3.5.15 \(\int \frac {-16 x^2-16 x^3+16 x^4+e^x (4 x-4 x^3+4 x^4)+e^{e^2} (-16 x^2-32 x^3+e^x (8 x+8 x^2-4 x^3))+(16 x^2+32 x^3+e^x (-8 x-8 x^2+4 x^3)) \log (x)}{e^{2 x} x^2-8 e^x x^3+16 x^4+e^{2 e^2} (e^{2 x}-8 e^x x+16 x^2)+e^{e^2} (-2 e^{2 x} x+16 e^x x^2-32 x^3)+(2 e^{2 x} x-16 e^x x^2+32 x^3+e^{e^2} (-2 e^{2 x}+16 e^x x-32 x^2)) \log (x)+(e^{2 x}-8 e^x x+16 x^2) \log ^2(x)} \, dx\) [415]

3.5.15.1 Optimal result

Integrand size = 252, antiderivative size = 31 \begin {dmath*} \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 )} \end {dmath*}

output

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

3.5.15.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \begin {dmath*} \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 )} \end {dmath*}

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]))
 

3.5.15.3 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.

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

\(\Big \downarrow \) 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
 

3.5.15.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.5.15.4 Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03

method	result	size
risch	\(-\frac {4 x^{2} \left (1+x \right )}{\left (4 x -{\mathrm e}^{x}\right ) \left ({\mathrm e}^{{\mathrm e}^{2}}-x -\ln \left (x \right )\right )}\)	\(32\)
parallelrisch	\(-\frac {4 x^{3}+4 x^{2}}{{\mathrm e}^{x} \ln \left (x \right )-4 x \ln \left (x \right )-{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{2}}+{\mathrm e}^{x} x +4 x \,{\mathrm e}^{{\mathrm e}^{2}}-4 x^{2}}\)	\(49\)

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

3.5.15.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \begin {dmath*} \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 )} \end {dmath*}

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=\
 

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))
 

3.5.15.6 Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \begin {dmath*} \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}} \end {dmath*}

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))
 

3.5.15.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \begin {dmath*} \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 )} \end {dmath*}

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=\
 

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))
 

3.5.15.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \begin {dmath*} \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 )}} \end {dmath*}

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=\
 

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))
 

3.5.15.9 Mupad [B] (verification not implemented)

Time = 14.66 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.55 \begin {dmath*} \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 )} \end {dmath*}

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)))