\(\int \frac {e^{1+x} (1-x)-5 e x^2+(e^{1+x}-6 x+5 e x^2) \log (\frac {-e^{1+x}+6 x-5 e x^2}{3 x})}{e^{1+x} x^2-6 x^3+5 e x^4+(2 e^{1+x} x-12 x^2+10 e x^3) \log (\frac {-e^{1+x}+6 x-5 e x^2}{3 x})+(e^{1+x}-6 x+5 e x^2) \log ^2(\frac {-e^{1+x}+6 x-5 e x^2}{3 x})} \, dx\) [1855]

Optimal result

Integrand size = 172, antiderivative size = 26 \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\frac {x}{x+\log \left (2-\frac {e \left (e^x+5 x^2\right )}{3 x}\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\frac {x}{x+\log \left (2-\frac {e^{1+x}}{3 x}-\frac {5 e x}{3}\right )} \] Input:

Integrate[(E^(1 + x)*(1 - x) - 5*E*x^2 + (E^(1 + x) - 6*x + 5*E*x^2)*Log[( 
-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)])/(E^(1 + x)*x^2 - 6*x^3 + 5*E*x^4 + (2* 
E^(1 + x)*x - 12*x^2 + 10*E*x^3)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)] + 
 (E^(1 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {7239, 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.

\(\displaystyle \int \frac {-5 e x^2+\left (5 e x^2-6 x+e^{x+1}\right ) \log \left (\frac {-5 e x^2+6 x-e^{x+1}}{3 x}\right )+e^{x+1} (1-x)}{5 e x^4-6 x^3+e^{x+1} x^2+\left (5 e x^2-6 x+e^{x+1}\right ) \log ^2\left (\frac {-5 e x^2+6 x-e^{x+1}}{3 x}\right )+\left (10 e x^3-12 x^2+2 e^{x+1} x\right ) \log \left (\frac {-5 e x^2+6 x-e^{x+1}}{3 x}\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7262

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

\(\Big \downarrow \) 17

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

Input:

Int[(E^(1 + x)*(1 - x) - 5*E*x^2 + (E^(1 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 
+ x) + 6*x - 5*E*x^2)/(3*x)])/(E^(1 + x)*x^2 - 6*x^3 + 5*E*x^4 + (2*E^(1 + 
 x)*x - 12*x^2 + 10*E*x^3)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)] + (E^(1 
 + x) - 6*x + 5*E*x^2)*Log[(-E^(1 + x) + 6*x - 5*E*x^2)/(3*x)]^2),x]
 

Output:

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

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 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 = 1.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(\frac {x}{\ln \left (-\frac {{\mathrm e} \,{\mathrm e}^{x}+5 x^{2} {\mathrm e}-6 x}{3 x}\right )+x}\)	\(29\)
risch	\(\frac {2 x}{-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )}{x}\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )}{x}\right )}^{2}-2 i \pi {\operatorname {csgn}\left (\frac {i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )}{x}\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )}{x}\right )}^{3}+2 i \pi -2 \ln \left (3\right )+2 \ln \left (5\right )+2 x -2 \ln \left (x \right )+2 \ln \left (x^{2} {\mathrm e}+\frac {{\mathrm e}^{1+x}}{5}-\frac {6 x}{5}\right )}\)	\(242\)

Input:

int(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+ 
6*x)/x)+(1-x)*exp(1)*exp(x)-5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp(1)-6*x 
)*ln(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp(x)+10*x^3* 
exp(1)-12*x^2)*ln(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*exp(1)*exp( 
x)+5*x^4*exp(1)-6*x^3),x,method=_RETURNVERBOSE)
 

Output:

x/(ln(-1/3*(exp(1)*exp(x)+5*x^2*exp(1)-6*x)/x)+x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\frac {x}{x + \log \left (-\frac {5 \, x^{2} e - 6 \, x + e^{\left (x + 1\right )}}{3 \, x}\right )} \] Input:

integrate(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2* 
exp(1)+6*x)/x)+(1-x)*exp(1)*exp(x)-5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp 
(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp(x) 
+10*x^3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*ex 
p(1)*exp(x)+5*x^4*exp(1)-6*x^3),x, algorithm="fricas")
 

Output:

x/(x + log(-1/3*(5*x^2*e - 6*x + e^(x + 1))/x))
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\frac {x}{x + \log {\left (\frac {- \frac {5 e x^{2}}{3} + 2 x - \frac {e e^{x}}{3}}{x} \right )}} \] Input:

integrate(((exp(1)*exp(x)+5*x**2*exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x**2 
*exp(1)+6*x)/x)+(1-x)*exp(1)*exp(x)-5*x**2*exp(1))/((exp(1)*exp(x)+5*x**2* 
exp(1)-6*x)*ln(1/3*(-exp(1)*exp(x)-5*x**2*exp(1)+6*x)/x)**2+(2*x*exp(1)*ex 
p(x)+10*x**3*exp(1)-12*x**2)*ln(1/3*(-exp(1)*exp(x)-5*x**2*exp(1)+6*x)/x)+ 
x**2*exp(1)*exp(x)+5*x**4*exp(1)-6*x**3),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\frac {x}{x - \log \left (3\right ) + \log \left (-5 \, x^{2} e + 6 \, x - e^{\left (x + 1\right )}\right ) - \log \left (x\right )} \] Input:

integrate(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2* 
exp(1)+6*x)/x)+(1-x)*exp(1)*exp(x)-5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp 
(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp(x) 
+10*x^3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*ex 
p(1)*exp(x)+5*x^4*exp(1)-6*x^3),x, algorithm="maxima")
 

Output:

x/(x - log(3) + log(-5*x^2*e + 6*x - e^(x + 1)) - log(x))
 

Giac [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\frac {x}{x + \log \left (-\frac {5 \, x^{2} e - 6 \, x + e^{\left (x + 1\right )}}{3 \, x}\right )} \] Input:

integrate(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2* 
exp(1)+6*x)/x)+(1-x)*exp(1)*exp(x)-5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp 
(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp(x) 
+10*x^3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*ex 
p(1)*exp(x)+5*x^4*exp(1)-6*x^3),x, algorithm="giac")
 

Output:

x/(x + log(-1/3*(5*x^2*e - 6*x + e^(x + 1))/x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\int -\frac {5\,x^2\,\mathrm {e}-\ln \left (-\frac {\frac {5\,x^2\,\mathrm {e}}{3}-2\,x+\frac {\mathrm {e}\,{\mathrm {e}}^x}{3}}{x}\right )\,\left (5\,x^2\,\mathrm {e}-6\,x+\mathrm {e}\,{\mathrm {e}}^x\right )+\mathrm {e}\,{\mathrm {e}}^x\,\left (x-1\right )}{{\ln \left (-\frac {\frac {5\,x^2\,\mathrm {e}}{3}-2\,x+\frac {\mathrm {e}\,{\mathrm {e}}^x}{3}}{x}\right )}^2\,\left (5\,x^2\,\mathrm {e}-6\,x+\mathrm {e}\,{\mathrm {e}}^x\right )+5\,x^4\,\mathrm {e}+\ln \left (-\frac {\frac {5\,x^2\,\mathrm {e}}{3}-2\,x+\frac {\mathrm {e}\,{\mathrm {e}}^x}{3}}{x}\right )\,\left (10\,x^3\,\mathrm {e}-12\,x^2+2\,x\,\mathrm {e}\,{\mathrm {e}}^x\right )-6\,x^3+x^2\,\mathrm {e}\,{\mathrm {e}}^x} \,d x \] Input:

int(-(5*x^2*exp(1) - log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)* 
(5*x^2*exp(1) - 6*x + exp(1)*exp(x)) + exp(1)*exp(x)*(x - 1))/(log(-((5*x^ 
2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)^2*(5*x^2*exp(1) - 6*x + exp(1)*e 
xp(x)) + 5*x^4*exp(1) + log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/ 
x)*(10*x^3*exp(1) - 12*x^2 + 2*x*exp(1)*exp(x)) - 6*x^3 + x^2*exp(1)*exp(x 
)),x)
 

Output:

int(-(5*x^2*exp(1) - log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)* 
(5*x^2*exp(1) - 6*x + exp(1)*exp(x)) + exp(1)*exp(x)*(x - 1))/(log(-((5*x^ 
2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/x)^2*(5*x^2*exp(1) - 6*x + exp(1)*e 
xp(x)) + 5*x^4*exp(1) + log(-((5*x^2*exp(1))/3 - 2*x + (exp(1)*exp(x))/3)/ 
x)*(10*x^3*exp(1) - 12*x^2 + 2*x*exp(1)*exp(x)) - 6*x^3 + x^2*exp(1)*exp(x 
)), x)
 

Reduce [F]

\[ \int \frac {e^{1+x} (1-x)-5 e x^2+\left (e^{1+x}-6 x+5 e x^2\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )}{e^{1+x} x^2-6 x^3+5 e x^4+\left (2 e^{1+x} x-12 x^2+10 e x^3\right ) \log \left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )+\left (e^{1+x}-6 x+5 e x^2\right ) \log ^2\left (\frac {-e^{1+x}+6 x-5 e x^2}{3 x}\right )} \, dx=\text {too large to display} \] Input:

int(((exp(1)*exp(x)+5*x^2*exp(1)-6*x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1) 
+6*x)/x)+(1-x)*exp(1)*exp(x)-5*x^2*exp(1))/((exp(1)*exp(x)+5*x^2*exp(1)-6* 
x)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)^2+(2*x*exp(1)*exp(x)+10*x^ 
3*exp(1)-12*x^2)*log(1/3*(-exp(1)*exp(x)-5*x^2*exp(1)+6*x)/x)+x^2*exp(1)*e 
xp(x)+5*x^4*exp(1)-6*x^3),x)
 

Output:

( - int(e**x/(e**x*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*e*x + 2*e**x 
*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*e*x**2 + e**x*e*x**3 + 5*log(( - 
e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*e*x**3 - 6*log(( - e**x*e - 5*e*x**2 + 
6*x)/(3*x))**2*x**2 + 10*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*e*x**4 - 
12*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*x**3 + 5*e*x**5 - 6*x**4),x)*lo 
g(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*e - int(e**x/(e**x*log(( - e**x*e - 
5*e*x**2 + 6*x)/(3*x))**2*e*x + 2*e**x*log(( - e**x*e - 5*e*x**2 + 6*x)/(3 
*x))*e*x**2 + e**x*e*x**3 + 5*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*e 
*x**3 - 6*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*x**2 + 10*log(( - e** 
x*e - 5*e*x**2 + 6*x)/(3*x))*e*x**4 - 12*log(( - e**x*e - 5*e*x**2 + 6*x)/ 
(3*x))*x**3 + 5*e*x**5 - 6*x**4),x)*e*x + 2*int(e**x/(e**x*log(( - e**x*e 
- 5*e*x**2 + 6*x)/(3*x))**2*e + 2*e**x*log(( - e**x*e - 5*e*x**2 + 6*x)/(3 
*x))*e*x + e**x*e*x**2 + 5*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*e*x* 
*2 - 6*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*x + 10*log(( - e**x*e - 
5*e*x**2 + 6*x)/(3*x))*e*x**3 - 12*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x)) 
*x**2 + 5*e*x**4 - 6*x**3),x)*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*e + 
2*int(e**x/(e**x*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))**2*e + 2*e**x*log 
(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*e*x + e**x*e*x**2 + 5*log(( - e**x*e 
- 5*e*x**2 + 6*x)/(3*x))**2*e*x**2 - 6*log(( - e**x*e - 5*e*x**2 + 6*x)/(3 
*x))**2*x + 10*log(( - e**x*e - 5*e*x**2 + 6*x)/(3*x))*e*x**3 - 12*log(...