Integrand size = 83, antiderivative size = 25 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=5-e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \]
\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx \]
Integrate[(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*(-5 + E^x*(2 - 2*x) + (x + 2*E^x*x)*Log[4/x]))/(25*x + 4*E^(2*x)*x - 10*x^2 + x^3 + E^x*(-20*x + 4*x ^2)),x]
Integrate[(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*(-5 + E^x*(2 - 2*x) + (x + 2*E^x*x)*Log[4/x]))/(25*x + 4*E^(2*x)*x - 10*x^2 + x^3 + E^x*(-20*x + 4*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.
\(\displaystyle \int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} \left (e^x (2-2 x)+\left (2 e^x x+x\right ) \log \left (\frac {4}{x}\right )-5\right )}{x^3-10 x^2+e^x \left (4 x^2-20 x\right )+4 e^{2 x} x+25 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} \left (e^x (2-2 x)+\left (2 e^x x+x\right ) \log \left (\frac {4}{x}\right )-5\right )}{\left (-x-2 e^x+5\right )^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} \left (-x+x \log \left (\frac {4}{x}\right )+1\right )}{x \left (x+2 e^x-5\right )}-\frac {(x-6) e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} \left (\log \left (\frac {4}{x}\right )-1\right )}{\left (x+2 e^x-5\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -6 \int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}}}{\left (x+2 e^x-5\right )^2}dx+\int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} x}{\left (x+2 e^x-5\right )^2}dx-\int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}}}{x+2 e^x-5}dx+\int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}}}{x \left (x+2 e^x-5\right )}dx+6 \int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} \log \left (\frac {4}{x}\right )}{\left (x+2 e^x-5\right )^2}dx-\int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} x \log \left (\frac {4}{x}\right )}{\left (x+2 e^x-5\right )^2}dx+\int \frac {e^{\frac {\log \left (\frac {4}{x}\right )-1}{x+2 e^x-5}} \log \left (\frac {4}{x}\right )}{x+2 e^x-5}dx\) |
Int[(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*(-5 + E^x*(2 - 2*x) + (x + 2*E^x *x)*Log[4/x]))/(25*x + 4*E^(2*x)*x - 10*x^2 + x^3 + E^x*(-20*x + 4*x^2)),x ]
3.7.8.3.1 Defintions of rubi rules used
Time = 3.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(-{\mathrm e}^{\frac {\ln \left (\frac {4}{x}\right )-1}{2 \,{\mathrm e}^{x}+x -5}}\) | \(22\) |
risch | \(-{\mathrm e}^{\frac {2 \ln \left (2\right )-\ln \left (x \right )-1}{2 \,{\mathrm e}^{x}+x -5}}\) | \(24\) |
int(((2*exp(x)*x+x)*ln(4/x)+(2-2*x)*exp(x)-5)*exp((ln(4/x)-1)/(2*exp(x)+x- 5))/(4*x*exp(x)^2+(4*x^2-20*x)*exp(x)+x^3-10*x^2+25*x),x,method=_RETURNVER BOSE)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )} \]
integrate(((2*exp(x)*x+x)*log(4/x)+(2-2*x)*exp(x)-5)*exp((log(4/x)-1)/(2*e xp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*x)*exp(x)+x^3-10*x^2+25*x),x, algorith m=\
Time = 0.49 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=- e^{\frac {\log {\left (\frac {4}{x} \right )} - 1}{x + 2 e^{x} - 5}} \]
integrate(((2*exp(x)*x+x)*ln(4/x)+(2-2*x)*exp(x)-5)*exp((ln(4/x)-1)/(2*exp (x)+x-5))/(4*x*exp(x)**2+(4*x**2-20*x)*exp(x)+x**3-10*x**2+25*x),x)
Time = 0.38 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-e^{\left (\frac {2 \, \log \left (2\right )}{x + 2 \, e^{x} - 5} - \frac {\log \left (x\right )}{x + 2 \, e^{x} - 5} - \frac {1}{x + 2 \, e^{x} - 5}\right )} \]
integrate(((2*exp(x)*x+x)*log(4/x)+(2-2*x)*exp(x)-5)*exp((log(4/x)-1)/(2*e xp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*x)*exp(x)+x^3-10*x^2+25*x),x, algorith m=\
\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int { -\frac {{\left (2 \, {\left (x - 1\right )} e^{x} - {\left (2 \, x e^{x} + x\right )} \log \left (\frac {4}{x}\right ) + 5\right )} e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )}}{x^{3} - 10 \, x^{2} + 4 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} - 5 \, x\right )} e^{x} + 25 \, x} \,d x } \]
integrate(((2*exp(x)*x+x)*log(4/x)+(2-2*x)*exp(x)-5)*exp((log(4/x)-1)/(2*e xp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*x)*exp(x)+x^3-10*x^2+25*x),x, algorith m=\
integrate(-(2*(x - 1)*e^x - (2*x*e^x + x)*log(4/x) + 5)*e^((log(4/x) - 1)/ (x + 2*e^x - 5))/(x^3 - 10*x^2 + 4*x*e^(2*x) + 4*(x^2 - 5*x)*e^x + 25*x), x)
Time = 9.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-{\mathrm {e}}^{-\frac {1}{x+2\,{\mathrm {e}}^x-5}}\,{\left (\frac {4}{x}\right )}^{\frac {1}{x+2\,{\mathrm {e}}^x-5}} \]