Integrand size = 153, antiderivative size = 35 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=2 x-x^2-\frac {\log (x+\log (x-(2-x) x))}{\log \left (4 e^{e^x}\right )} \]
Time = 0.48 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=2 x-x^2-\frac {\log (x+\log ((-1+x) x))}{\log \left (4 e^{e^x}\right )} \]
Integrate[((1 - x - x^2)*Log[4*E^E^x] + Log[4*E^E^x]^2*(-2*x^2 + 4*x^3 - 2 *x^4 + (-2*x + 4*x^2 - 2*x^3)*Log[-x + x^2]) + (E^x*(-x^2 + x^3) + E^x*(-x + x^2)*Log[-x + x^2])*Log[x + Log[-x + x^2]])/(Log[4*E^E^x]^2*(-x^2 + x^3 + (-x + x^2)*Log[-x + 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.
\(\displaystyle \int \frac {\left (-x^2-x+1\right ) \log \left (4 e^{e^x}\right )+\left (e^x \left (x^2-x\right ) \log \left (x^2-x\right )+e^x \left (x^3-x^2\right )\right ) \log \left (\log \left (x^2-x\right )+x\right )+\left (-2 x^4+4 x^3-2 x^2+\left (-2 x^3+4 x^2-2 x\right ) \log \left (x^2-x\right )\right ) \log ^2\left (4 e^{e^x}\right )}{\log ^2\left (4 e^{e^x}\right ) \left (x^3-x^2+\left (x^2-x\right ) \log \left (x^2-x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {x^2+x-1}{(x-1) x \log \left (4 e^{e^x}\right ) (x+\log ((x-1) x))}-2 x+\frac {e^x \log (x+\log ((x-1) x))}{\log ^2\left (4 e^{e^x}\right )}+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^x \log (x+\log ((x-1) x))}{\log ^2\left (4 e^{e^x}\right )}dx-\int \frac {1}{\log \left (4 e^{e^x}\right ) (x+\log ((x-1) x))}dx-\int \frac {1}{(x-1) \log \left (4 e^{e^x}\right ) (x+\log ((x-1) x))}dx-\int \frac {1}{x \log \left (4 e^{e^x}\right ) (x+\log ((x-1) x))}dx-x^2+2 x\) |
Int[((1 - x - x^2)*Log[4*E^E^x] + Log[4*E^E^x]^2*(-2*x^2 + 4*x^3 - 2*x^4 + (-2*x + 4*x^2 - 2*x^3)*Log[-x + x^2]) + (E^x*(-x^2 + x^3) + E^x*(-x + x^2 )*Log[-x + x^2])*Log[x + Log[-x + x^2]])/(Log[4*E^E^x]^2*(-x^2 + x^3 + (-x + x^2)*Log[-x + x^2])),x]
3.30.37.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.51 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.37
\[-x^{2}+2 x -\frac {2 i \ln \left (\ln \left (-1+x \right )+\ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (-1+x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (-1+x \right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (-1+x \right )\right )+\operatorname {csgn}\left (i \left (-1+x \right )\right )\right )}{2}+x \right )}{4 i \ln \left (2\right )+2 i \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )}\]
int((((-2*x^3+4*x^2-2*x)*ln(x^2-x)-2*x^4+4*x^3-2*x^2)*ln(4*exp(exp(x)))^2+ (-x^2-x+1)*ln(4*exp(exp(x)))+((x^2-x)*exp(x)*ln(x^2-x)+(x^3-x^2)*exp(x))*l n(ln(x^2-x)+x))/((x^2-x)*ln(x^2-x)+x^3-x^2)/ln(4*exp(exp(x)))^2,x)
-x^2+2*x-2*I*ln(ln(-1+x)+ln(x)-1/2*I*Pi*csgn(I*x*(-1+x))*(-csgn(I*x*(-1+x) )+csgn(I*x))*(-csgn(I*x*(-1+x))+csgn(I*(-1+x)))+x)/(4*I*ln(2)+2*I*ln(exp(e xp(x))))
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=-\frac {{\left (x^{2} - 2 \, x\right )} e^{x} + 2 \, {\left (x^{2} - 2 \, x\right )} \log \left (2\right ) + \log \left (x + \log \left (x^{2} - x\right )\right )}{e^{x} + 2 \, \log \left (2\right )} \]
integrate((((-2*x^3+4*x^2-2*x)*log(x^2-x)-2*x^4+4*x^3-2*x^2)*log(4*exp(exp (x)))^2+(-x^2-x+1)*log(4*exp(exp(x)))+((x^2-x)*exp(x)*log(x^2-x)+(x^3-x^2) *exp(x))*log(log(x^2-x)+x))/((x^2-x)*log(x^2-x)+x^3-x^2)/log(4*exp(exp(x)) )^2,x, algorithm=\
Time = 3.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.69 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=- x^{2} + 2 x - \frac {\log {\left (x + \log {\left (x^{2} - x \right )} \right )}}{e^{x} + 2 \log {\left (2 \right )}} \]
integrate((((-2*x**3+4*x**2-2*x)*ln(x**2-x)-2*x**4+4*x**3-2*x**2)*ln(4*exp (exp(x)))**2+(-x**2-x+1)*ln(4*exp(exp(x)))+((x**2-x)*exp(x)*ln(x**2-x)+(x* *3-x**2)*exp(x))*ln(ln(x**2-x)+x))/((x**2-x)*ln(x**2-x)+x**3-x**2)/ln(4*ex p(exp(x)))**2,x)
Time = 0.38 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=-\frac {2 \, x^{2} \log \left (2\right ) + {\left (x^{2} - 2 \, x\right )} e^{x} - 4 \, x \log \left (2\right ) + \log \left (x + \log \left (x - 1\right ) + \log \left (x\right )\right )}{e^{x} + 2 \, \log \left (2\right )} \]
integrate((((-2*x^3+4*x^2-2*x)*log(x^2-x)-2*x^4+4*x^3-2*x^2)*log(4*exp(exp (x)))^2+(-x^2-x+1)*log(4*exp(exp(x)))+((x^2-x)*exp(x)*log(x^2-x)+(x^3-x^2) *exp(x))*log(log(x^2-x)+x))/((x^2-x)*log(x^2-x)+x^3-x^2)/log(4*exp(exp(x)) )^2,x, algorithm=\
-(2*x^2*log(2) + (x^2 - 2*x)*e^x - 4*x*log(2) + log(x + log(x - 1) + log(x )))/(e^x + 2*log(2))
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (30) = 60\).
Time = 0.38 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.74 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=-\frac {x^{2} e^{x} + 2 \, x^{2} \log \left (2\right ) - 2 \, x e^{x} - 4 \, x \log \left (2\right ) + 4 \, e^{x} \log \left (e^{x} + 2 \, \log \left (2\right )\right ) + 8 \, \log \left (2\right ) \log \left (e^{x} + 2 \, \log \left (2\right )\right ) - 4 \, e^{x} \log \left (-e^{x} - 2 \, \log \left (2\right )\right ) - 8 \, \log \left (2\right ) \log \left (-e^{x} - 2 \, \log \left (2\right )\right ) + \log \left (x + \log \left (x - 1\right ) + \log \left (x\right )\right )}{e^{x} + 2 \, \log \left (2\right )} \]
integrate((((-2*x^3+4*x^2-2*x)*log(x^2-x)-2*x^4+4*x^3-2*x^2)*log(4*exp(exp (x)))^2+(-x^2-x+1)*log(4*exp(exp(x)))+((x^2-x)*exp(x)*log(x^2-x)+(x^3-x^2) *exp(x))*log(log(x^2-x)+x))/((x^2-x)*log(x^2-x)+x^3-x^2)/log(4*exp(exp(x)) )^2,x, algorithm=\
-(x^2*e^x + 2*x^2*log(2) - 2*x*e^x - 4*x*log(2) + 4*e^x*log(e^x + 2*log(2) ) + 8*log(2)*log(e^x + 2*log(2)) - 4*e^x*log(-e^x - 2*log(2)) - 8*log(2)*l og(-e^x - 2*log(2)) + log(x + log(x - 1) + log(x)))/(e^x + 2*log(2))
Time = 9.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {\left (1-x-x^2\right ) \log \left (4 e^{e^x}\right )+\log ^2\left (4 e^{e^x}\right ) \left (-2 x^2+4 x^3-2 x^4+\left (-2 x+4 x^2-2 x^3\right ) \log \left (-x+x^2\right )\right )+\left (e^x \left (-x^2+x^3\right )+e^x \left (-x+x^2\right ) \log \left (-x+x^2\right )\right ) \log \left (x+\log \left (-x+x^2\right )\right )}{\log ^2\left (4 e^{e^x}\right ) \left (-x^2+x^3+\left (-x+x^2\right ) \log \left (-x+x^2\right )\right )} \, dx=2\,x-\frac {\ln \left (x+\ln \left (x^2-x\right )\right )}{\ln \left (4\right )+{\mathrm {e}}^x}-x^2 \]