\(\int \frac {(-4-4 x+2 e^x x^2) \log (\frac {e^x x}{2+e^x (x+x^2)})}{90 e^4 x+e^{4+x} (45 x^2+45 x^3)+(60 e^4 x+e^{4+x} (30 x^2+30 x^3)) \log ^2(\frac {e^x x}{2+e^x (x+x^2)})+(10 e^4 x+e^{4+x} (5 x^2+5 x^3)) \log ^4(\frac {e^x x}{2+e^x (x+x^2)})} \, dx\) [2089]

Optimal result

Integrand size = 151, antiderivative size = 30 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {1}{5 e^4 \left (3+\log ^2\left (\frac {x}{2 e^{-x}+x+x^2}\right )\right )} \] Output:

1/5/(3+ln(x/(x+x^2+2/exp(x)))^2)/exp(4)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {1}{5 e^4 \left (3+\log ^2\left (\frac {e^x x}{2+e^x x (1+x)}\right )\right )} \] Input:

Integrate[((-4 - 4*x + 2*E^x*x^2)*Log[(E^x*x)/(2 + E^x*(x + x^2))])/(90*E^ 
4*x + E^(4 + x)*(45*x^2 + 45*x^3) + (60*E^4*x + E^(4 + x)*(30*x^2 + 30*x^3 
))*Log[(E^x*x)/(2 + E^x*(x + x^2))]^2 + (10*E^4*x + E^(4 + x)*(5*x^2 + 5*x 
^3))*Log[(E^x*x)/(2 + E^x*(x + x^2))]^4),x]
 

Output:

1/(5*E^4*(3 + Log[(E^x*x)/(2 + E^x*x*(1 + x))]^2))
 

Rubi [A] (verified)

Time = 8.75 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {7239, 27, 25, 7262, 17}

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.

\(\displaystyle \int \frac {\left (2 e^x x^2-4 x-4\right ) \log \left (\frac {e^x x}{e^x \left (x^2+x\right )+2}\right )}{e^{x+4} \left (45 x^3+45 x^2\right )+\left (e^{x+4} \left (5 x^3+5 x^2\right )+10 e^4 x\right ) \log ^4\left (\frac {e^x x}{e^x \left (x^2+x\right )+2}\right )+\left (e^{x+4} \left (30 x^3+30 x^2\right )+60 e^4 x\right ) \log ^2\left (\frac {e^x x}{e^x \left (x^2+x\right )+2}\right )+90 e^4 x} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7262

\(\displaystyle \frac {\int \frac {1}{\left (1+\frac {3}{\log ^2\left (\frac {e^x x}{e^x x (x+1)+2}\right )}\right )^2}d\frac {3}{\log ^2\left (\frac {e^x x}{e^x x (x+1)+2}\right )}}{15 e^4}\)

\(\Big \downarrow \) 17

\(\displaystyle -\frac {1}{15 e^4 \left (\frac {3}{\log ^2\left (\frac {e^x x}{e^x x (x+1)+2}\right )}+1\right )}\)

Input:

Int[((-4 - 4*x + 2*E^x*x^2)*Log[(E^x*x)/(2 + E^x*(x + x^2))])/(90*E^4*x + 
E^(4 + x)*(45*x^2 + 45*x^3) + (60*E^4*x + E^(4 + x)*(30*x^2 + 30*x^3))*Log 
[(E^x*x)/(2 + E^x*(x + x^2))]^2 + (10*E^4*x + E^(4 + x)*(5*x^2 + 5*x^3))*L 
og[(E^x*x)/(2 + E^x*(x + x^2))]^4),x]
 

Output:

-1/15*1/(E^4*(1 + 3/Log[(E^x*x)/(2 + E^x*x*(1 + x))]^2))
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

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 7239

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

rule 7262

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c 
= Simplify[u/(p*w*D[v, x] - q*v*D[w, x])]}, Simp[c*p   Subst[Int[(b + a*x^p 
)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}, x] 
 && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]
 

Maple [A] (verified)

Time = 10.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{-4}}{5 \ln \left (\frac {{\mathrm e}^{x} x}{{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x +2}\right )^{2}+15}\)	\(32\)
risch	\(\text {Expression too large to display}\)	\(2968\)

Input:

int((2*exp(x)*x^2-4*x-4)*ln(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5*x^2)*e 
xp(4)*exp(x)+10*x*exp(4))*ln(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x^3+30*x^ 
2)*exp(4)*exp(x)+60*x*exp(4))*ln(x*exp(x)/((x^2+x)*exp(x)+2))^2+(45*x^3+45 
*x^2)*exp(4)*exp(x)+90*x*exp(4)),x,method=_RETURNVERBOSE)
 

Output:

1/5/(ln(exp(x)*x/(exp(x)*x^2+exp(x)*x+2))^2+3)/exp(4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {1}{5 \, {\left (e^{4} \log \left (\frac {x e^{\left (x + 4\right )}}{{\left (x^{2} + x\right )} e^{\left (x + 4\right )} + 2 \, e^{4}}\right )^{2} + 3 \, e^{4}\right )}} \] Input:

integrate((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5 
*x^2)*exp(4)*exp(x)+10*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x 
^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2+( 
45*x^3+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x, algorithm="fricas")
 

Output:

1/5/(e^4*log(x*e^(x + 4)/((x^2 + x)*e^(x + 4) + 2*e^4))^2 + 3*e^4)
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {1}{5 e^{4} \log {\left (\frac {x e^{x}}{\left (x^{2} + x\right ) e^{x} + 2} \right )}^{2} + 15 e^{4}} \] Input:

integrate((2*exp(x)*x**2-4*x-4)*ln(x*exp(x)/((x**2+x)*exp(x)+2))/(((5*x**3 
+5*x**2)*exp(4)*exp(x)+10*x*exp(4))*ln(x*exp(x)/((x**2+x)*exp(x)+2))**4+(( 
30*x**3+30*x**2)*exp(4)*exp(x)+60*x*exp(4))*ln(x*exp(x)/((x**2+x)*exp(x)+2 
))**2+(45*x**3+45*x**2)*exp(4)*exp(x)+90*x*exp(4)),x)
 

Output:

1/(5*exp(4)*log(x*exp(x)/((x**2 + x)*exp(x) + 2))**2 + 15*exp(4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (26) = 52\).

Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {1}{5 \, {\left (x^{2} e^{4} + e^{4} \log \left ({\left (x^{2} + x\right )} e^{x} + 2\right )^{2} + 2 \, x e^{4} \log \left (x\right ) + e^{4} \log \left (x\right )^{2} - 2 \, {\left (x e^{4} + e^{4} \log \left (x\right )\right )} \log \left ({\left (x^{2} + x\right )} e^{x} + 2\right ) + 3 \, e^{4}\right )}} \] Input:

integrate((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5 
*x^2)*exp(4)*exp(x)+10*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x 
^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2+( 
45*x^3+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x, algorithm="maxima")
 

Output:

1/5/(x^2*e^4 + e^4*log((x^2 + x)*e^x + 2)^2 + 2*x*e^4*log(x) + e^4*log(x)^ 
2 - 2*(x*e^4 + e^4*log(x))*log((x^2 + x)*e^x + 2) + 3*e^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (26) = 52\).

Time = 45.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.97 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {2}{5 \, {\left (x^{2} e^{4} + 2 \, x e^{4} \log \left (\frac {x}{x^{2} e^{x} + x e^{x} + 2}\right ) + e^{4} \log \left (\frac {x}{x^{2} e^{x} + x e^{x} + 2}\right )^{2} + 3 \, e^{4}\right )}} \] Input:

integrate((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5 
*x^2)*exp(4)*exp(x)+10*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x 
^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2+( 
45*x^3+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x, algorithm="giac")
 

Output:

2/5/(x^2*e^4 + 2*x*e^4*log(x/(x^2*e^x + x*e^x + 2)) + e^4*log(x/(x^2*e^x + 
 x*e^x + 2))^2 + 3*e^4)
 

Mupad [B] (verification not implemented)

Time = 3.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {{\mathrm {e}}^{-4}}{5\,\left ({\ln \left (\frac {x\,{\mathrm {e}}^x}{{\mathrm {e}}^x\,\left (x^2+x\right )+2}\right )}^2+3\right )} \] Input:

int(-(log((x*exp(x))/(exp(x)*(x + x^2) + 2))*(4*x - 2*x^2*exp(x) + 4))/(90 
*x*exp(4) + log((x*exp(x))/(exp(x)*(x + x^2) + 2))^4*(10*x*exp(4) + exp(4) 
*exp(x)*(5*x^2 + 5*x^3)) + log((x*exp(x))/(exp(x)*(x + x^2) + 2))^2*(60*x* 
exp(4) + exp(4)*exp(x)*(30*x^2 + 30*x^3)) + exp(4)*exp(x)*(45*x^2 + 45*x^3 
)),x)
 

Output:

exp(-4)/(5*(log((x*exp(x))/(exp(x)*(x + x^2) + 2))^2 + 3))
 

Reduce [F]

\[ \int \frac {\left (-4-4 x+2 e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )}{90 e^4 x+e^{4+x} \left (45 x^2+45 x^3\right )+\left (60 e^4 x+e^{4+x} \left (30 x^2+30 x^3\right )\right ) \log ^2\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )+\left (10 e^4 x+e^{4+x} \left (5 x^2+5 x^3\right )\right ) \log ^4\left (\frac {e^x x}{2+e^x \left (x+x^2\right )}\right )} \, dx=\frac {-\frac {4 \left (\int \frac {\mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )}{e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x^{3}+e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x^{2}+6 e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x^{3}+6 e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x^{2}+9 e^{x} x^{3}+9 e^{x} x^{2}+2 \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x +12 \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x +18 x}d x \right )}{5}-\frac {4 \left (\int \frac {\mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )}{e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x^{2}+e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x +6 e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x^{2}+6 e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x +9 e^{x} x^{2}+9 e^{x} x +2 \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4}+12 \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2}+18}d x \right )}{5}+\frac {2 \left (\int \frac {e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right ) x}{e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x^{2}+e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4} x +6 e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x^{2}+6 e^{x} \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2} x +9 e^{x} x^{2}+9 e^{x} x +2 \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{4}+12 \mathrm {log}\left (\frac {e^{x} x}{e^{x} x^{2}+e^{x} x +2}\right )^{2}+18}d x \right )}{5}}{e^{4}} \] Input:

int((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5*x^2)* 
exp(4)*exp(x)+10*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x^3+30* 
x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2+(45*x^3 
+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x)
 

Output:

(2*( - 2*int(log((e**x*x)/(e**x*x**2 + e**x*x + 2))/(e**x*log((e**x*x)/(e* 
*x*x**2 + e**x*x + 2))**4*x**3 + e**x*log((e**x*x)/(e**x*x**2 + e**x*x + 2 
))**4*x**2 + 6*e**x*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**2*x**3 + 6*e** 
x*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**2*x**2 + 9*e**x*x**3 + 9*e**x*x* 
*2 + 2*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**4*x + 12*log((e**x*x)/(e**x 
*x**2 + e**x*x + 2))**2*x + 18*x),x) - 2*int(log((e**x*x)/(e**x*x**2 + e** 
x*x + 2))/(e**x*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**4*x**2 + e**x*log( 
(e**x*x)/(e**x*x**2 + e**x*x + 2))**4*x + 6*e**x*log((e**x*x)/(e**x*x**2 + 
 e**x*x + 2))**2*x**2 + 6*e**x*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**2*x 
 + 9*e**x*x**2 + 9*e**x*x + 2*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**4 + 
12*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**2 + 18),x) + int((e**x*log((e** 
x*x)/(e**x*x**2 + e**x*x + 2))*x)/(e**x*log((e**x*x)/(e**x*x**2 + e**x*x + 
 2))**4*x**2 + e**x*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**4*x + 6*e**x*l 
og((e**x*x)/(e**x*x**2 + e**x*x + 2))**2*x**2 + 6*e**x*log((e**x*x)/(e**x* 
x**2 + e**x*x + 2))**2*x + 9*e**x*x**2 + 9*e**x*x + 2*log((e**x*x)/(e**x*x 
**2 + e**x*x + 2))**4 + 12*log((e**x*x)/(e**x*x**2 + e**x*x + 2))**2 + 18) 
,x)))/(5*e**4)