Integrand size = 272, antiderivative size = 26 \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=-\log (x)+\log \left (\log \left (-1+x-\log \left (x \log \left (-\frac {2}{5}+e^x-x\right )\right )\right )\right ) \]
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=-\log (x)+\log \left (\log \left (-1+x-\log \left (x \log \left (-\frac {2}{5}+e^x-x\right )\right )\right )\right ) \]
Integrate[(-5*x + 5*E^x*x + (-2 + E^x*(5 - 5*x) - 3*x + 5*x^2)*Log[(-2 + 5 *E^x - 5*x)/5] + ((2 + 3*x - 5*x^2 + E^x*(-5 + 5*x))*Log[(-2 + 5*E^x - 5*x )/5] + (2 - 5*E^x + 5*x)*Log[(-2 + 5*E^x - 5*x)/5]*Log[x*Log[(-2 + 5*E^x - 5*x)/5]])*Log[-1 + x - Log[x*Log[(-2 + 5*E^x - 5*x)/5]]])/(((-2*x - 3*x^2 + 5*x^3 + E^x*(5*x - 5*x^2))*Log[(-2 + 5*E^x - 5*x)/5] + (-2*x + 5*E^x*x - 5*x^2)*Log[(-2 + 5*E^x - 5*x)/5]*Log[x*Log[(-2 + 5*E^x - 5*x)/5]])*Log[- 1 + x - Log[x*Log[(-2 + 5*E^x - 5*x)/5]]]),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.
| \(\displaystyle \int \frac {\left (5 x^2-3 x+e^x (5-5 x)-2\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )+\left (\left (-5 x^2+3 x+e^x (5 x-5)+2\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )+\left (5 x-5 e^x+2\right ) \log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right ) \log \left (x-\log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right )-1\right )+5 e^x x-5 x}{\left (\left (-5 x^2+5 e^x x-2 x\right ) \log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )+\left (5 x^3-3 x^2+e^x \left (5 x-5 x^2\right )-2 x\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right ) \log \left (x-\log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right )-1\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-\left (5 x^2-3 x+e^x (5-5 x)-2\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )-\left (\left (-5 x^2+3 x+e^x (5 x-5)+2\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )+\left (5 x-5 e^x+2\right ) \log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right ) \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right ) \log \left (x-\log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right )-1\right )-5 e^x x+5 x}{x \left (5 x-5 e^x+2\right ) \log \left (-x+e^x-\frac {2}{5}\right ) \left (-x+\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )+1\right ) \log \left (x-\log \left (x \log \left (\frac {1}{5} \left (-5 x+5 e^x-2\right )\right )\right )-1\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {5 x-3}{\left (5 x-5 e^x+2\right ) \log \left (-x+e^x-\frac {2}{5}\right ) \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}+\frac {-x+x \log \left (-x+e^x-\frac {2}{5}\right )-x \log \left (-x+e^x-\frac {2}{5}\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )-\log \left (-x+e^x-\frac {2}{5}\right )+\log \left (-x+e^x-\frac {2}{5}\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )+\log \left (-x+e^x-\frac {2}{5}\right ) \log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}{x \log \left (-x+e^x-\frac {2}{5}\right ) \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {1}{\left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}dx-\int \frac {1}{x \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}dx-\int \frac {1}{\log \left (-x+e^x-\frac {2}{5}\right ) \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}dx+3 \int \frac {1}{\left (-5 x+5 e^x-2\right ) \log \left (-x+e^x-\frac {2}{5}\right ) \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}dx-5 \int \frac {x}{\left (-5 x+5 e^x-2\right ) \log \left (-x+e^x-\frac {2}{5}\right ) \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right ) \log \left (x-\log \left (x \log \left (-x+e^x-\frac {2}{5}\right )\right )-1\right )}dx-\log (x)\) |
Int[(-5*x + 5*E^x*x + (-2 + E^x*(5 - 5*x) - 3*x + 5*x^2)*Log[(-2 + 5*E^x - 5*x)/5] + ((2 + 3*x - 5*x^2 + E^x*(-5 + 5*x))*Log[(-2 + 5*E^x - 5*x)/5] + (2 - 5*E^x + 5*x)*Log[(-2 + 5*E^x - 5*x)/5]*Log[x*Log[(-2 + 5*E^x - 5*x)/ 5]])*Log[-1 + x - Log[x*Log[(-2 + 5*E^x - 5*x)/5]]])/(((-2*x - 3*x^2 + 5*x ^3 + E^x*(5*x - 5*x^2))*Log[(-2 + 5*E^x - 5*x)/5] + (-2*x + 5*E^x*x - 5*x^ 2)*Log[(-2 + 5*E^x - 5*x)/5]*Log[x*Log[(-2 + 5*E^x - 5*x)/5]])*Log[-1 + x - Log[x*Log[(-2 + 5*E^x - 5*x)/5]]]),x]
3.25.5.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.54
\[-\ln \left (x \right )+\ln \left (\ln \left (-\ln \left (x \right )-\ln \left (\ln \left ({\mathrm e}^{x}-x -\frac {2}{5}\right )\right )+\frac {i \pi \,\operatorname {csgn}\left (i x \ln \left ({\mathrm e}^{x}-x -\frac {2}{5}\right )\right ) \left (-\operatorname {csgn}\left (i x \ln \left ({\mathrm e}^{x}-x -\frac {2}{5}\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \ln \left ({\mathrm e}^{x}-x -\frac {2}{5}\right )\right )+\operatorname {csgn}\left (i \ln \left ({\mathrm e}^{x}-x -\frac {2}{5}\right )\right )\right )}{2}+x -1\right )\right )\]
int((((-5*exp(x)+5*x+2)*ln(exp(x)-x-2/5)*ln(x*ln(exp(x)-x-2/5))+((5*x-5)*e xp(x)-5*x^2+3*x+2)*ln(exp(x)-x-2/5))*ln(-ln(x*ln(exp(x)-x-2/5))+x-1)+((-5* x+5)*exp(x)+5*x^2-3*x-2)*ln(exp(x)-x-2/5)+5*exp(x)*x-5*x)/((5*exp(x)*x-5*x ^2-2*x)*ln(exp(x)-x-2/5)*ln(x*ln(exp(x)-x-2/5))+((-5*x^2+5*x)*exp(x)+5*x^3 -3*x^2-2*x)*ln(exp(x)-x-2/5))/ln(-ln(x*ln(exp(x)-x-2/5))+x-1),x)
-ln(x)+ln(ln(-ln(x)-ln(ln(exp(x)-x-2/5))+1/2*I*Pi*csgn(I*x*ln(exp(x)-x-2/5 ))*(-csgn(I*x*ln(exp(x)-x-2/5))+csgn(I*x))*(-csgn(I*x*ln(exp(x)-x-2/5))+cs gn(I*ln(exp(x)-x-2/5)))+x-1))
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=-\log \left (x\right ) + \log \left (\log \left (x - \log \left (x \log \left (-x + e^{x} - \frac {2}{5}\right )\right ) - 1\right )\right ) \]
integrate((((-5*exp(x)+5*x+2)*log(exp(x)-x-2/5)*log(x*log(exp(x)-x-2/5))+( (5*x-5)*exp(x)-5*x^2+3*x+2)*log(exp(x)-x-2/5))*log(-log(x*log(exp(x)-x-2/5 ))+x-1)+((-5*x+5)*exp(x)+5*x^2-3*x-2)*log(exp(x)-x-2/5)+5*exp(x)*x-5*x)/(( 5*exp(x)*x-5*x^2-2*x)*log(exp(x)-x-2/5)*log(x*log(exp(x)-x-2/5))+((-5*x^2+ 5*x)*exp(x)+5*x^3-3*x^2-2*x)*log(exp(x)-x-2/5))/log(-log(x*log(exp(x)-x-2/ 5))+x-1),x, algorithm=\
Timed out. \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=\text {Timed out} \]
integrate((((-5*exp(x)+5*x+2)*ln(exp(x)-x-2/5)*ln(x*ln(exp(x)-x-2/5))+((5* x-5)*exp(x)-5*x**2+3*x+2)*ln(exp(x)-x-2/5))*ln(-ln(x*ln(exp(x)-x-2/5))+x-1 )+((-5*x+5)*exp(x)+5*x**2-3*x-2)*ln(exp(x)-x-2/5)+5*exp(x)*x-5*x)/((5*exp( x)*x-5*x**2-2*x)*ln(exp(x)-x-2/5)*ln(x*ln(exp(x)-x-2/5))+((-5*x**2+5*x)*ex p(x)+5*x**3-3*x**2-2*x)*ln(exp(x)-x-2/5))/ln(-ln(x*ln(exp(x)-x-2/5))+x-1), x)
Time = 0.53 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=-\log \left (x\right ) + \log \left (\log \left (x - \log \left (x\right ) - \log \left (-\log \left (5\right ) + \log \left (-5 \, x + 5 \, e^{x} - 2\right )\right ) - 1\right )\right ) \]
integrate((((-5*exp(x)+5*x+2)*log(exp(x)-x-2/5)*log(x*log(exp(x)-x-2/5))+( (5*x-5)*exp(x)-5*x^2+3*x+2)*log(exp(x)-x-2/5))*log(-log(x*log(exp(x)-x-2/5 ))+x-1)+((-5*x+5)*exp(x)+5*x^2-3*x-2)*log(exp(x)-x-2/5)+5*exp(x)*x-5*x)/(( 5*exp(x)*x-5*x^2-2*x)*log(exp(x)-x-2/5)*log(x*log(exp(x)-x-2/5))+((-5*x^2+ 5*x)*exp(x)+5*x^3-3*x^2-2*x)*log(exp(x)-x-2/5))/log(-log(x*log(exp(x)-x-2/ 5))+x-1),x, algorithm=\
Time = 2.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=-\log \left (x\right ) + \log \left (\log \left (x - \log \left (x\right ) - \log \left (-\log \left (5\right ) + \log \left (-5 \, x + 5 \, e^{x} - 2\right )\right ) - 1\right )\right ) \]
integrate((((-5*exp(x)+5*x+2)*log(exp(x)-x-2/5)*log(x*log(exp(x)-x-2/5))+( (5*x-5)*exp(x)-5*x^2+3*x+2)*log(exp(x)-x-2/5))*log(-log(x*log(exp(x)-x-2/5 ))+x-1)+((-5*x+5)*exp(x)+5*x^2-3*x-2)*log(exp(x)-x-2/5)+5*exp(x)*x-5*x)/(( 5*exp(x)*x-5*x^2-2*x)*log(exp(x)-x-2/5)*log(x*log(exp(x)-x-2/5))+((-5*x^2+ 5*x)*exp(x)+5*x^3-3*x^2-2*x)*log(exp(x)-x-2/5))/log(-log(x*log(exp(x)-x-2/ 5))+x-1),x, algorithm=\
Time = 12.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-5 x+5 e^x x+\left (-2+e^x (5-5 x)-3 x+5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (\left (2+3 x-5 x^2+e^x (-5+5 x)\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (2-5 e^x+5 x\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )}{\left (\left (-2 x-3 x^2+5 x^3+e^x \left (5 x-5 x^2\right )\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )+\left (-2 x+5 e^x x-5 x^2\right ) \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right ) \log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right ) \log \left (-1+x-\log \left (x \log \left (\frac {1}{5} \left (-2+5 e^x-5 x\right )\right )\right )\right )} \, dx=\ln \left (\ln \left (x-\ln \left (x\,\ln \left ({\mathrm {e}}^x-x-\frac {2}{5}\right )\right )-1\right )\right )-\ln \left (x\right ) \]
int((5*x - log(x - log(x*log(exp(x) - x - 2/5)) - 1)*(log(exp(x) - x - 2/5 )*(3*x + exp(x)*(5*x - 5) - 5*x^2 + 2) + log(exp(x) - x - 2/5)*log(x*log(e xp(x) - x - 2/5))*(5*x - 5*exp(x) + 2)) - 5*x*exp(x) + log(exp(x) - x - 2/ 5)*(3*x + exp(x)*(5*x - 5) - 5*x^2 + 2))/(log(x - log(x*log(exp(x) - x - 2 /5)) - 1)*(log(exp(x) - x - 2/5)*(2*x - exp(x)*(5*x - 5*x^2) + 3*x^2 - 5*x ^3) + log(exp(x) - x - 2/5)*log(x*log(exp(x) - x - 2/5))*(2*x - 5*x*exp(x) + 5*x^2))),x)