Integrand size = 161, antiderivative size = 31 \[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=e^{-x+\frac {\log \left (\frac {6}{x (-e+x)}\right )}{\log (\log (1-x))}} x \] Output:
x*exp(ln(6/x/(x-exp(1)))/ln(ln(1-x))-x)
\[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=\int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx \] Input:
Integrate[(E^((Log[-6/(E*x - x^2)] - x*Log[Log[1 - x]])/Log[Log[1 - x]])*( (-(E*x) + x^2)*Log[-6/(E*x - x^2)] + (E*(1 - x) - 2*x + 2*x^2)*Log[1 - x]* Log[Log[1 - x]] + (x - 2*x^2 + x^3 + E*(-1 + 2*x - x^2))*Log[1 - x]*Log[Lo g[1 - x]]^2))/((E*(-1 + x) + x - x^2)*Log[1 - x]*Log[Log[1 - x]]^2),x]
Output:
Integrate[(E^((Log[-6/(E*x - x^2)] - x*Log[Log[1 - x]])/Log[Log[1 - x]])*( (-(E*x) + x^2)*Log[-6/(E*x - x^2)] + (E*(1 - x) - 2*x + 2*x^2)*Log[1 - x]* Log[Log[1 - x]] + (x - 2*x^2 + x^3 + E*(-1 + 2*x - x^2))*Log[1 - x]*Log[Lo g[1 - x]]^2))/((E*(-1 + x) + x - x^2)*Log[1 - x]*Log[Log[1 - x]]^2), 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.
| \(\displaystyle \int \frac {\left (\left (2 x^2-2 x+e (1-x)\right ) \log (1-x) \log (\log (1-x))+\left (x^2-e x\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (x^3-2 x^2+e \left (-x^2+2 x-1\right )+x\right ) \log (1-x) \log ^2(\log (1-x))\right ) \exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )}{\left (-x^2+x+e (x-1)\right ) \log (1-x) \log ^2(\log (1-x))} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (-\left (2 x^2-2 x+e (1-x)\right ) \log (1-x) \log (\log (1-x))-\left (x^2-e x\right ) \log \left (-\frac {6}{e x-x^2}\right )-\left (\left (x^3-2 x^2+e \left (-x^2+2 x-1\right )+x\right ) \log (1-x) \log ^2(\log (1-x))\right )\right ) \exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )}{\left (x^2-(1+e) x+e\right ) \log (1-x) \log ^2(\log (1-x))}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {x \log \left (-\frac {6}{(e-x) x}\right ) \exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )}{(1-x) \log (1-x) \log ^2(\log (1-x))}+x \left (-\exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )\right )+\exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )+\frac {(2 x-e) \exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )}{(e-x) \log (\log (1-x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \frac {\exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right ) \log \left (-\frac {6}{(e-x) x}\right )}{\log (1-x) \log ^2(\log (1-x))}dx+\int \frac {\exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right ) \log \left (-\frac {6}{(e-x) x}\right )}{(1-x) \log (1-x) \log ^2(\log (1-x))}dx+\int \exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )dx-\int \exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right ) xdx-2 \int \frac {\exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}\right )}{\log (\log (1-x))}dx+\int \frac {\exp \left (\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}+1\right )}{(e-x) \log (\log (1-x))}dx\) |
Input:
Int[(E^((Log[-6/(E*x - x^2)] - x*Log[Log[1 - x]])/Log[Log[1 - x]])*((-(E*x ) + x^2)*Log[-6/(E*x - x^2)] + (E*(1 - x) - 2*x + 2*x^2)*Log[1 - x]*Log[Lo g[1 - x]] + (x - 2*x^2 + x^3 + E*(-1 + 2*x - x^2))*Log[1 - x]*Log[Log[1 - x]]^2))/((E*(-1 + x) + x - x^2)*Log[1 - x]*Log[Log[1 - x]]^2),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 6.19
\[x \,{\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi \operatorname {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{{\mathrm e}-x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}-x}\right )-2 i \operatorname {csgn}\left (\frac {i}{x \left ({\mathrm e}-x \right )}\right )^{2} \pi +2 i \pi -2 x \ln \left (\ln \left (1-x \right )\right )-2 \ln \left (x \right )+2 \ln \left (3\right )+2 \ln \left (2\right )-2 \ln \left ({\mathrm e}-x \right )}{2 \ln \left (\ln \left (1-x \right )\right )}}\]
Input:
int((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*ln(1-x)*ln(ln(1-x))^2+((1-x)*exp(1) +2*x^2-2*x)*ln(1-x)*ln(ln(1-x))+(-x*exp(1)+x^2)*ln(-6/(x*exp(1)-x^2)))*exp ((-x*ln(ln(1-x))+ln(-6/(x*exp(1)-x^2)))/ln(ln(1-x)))/((-1+x)*exp(1)-x^2+x) /ln(1-x)/ln(ln(1-x))^2,x)
Output:
x*exp(1/2*(I*Pi*csgn(I/x/(exp(1)-x))^3+I*Pi*csgn(I/x/(exp(1)-x))^2*csgn(I/ x)+I*Pi*csgn(I/x/(exp(1)-x))^2*csgn(I/(exp(1)-x))-I*Pi*csgn(I/x/(exp(1)-x) )*csgn(I/x)*csgn(I/(exp(1)-x))-2*I*csgn(I/x/(exp(1)-x))^2*Pi+2*I*Pi-2*x*ln (ln(1-x))-2*ln(x)+2*ln(3)+2*ln(2)-2*ln(exp(1)-x))/ln(ln(1-x)))
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=x e^{\left (-\frac {x \log \left (\log \left (-x + 1\right )\right ) - \log \left (\frac {6}{x^{2} - x e}\right )}{\log \left (\log \left (-x + 1\right )\right )}\right )} \] Input:
integrate((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(1-x)*log(log(1-x))^2+((1- x)*exp(1)+2*x^2-2*x)*log(1-x)*log(log(1-x))+(-exp(1)*x+x^2)*log(-6/(exp(1) *x-x^2)))*exp((-x*log(log(1-x))+log(-6/(exp(1)*x-x^2)))/log(log(1-x)))/((- 1+x)*exp(1)-x^2+x)/log(1-x)/log(log(1-x))^2,x, algorithm="fricas")
Output:
x*e^(-(x*log(log(-x + 1)) - log(6/(x^2 - x*e)))/log(log(-x + 1)))
Time = 57.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=x e^{\frac {- x \log {\left (\log {\left (1 - x \right )} \right )} + \log {\left (- \frac {6}{- x^{2} + e x} \right )}}{\log {\left (\log {\left (1 - x \right )} \right )}}} \] Input:
integrate((((-x**2+2*x-1)*exp(1)+x**3-2*x**2+x)*ln(1-x)*ln(ln(1-x))**2+((1 -x)*exp(1)+2*x**2-2*x)*ln(1-x)*ln(ln(1-x))+(-exp(1)*x+x**2)*ln(-6/(exp(1)* x-x**2)))*exp((-x*ln(ln(1-x))+ln(-6/(exp(1)*x-x**2)))/ln(ln(1-x)))/((-1+x) *exp(1)-x**2+x)/ln(1-x)/ln(ln(1-x))**2,x)
Output:
x*exp((-x*log(log(1 - x)) + log(-6/(-x**2 + E*x)))/log(log(1 - x)))
\[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=\int { -\frac {{\left ({\left (x^{3} - 2 \, x^{2} - {\left (x^{2} - 2 \, x + 1\right )} e + x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right )^{2} + {\left (2 \, x^{2} - {\left (x - 1\right )} e - 2 \, x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right ) + {\left (x^{2} - x e\right )} \log \left (\frac {6}{x^{2} - x e}\right )\right )} e^{\left (-\frac {x \log \left (\log \left (-x + 1\right )\right ) - \log \left (\frac {6}{x^{2} - x e}\right )}{\log \left (\log \left (-x + 1\right )\right )}\right )}}{{\left (x^{2} - {\left (x - 1\right )} e - x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right )^{2}} \,d x } \] Input:
integrate((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(1-x)*log(log(1-x))^2+((1- x)*exp(1)+2*x^2-2*x)*log(1-x)*log(log(1-x))+(-exp(1)*x+x^2)*log(-6/(exp(1) *x-x^2)))*exp((-x*log(log(1-x))+log(-6/(exp(1)*x-x^2)))/log(log(1-x)))/((- 1+x)*exp(1)-x^2+x)/log(1-x)/log(log(1-x))^2,x, algorithm="maxima")
Output:
-integrate(((x^3 - 2*x^2 - (x^2 - 2*x + 1)*e + x)*log(-x + 1)*log(log(-x + 1))^2 + (2*x^2 - (x - 1)*e - 2*x)*log(-x + 1)*log(log(-x + 1)) + (x^2 - x *e)*log(6/(x^2 - x*e)))*e^(-(x*log(log(-x + 1)) - log(6/(x^2 - x*e)))/log( log(-x + 1)))/((x^2 - (x - 1)*e - x)*log(-x + 1)*log(log(-x + 1))^2), x)
\[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=\int { -\frac {{\left ({\left (x^{3} - 2 \, x^{2} - {\left (x^{2} - 2 \, x + 1\right )} e + x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right )^{2} + {\left (2 \, x^{2} - {\left (x - 1\right )} e - 2 \, x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right ) + {\left (x^{2} - x e\right )} \log \left (\frac {6}{x^{2} - x e}\right )\right )} e^{\left (-\frac {x \log \left (\log \left (-x + 1\right )\right ) - \log \left (\frac {6}{x^{2} - x e}\right )}{\log \left (\log \left (-x + 1\right )\right )}\right )}}{{\left (x^{2} - {\left (x - 1\right )} e - x\right )} \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right )\right )^{2}} \,d x } \] Input:
integrate((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(1-x)*log(log(1-x))^2+((1- x)*exp(1)+2*x^2-2*x)*log(1-x)*log(log(1-x))+(-exp(1)*x+x^2)*log(-6/(exp(1) *x-x^2)))*exp((-x*log(log(1-x))+log(-6/(exp(1)*x-x^2)))/log(log(1-x)))/((- 1+x)*exp(1)-x^2+x)/log(1-x)/log(log(1-x))^2,x, algorithm="giac")
Output:
undef
Time = 0.98 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=x\,{\mathrm {e}}^{-x}\,{\left (-\frac {6}{x\,\mathrm {e}-x^2}\right )}^{\frac {1}{\ln \left (\ln \left (1-x\right )\right )}} \] Input:
int(-(exp((log(-6/(x*exp(1) - x^2)) - x*log(log(1 - x)))/log(log(1 - x)))* (log(-6/(x*exp(1) - x^2))*(x*exp(1) - x^2) + log(log(1 - x))*log(1 - x)*(2 *x + exp(1)*(x - 1) - 2*x^2) - log(log(1 - x))^2*log(1 - x)*(x - exp(1)*(x ^2 - 2*x + 1) - 2*x^2 + x^3)))/(log(log(1 - x))^2*log(1 - x)*(x + exp(1)*( x - 1) - x^2)),x)
Output:
x*exp(-x)*(-6/(x*exp(1) - x^2))^(1/log(log(1 - x)))
Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {\log \left (-\frac {6}{e x-x^2}\right )-x \log (\log (1-x))}{\log (\log (1-x))}} \left (\left (-e x+x^2\right ) \log \left (-\frac {6}{e x-x^2}\right )+\left (e (1-x)-2 x+2 x^2\right ) \log (1-x) \log (\log (1-x))+\left (x-2 x^2+x^3+e \left (-1+2 x-x^2\right )\right ) \log (1-x) \log ^2(\log (1-x))\right )}{\left (e (-1+x)+x-x^2\right ) \log (1-x) \log ^2(\log (1-x))} \, dx=\frac {e^{\frac {\mathrm {log}\left (-\frac {6}{e x -x^{2}}\right )}{\mathrm {log}\left (\mathrm {log}\left (1-x \right )\right )}} x}{e^{x}} \] Input:
int((((-x^2+2*x-1)*exp(1)+x^3-2*x^2+x)*log(1-x)*log(log(1-x))^2+((1-x)*exp (1)+2*x^2-2*x)*log(1-x)*log(log(1-x))+(-exp(1)*x+x^2)*log(-6/(exp(1)*x-x^2 )))*exp((-x*log(log(1-x))+log(-6/(exp(1)*x-x^2)))/log(log(1-x)))/((-1+x)*e xp(1)-x^2+x)/log(1-x)/log(log(1-x))^2,x)
Output:
(e**(log(( - 6)/(e*x - x**2))/log(log( - x + 1)))*x)/e**x