left (1+e^{e^2}\right ) x^2+2 \log (x) x-2 e^{e^2} x+2 \log (x)-2 e^{e^2}-1\right )}{\left (e^x-4 x\right ) \left (-x-\log (x)+e^{e^2}\right )^2}-\frac {4 x^2 \left (x^2-1\right )}{\left (e^x-4 x\right )^2 \left (x+\log (x)-e^{e^2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (-\int \frac {x^4}{\left (e^x-4 x\right ) \left (x+\log (x)-e^{e^2}\right )^2}dx-4 \int \frac {x^4}{\left (e^x-4 x\right )^2 \left (x+\log (x)-e^{e^2}\right )}dx+\left (1+e^{e^2}\right ) \int \frac {x^3}{\left (e^x-4 x\right ) \left (-x-\log (x)+e^{e^2}\right )^2}dx-\int \frac {x^3 \log (x)}{\left (e^x-4 x\right ) \left (x+\log (x)-e^{e^2}\right )^2}dx-2 e^{e^2} \int \frac {x^2}{\left (e^x-4 x\right ) \left (-x-\log (x)+e^{e^2}\right )^2}dx+2 \int \frac {x^2 \log (x)}{\left (e^x-4 x\right ) \left (x+\log (x)-e^{e^2}\right )^2}dx+4 \int \frac {x^2}{\left (e^x-4 x\right )^2 \left (x+\log (x)-e^{e^2}\right )}dx-\left (\left (1+2 e^{e^2}\right ) \int \frac {x}{\left (e^x-4 x\right ) \left (-x-\log (x)+e^{e^2}\right )^2}dx\right )+2 \int \frac {x \log (x)}{\left (e^x-4 x\right ) \left (x+\log (x)-e^{e^2}\right )^2}dx\right )\) |
Input:
Int[(-16*x^2 - 16*x^3 + 16*x^4 + E^x*(4*x - 4*x^3 + 4*x^4) + E^E^2*(-16*x^ 2 - 32*x^3 + E^x*(8*x + 8*x^2 - 4*x^3)) + (16*x^2 + 32*x^3 + E^x*(-8*x - 8 *x^2 + 4*x^3))*Log[x])/(E^(2*x)*x^2 - 8*E^x*x^3 + 16*x^4 + E^(2*E^2)*(E^(2 *x) - 8*E^x*x + 16*x^2) + E^E^2*(-2*E^(2*x)*x + 16*E^x*x^2 - 32*x^3) + (2* E^(2*x)*x - 16*E^x*x^2 + 32*x^3 + E^E^2*(-2*E^(2*x) + 16*E^x*x - 32*x^2))* Log[x] + (E^(2*x) - 8*E^x*x + 16*x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 2.64 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {4 \left (1+x \right ) x^{2}}{\left (4 x -{\mathrm e}^{x}\right ) \left ({\mathrm e}^{{\mathrm e}^{2}}-x -\ln \left (x \right )\right )}\) | \(32\) |
parallelrisch | \(-\frac {4 x^{3}+4 x^{2}}{{\mathrm e}^{x} \ln \left (x \right )-{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{2}}+{\mathrm e}^{x} x -4 x \ln \left (x \right )+4 x \,{\mathrm e}^{{\mathrm e}^{2}}-4 x^{2}}\) | \(49\) |
Input:
int((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*ln(x)+((-4*x^3+8*x^2+8*x)*ex p(x)-32*x^3-16*x^2)*exp(exp(2))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16*x^3-16* x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*ln(x)^2+((-2*exp(x)^2+16*exp(x)*x-32*x^ 2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*ln(x)+(exp(x)^2-8*exp(x) *x+16*x^2)*exp(exp(2))^2+(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp(exp(2))+ exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x,method=_RETURNVERBOSE)
Output:
-4*(1+x)*x^2/(4*x-exp(x))/(exp(exp(2))-x-ln(x))
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4 \, {\left (x^{3} + x^{2}\right )}}{4 \, x^{2} - x e^{x} - {\left (4 \, x - e^{x}\right )} e^{\left (e^{2}\right )} + {\left (4 \, x - e^{x}\right )} \log \left (x\right )} \] Input:
integrate((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+ 8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16* x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(x) *x-32*x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2 -8*exp(x)*x+16*x^2)*exp(exp(2))^2+(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp (exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x, algorithm="fricas")
Output:
4*(x^3 + x^2)/(4*x^2 - x*e^x - (4*x - e^x)*e^(e^2) + (4*x - e^x)*log(x))
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {- 4 x^{3} - 4 x^{2}}{- 4 x^{2} - 4 x \log {\left (x \right )} + 4 x e^{e^{2}} + \left (x + \log {\left (x \right )} - e^{e^{2}}\right ) e^{x}} \] Input:
integrate((((4*x**3-8*x**2-8*x)*exp(x)+32*x**3+16*x**2)*ln(x)+((-4*x**3+8* x**2+8*x)*exp(x)-32*x**3-16*x**2)*exp(exp(2))+(4*x**4-4*x**3+4*x)*exp(x)+1 6*x**4-16*x**3-16*x**2)/((exp(x)**2-8*exp(x)*x+16*x**2)*ln(x)**2+((-2*exp( x)**2+16*exp(x)*x-32*x**2)*exp(exp(2))+2*x*exp(x)**2-16*exp(x)*x**2+32*x** 3)*ln(x)+(exp(x)**2-8*exp(x)*x+16*x**2)*exp(exp(2))**2+(-2*x*exp(x)**2+16* exp(x)*x**2-32*x**3)*exp(exp(2))+exp(x)**2*x**2-8*exp(x)*x**3+16*x**4),x)
Output:
(-4*x**3 - 4*x**2)/(-4*x**2 - 4*x*log(x) + 4*x*exp(exp(2)) + (x + log(x) - exp(exp(2)))*exp(x))
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4 \, {\left (x^{3} + x^{2}\right )}}{4 \, x^{2} - {\left (x - e^{\left (e^{2}\right )} + \log \left (x\right )\right )} e^{x} - 4 \, x e^{\left (e^{2}\right )} + 4 \, x \log \left (x\right )} \] Input:
integrate((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+ 8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16* x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(x) *x-32*x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2 -8*exp(x)*x+16*x^2)*exp(exp(2))^2+(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp (exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x, algorithm="maxima")
Output:
4*(x^3 + x^2)/(4*x^2 - (x - e^(e^2) + log(x))*e^x - 4*x*e^(e^2) + 4*x*log( x))
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4 \, {\left (x^{3} + x^{2}\right )}}{4 \, x^{2} - x e^{x} - 4 \, x e^{\left (e^{2}\right )} + 4 \, x \log \left (x\right ) - e^{x} \log \left (x\right ) + e^{\left (x + e^{2}\right )}} \] Input:
integrate((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+ 8*x)*exp(x)-32*x^3-16*x^2)*exp(exp(2))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16* x^3-16*x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(x) *x-32*x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2 -8*exp(x)*x+16*x^2)*exp(exp(2))^2+(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp (exp(2))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x, algorithm="giac")
Output:
4*(x^3 + x^2)/(4*x^2 - x*e^x - 4*x*e^(e^2) + 4*x*log(x) - e^x*log(x) + e^( x + e^2))
Time = 4.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.55 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4\,x^2\,\left (4\,x-2\,{\mathrm {e}}^{x+{\mathrm {e}}^2}-{\mathrm {e}}^x+2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{x+{\mathrm {e}}^2}-2\,x\,{\mathrm {e}}^x+8\,x^2+4\,x^3+x^2\,{\mathrm {e}}^{x+{\mathrm {e}}^2}+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x\right )}{{\left (4\,x-{\mathrm {e}}^x\right )}^2\,\left (x+1\right )\,\left (x-{\mathrm {e}}^{{\mathrm {e}}^2}+\ln \left (x\right )\right )} \] Input:
int(-(16*x^2 + 16*x^3 - 16*x^4 + exp(exp(2))*(16*x^2 + 32*x^3 - exp(x)*(8* x + 8*x^2 - 4*x^3)) - exp(x)*(4*x - 4*x^3 + 4*x^4) - log(x)*(16*x^2 + 32*x ^3 - exp(x)*(8*x + 8*x^2 - 4*x^3)))/(exp(2*exp(2))*(exp(2*x) - 8*x*exp(x) + 16*x^2) - 8*x^3*exp(x) - log(x)*(exp(exp(2))*(2*exp(2*x) - 16*x*exp(x) + 32*x^2) - 2*x*exp(2*x) + 16*x^2*exp(x) - 32*x^3) + log(x)^2*(exp(2*x) - 8 *x*exp(x) + 16*x^2) + x^2*exp(2*x) + 16*x^4 - exp(exp(2))*(2*x*exp(2*x) - 16*x^2*exp(x) + 32*x^3)),x)
Output:
(4*x^2*(4*x - 2*exp(x + exp(2)) - exp(x) + 2*exp(exp(2))*exp(x) - x^2*exp( x) - 2*x*exp(x + exp(2)) - 2*x*exp(x) + 8*x^2 + 4*x^3 + x^2*exp(x + exp(2) ) + 2*x*exp(exp(2))*exp(x) - x^2*exp(exp(2))*exp(x)))/((4*x - exp(x))^2*(x + 1)*(x - exp(exp(2)) + log(x)))
Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {-16 x^2-16 x^3+16 x^4+e^x \left (4 x-4 x^3+4 x^4\right )+e^{e^2} \left (-16 x^2-32 x^3+e^x \left (8 x+8 x^2-4 x^3\right )\right )+\left (16 x^2+32 x^3+e^x \left (-8 x-8 x^2+4 x^3\right )\right ) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} \left (e^{2 x}-8 e^x x+16 x^2\right )+e^{e^2} \left (-2 e^{2 x} x+16 e^x x^2-32 x^3\right )+\left (2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} \left (-2 e^{2 x}+16 e^x x-32 x^2\right )\right ) \log (x)+\left (e^{2 x}-8 e^x x+16 x^2\right ) \log ^2(x)} \, dx=\frac {4 x^{2} \left (x +1\right )}{e^{e^{2}+x}-4 e^{e^{2}} x -e^{x} \mathrm {log}\left (x \right )-e^{x} x +4 \,\mathrm {log}\left (x \right ) x +4 x^{2}} \] Input:
int((((4*x^3-8*x^2-8*x)*exp(x)+32*x^3+16*x^2)*log(x)+((-4*x^3+8*x^2+8*x)*e xp(x)-32*x^3-16*x^2)*exp(exp(2))+(4*x^4-4*x^3+4*x)*exp(x)+16*x^4-16*x^3-16 *x^2)/((exp(x)^2-8*exp(x)*x+16*x^2)*log(x)^2+((-2*exp(x)^2+16*exp(x)*x-32* x^2)*exp(exp(2))+2*x*exp(x)^2-16*exp(x)*x^2+32*x^3)*log(x)+(exp(x)^2-8*exp (x)*x+16*x^2)*exp(exp(2))^2+(-2*x*exp(x)^2+16*exp(x)*x^2-32*x^3)*exp(exp(2 ))+exp(x)^2*x^2-8*exp(x)*x^3+16*x^4),x)
Output:
(4*x**2*(x + 1))/(e**(e**2 + x) - 4*e**(e**2)*x - e**x*log(x) - e**x*x + 4 *log(x)*x + 4*x**2)