Integrand size = 101, antiderivative size = 26 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=\frac {2-x}{x \log \left (e^4+e^x-x-\log (x)\right )} \]
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=\frac {2-x}{x \log \left (e^4+e^x-x-\log (x)\right )} \]
Integrate[(-2 - x + x^2 + E^x*(2*x - x^2) + (2*E^4 + 2*E^x - 2*x - 2*Log[x ])*Log[E^4 + E^x - x - Log[x]])/((-(E^4*x^2) - E^x*x^2 + x^3 + x^2*Log[x]) *Log[E^4 + E^x - x - Log[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.
\(\displaystyle \int \frac {x^2+e^x \left (2 x-x^2\right )-x+\left (-2 x+2 e^x-2 \log (x)+2 e^4\right ) \log \left (-x+e^x-\log (x)+e^4\right )-2}{\left (x^3-e^x x^2-e^4 x^2+x^2 \log (x)\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(2-x) \left (-x^2+\left (1+e^4\right ) x-x \log (x)+1\right )}{x^2 \left (-x+e^x-\log (x)+e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}+\frac {x^2-2 x-2 \log \left (-x+e^x-\log (x)+e^4\right )}{x^2 \log ^2\left (-x+e^x-\log (x)+e^4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{x^2 \left (x-e^x+\log (x)-e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx-2 \int \frac {1}{x^2 \log \left (-x+e^x-\log (x)+e^4\right )}dx+\int \frac {1}{\log ^2\left (-x+e^x-\log (x)+e^4\right )}dx-2 \int \frac {1}{x \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx-\left (1+e^4\right ) \int \frac {1}{\left (-x+e^x-\log (x)+e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx-2 \int \frac {1}{\left (-x+e^x-\log (x)+e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx+2 \left (1+e^4\right ) \int \frac {1}{x \left (-x+e^x-\log (x)+e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx+\int \frac {\log (x)}{\left (-x+e^x-\log (x)+e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx+\int \frac {1}{x \left (x-e^x+\log (x)-e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx-\int \frac {x}{\left (x-e^x+\log (x)-e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx+2 \int \frac {\log (x)}{x \left (x-e^x+\log (x)-e^4\right ) \log ^2\left (-x+e^x-\log (x)+e^4\right )}dx\) |
Int[(-2 - x + x^2 + E^x*(2*x - x^2) + (2*E^4 + 2*E^x - 2*x - 2*Log[x])*Log [E^4 + E^x - x - Log[x]])/((-(E^4*x^2) - E^x*x^2 + x^3 + x^2*Log[x])*Log[E ^4 + E^x - x - Log[x]]^2),x]
3.26.31.3.1 Defintions of rubi rules used
Time = 4.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {-2+x}{x \ln \left (-\ln \left (x \right )+{\mathrm e}^{x}+{\mathrm e}^{4}-x \right )}\) | \(24\) |
parallelrisch | \(\frac {2-x}{x \ln \left (-\ln \left (x \right )+{\mathrm e}^{x}+{\mathrm e}^{4}-x \right )}\) | \(25\) |
int(((-2*ln(x)+2*exp(x)+2*exp(4)-2*x)*ln(-ln(x)+exp(x)+exp(4)-x)+(-x^2+2*x )*exp(x)+x^2-x-2)/(x^2*ln(x)-exp(x)*x^2-x^2*exp(4)+x^3)/ln(-ln(x)+exp(x)+e xp(4)-x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=-\frac {x - 2}{x \log \left (-x + e^{4} + e^{x} - \log \left (x\right )\right )} \]
integrate(((-2*log(x)+2*exp(x)+2*exp(4)-2*x)*log(-log(x)+exp(x)+exp(4)-x)+ (-x^2+2*x)*exp(x)+x^2-x-2)/(x^2*log(x)-exp(x)*x^2-x^2*exp(4)+x^3)/log(-log (x)+exp(x)+exp(4)-x)^2,x, algorithm=\
Time = 0.48 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=\frac {2 - x}{x \log {\left (- x + e^{x} - \log {\left (x \right )} + e^{4} \right )}} \]
integrate(((-2*ln(x)+2*exp(x)+2*exp(4)-2*x)*ln(-ln(x)+exp(x)+exp(4)-x)+(-x **2+2*x)*exp(x)+x**2-x-2)/(x**2*ln(x)-exp(x)*x**2-x**2*exp(4)+x**3)/ln(-ln (x)+exp(x)+exp(4)-x)**2,x)
Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=-\frac {x - 2}{x \log \left (-x + e^{4} + e^{x} - \log \left (x\right )\right )} \]
integrate(((-2*log(x)+2*exp(x)+2*exp(4)-2*x)*log(-log(x)+exp(x)+exp(4)-x)+ (-x^2+2*x)*exp(x)+x^2-x-2)/(x^2*log(x)-exp(x)*x^2-x^2*exp(4)+x^3)/log(-log (x)+exp(x)+exp(4)-x)^2,x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=-\frac {x - 2}{x \log \left (-x + e^{4} + e^{x} - \log \left (x\right )\right )} \]
integrate(((-2*log(x)+2*exp(x)+2*exp(4)-2*x)*log(-log(x)+exp(x)+exp(4)-x)+ (-x^2+2*x)*exp(x)+x^2-x-2)/(x^2*log(x)-exp(x)*x^2-x^2*exp(4)+x^3)/log(-log (x)+exp(x)+exp(4)-x)^2,x, algorithm=\
Time = 12.84 (sec) , antiderivative size = 140, normalized size of antiderivative = 5.38 \[ \int \frac {-2-x+x^2+e^x \left (2 x-x^2\right )+\left (2 e^4+2 e^x-2 x-2 \log (x)\right ) \log \left (e^4+e^x-x-\log (x)\right )}{\left (-e^4 x^2-e^x x^2+x^3+x^2 \log (x)\right ) \log ^2\left (e^4+e^x-x-\log (x)\right )} \, dx=\frac {2\,\left (2\,x+x\,{\mathrm {e}}^4+x^2\,{\mathrm {e}}^4+x^3\,{\mathrm {e}}^4+x^2-x^4+1\right )}{x^2\,\left (x-x\,{\mathrm {e}}^x+1\right )\,\left (x^2+x+1\right )}-\frac {2}{x^2}-\frac {2\,\ln \left (x\right )}{x\,\left (x-x\,{\mathrm {e}}^x+1\right )}-\frac {\frac {x-2}{x}-\frac {2\,\ln \left ({\mathrm {e}}^4-x+{\mathrm {e}}^x-\ln \left (x\right )\right )\,\left (x-{\mathrm {e}}^4-{\mathrm {e}}^x+\ln \left (x\right )\right )}{x\,\left (x-x\,{\mathrm {e}}^x+1\right )}}{\ln \left ({\mathrm {e}}^4-x+{\mathrm {e}}^x-\ln \left (x\right )\right )} \]
int((x + log(exp(4) - x + exp(x) - log(x))*(2*x - 2*exp(4) - 2*exp(x) + 2* log(x)) - exp(x)*(2*x - x^2) - x^2 + 2)/(log(exp(4) - x + exp(x) - log(x)) ^2*(x^2*exp(x) - x^2*log(x) + x^2*exp(4) - x^3)),x)