Integrand size = 109, antiderivative size = 21 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x+\log ^2\left (4 \left (-4+\frac {2}{\log \left (e^x-x\right )}\right )\right ) \]
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x+\log ^2\left (8 \left (-2+\frac {1}{\log \left (e^x-x\right )}\right )\right ) \]
Integrate[((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2 + (-2 + 2*E^x)*Log[(8 - 16*Log[E^x - x])/Log[E^x - x]])/((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^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.
\(\displaystyle \int \frac {\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (x-e^x\right ) \log \left (e^x-x\right )+\left (2 e^x-2\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (x-e^x\right ) \log \left (e^x-x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )\right )-\left (x-e^x\right ) \log \left (e^x-x\right )-\left (2 e^x-2\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (e^x-x\right ) \left (1-2 \log \left (e^x-x\right )\right ) \log \left (e^x-x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \log ^2\left (e^x-x\right )-\log \left (e^x-x\right )+2 \log \left (8 \left (\frac {1}{\log \left (e^x-x\right )}-2\right )\right )}{\log \left (e^x-x\right ) \left (2 \log \left (e^x-x\right )-1\right )}+\frac {2 (x-1) \log \left (8 \left (\frac {1}{\log \left (e^x-x\right )}-2\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (2 \log \left (e^x-x\right )-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {\log \left (8 \left (\frac {1}{\log \left (e^x-x\right )}-2\right )\right )}{\log \left (e^x-x\right ) \left (2 \log \left (e^x-x\right )-1\right )}dx-2 \int \frac {\log \left (8 \left (\frac {1}{\log \left (e^x-x\right )}-2\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (2 \log \left (e^x-x\right )-1\right )}dx+2 \int \frac {x \log \left (8 \left (\frac {1}{\log \left (e^x-x\right )}-2\right )\right )}{\left (e^x-x\right ) \log \left (e^x-x\right ) \left (2 \log \left (e^x-x\right )-1\right )}dx+x\) |
Int[((-E^x + x)*Log[E^x - x] + (2*E^x - 2*x)*Log[E^x - x]^2 + (-2 + 2*E^x) *Log[(8 - 16*Log[E^x - x])/Log[E^x - x]])/((-E^x + x)*Log[E^x - x] + (2*E^ x - 2*x)*Log[E^x - x]^2),x]
3.6.98.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.97 (sec) , antiderivative size = 567, normalized size of antiderivative = 27.00
int(((2*exp(x)-2)*ln((-16*ln(exp(x)-x)+8)/ln(exp(x)-x))+(2*exp(x)-2*x)*ln( exp(x)-x)^2+(x-exp(x))*ln(exp(x)-x))/((2*exp(x)-2*x)*ln(exp(x)-x)^2+(x-exp (x))*ln(exp(x)-x)),x,method=_RETURNVERBOSE)
ln(ln(exp(x)-x)-1/2)^2-2*ln(ln(exp(x)-x))*ln(ln(exp(x)-x)-1/2)+ln(ln(exp(x )-x))^2+I*Pi*ln(ln(exp(x)-x)-1/2)*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp (x)-x)*(ln(exp(x)-x)-1/2))^2+I*Pi*ln(ln(exp(x)-x))*csgn(I*(ln(exp(x)-x)-1/ 2))*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))-I*Pi*ln(l n(exp(x)-x))*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^ 2-I*Pi*ln(ln(exp(x)-x))*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x)*(ln (exp(x)-x)-1/2))^2-2*I*Pi*ln(ln(exp(x)-x))+I*Pi*ln(ln(exp(x)-x)-1/2)*csgn( I/ln(exp(x)-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-I*Pi*ln(ln(exp(x )-x))*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^3+2*I*Pi*ln(ln(exp(x)-x))*cs gn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2+I*Pi*ln(ln(exp(x)-x)-1/2)*csgn(I/l n(exp(x)-x)*(ln(exp(x)-x)-1/2))^3+2*I*Pi*ln(ln(exp(x)-x)-1/2)-2*I*Pi*ln(ln (exp(x)-x)-1/2)*csgn(I/ln(exp(x)-x)*(ln(exp(x)-x)-1/2))^2-I*Pi*ln(ln(exp(x )-x)-1/2)*csgn(I*(ln(exp(x)-x)-1/2))*csgn(I/ln(exp(x)-x))*csgn(I/ln(exp(x) -x)*(ln(exp(x)-x)-1/2))-8*ln(2)*ln(ln(exp(x)-x))+8*ln(2)*ln(ln(exp(x)-x)-1 /2)+x
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=\log \left (-\frac {8 \, {\left (2 \, \log \left (-x + e^{x}\right ) - 1\right )}}{\log \left (-x + e^{x}\right )}\right )^{2} + x \]
integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x) -2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x) -x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm=\
Time = 0.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=x + \log {\left (\frac {8 - 16 \log {\left (- x + e^{x} \right )}}{\log {\left (- x + e^{x} \right )}} \right )}^{2} \]
integrate(((2*exp(x)-2)*ln((-16*ln(exp(x)-x)+8)/ln(exp(x)-x))+(2*exp(x)-2* x)*ln(exp(x)-x)**2+(x-exp(x))*ln(exp(x)-x))/((2*exp(x)-2*x)*ln(exp(x)-x)** 2+(x-exp(x))*ln(exp(x)-x)),x)
Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.52 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=-2 \, {\left (-i \, \pi - 3 \, \log \left (2\right ) + \log \left (\log \left (-x + e^{x}\right )\right )\right )} \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) + \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} - 2 \, {\left (i \, \pi + 3 \, \log \left (2\right )\right )} \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \]
integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x) -2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x) -x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm=\
-2*(-I*pi - 3*log(2) + log(log(-x + e^x)))*log(2*log(-x + e^x) - 1) + log( 2*log(-x + e^x) - 1)^2 - 2*(I*pi + 3*log(2))*log(log(-x + e^x)) + log(log( -x + e^x))^2 + x
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (18) = 36\).
Time = 0.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.62 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx=-\log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right )^{2} + 2 \, \log \left (2 \, \log \left (-x + e^{x}\right ) - 1\right ) \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) - 2 \, \log \left (-16 \, \log \left (-x + e^{x}\right ) + 8\right ) \log \left (\log \left (-x + e^{x}\right )\right ) + \log \left (\log \left (-x + e^{x}\right )\right )^{2} + x \]
integrate(((2*exp(x)-2)*log((-16*log(exp(x)-x)+8)/log(exp(x)-x))+(2*exp(x) -2*x)*log(exp(x)-x)^2+(x-exp(x))*log(exp(x)-x))/((2*exp(x)-2*x)*log(exp(x) -x)^2+(x-exp(x))*log(exp(x)-x)),x, algorithm=\
-log(2*log(-x + e^x) - 1)^2 + 2*log(2*log(-x + e^x) - 1)*log(-16*log(-x + e^x) + 8) - 2*log(-16*log(-x + e^x) + 8)*log(log(-x + e^x)) + log(log(-x + e^x))^2 + x
Time = 9.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )+\left (-2+2 e^x\right ) \log \left (\frac {8-16 \log \left (e^x-x\right )}{\log \left (e^x-x\right )}\right )}{\left (-e^x+x\right ) \log \left (e^x-x\right )+\left (2 e^x-2 x\right ) \log ^2\left (e^x-x\right )} \, dx={\ln \left (\frac {8}{\ln \left ({\mathrm {e}}^x-x\right )}-16\right )}^2+x \]