Integrand size = 91, antiderivative size = 28 \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=\log \left (x+\frac {e^x (4-x) x \log \left (\frac {3}{x}\right )}{-x+x^2}\right ) \]
Time = 1.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=-\log (1-x)+\log \left (-x+x^2+4 e^x \log \left (\frac {3}{x}\right )-e^x x \log \left (\frac {3}{x}\right )\right ) \]
Integrate[(-x + 2*x^2 - x^3 + E^x*(-4 + 5*x - x^2) + E^x*(7*x - 5*x^2 + x^ 3)*Log[3/x])/(-x^2 + 2*x^3 - x^4 + E^x*(4*x - 5*x^2 + x^3)*Log[3/x]),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.
\(\displaystyle \int \frac {-x^3+2 x^2+e^x \left (-x^2+5 x-4\right )+e^x \left (x^3-5 x^2+7 x\right ) \log \left (\frac {3}{x}\right )-x}{-x^4+2 x^3-x^2+e^x \left (x^3-5 x^2+4 x\right ) \log \left (\frac {3}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^3-2 x^2-e^x \left (-x^2+5 x-4\right )-e^x \left (x^3-5 x^2+7 x\right ) \log \left (\frac {3}{x}\right )+x}{(1-x) x \left (-x^2+x+e^x x \log \left (\frac {3}{x}\right )-4 e^x \log \left (\frac {3}{x}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^3 \log \left (\frac {3}{x}\right )-x^2-5 x^2 \log \left (\frac {3}{x}\right )+5 x+7 x \log \left (\frac {3}{x}\right )-4}{(x-4) (x-1) x \log \left (\frac {3}{x}\right )}-\frac {x^3 \log \left (\frac {3}{x}\right )-x^2-6 x^2 \log \left (\frac {3}{x}\right )+5 x+12 x \log \left (\frac {3}{x}\right )-4 \log \left (\frac {3}{x}\right )-4}{(x-4) \log \left (\frac {3}{x}\right ) \left (x^2-x-e^x x \log \left (\frac {3}{x}\right )+4 e^x \log \left (\frac {3}{x}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {1}{x^2-e^x \log \left (\frac {3}{x}\right ) x-x+4 e^x \log \left (\frac {3}{x}\right )}dx-12 \int \frac {1}{(x-4) \left (x^2-e^x \log \left (\frac {3}{x}\right ) x-x+4 e^x \log \left (\frac {3}{x}\right )\right )}dx+2 \int \frac {x}{x^2-e^x \log \left (\frac {3}{x}\right ) x-x+4 e^x \log \left (\frac {3}{x}\right )}dx-\int \frac {x^2}{x^2-e^x \log \left (\frac {3}{x}\right ) x-x+4 e^x \log \left (\frac {3}{x}\right )}dx+\int \frac {x}{\log \left (\frac {3}{x}\right ) \left (x^2-e^x \log \left (\frac {3}{x}\right ) x-x+4 e^x \log \left (\frac {3}{x}\right )\right )}dx+\int \frac {1}{\log \left (\frac {3}{x}\right ) \left (-x^2+e^x \log \left (\frac {3}{x}\right ) x+x-4 e^x \log \left (\frac {3}{x}\right )\right )}dx+x-\log (1-x)+\log (4-x)+\log \left (\log \left (\frac {3}{x}\right )\right )\) |
Int[(-x + 2*x^2 - x^3 + E^x*(-4 + 5*x - x^2) + E^x*(7*x - 5*x^2 + x^3)*Log [3/x])/(-x^2 + 2*x^3 - x^4 + E^x*(4*x - 5*x^2 + x^3)*Log[3/x]),x]
3.19.22.3.1 Defintions of rubi rules used
Time = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
method | result | size |
norman | \(-\ln \left (-1+x \right )+\ln \left (-x \,{\mathrm e}^{x} \ln \left (\frac {3}{x}\right )+x^{2}+4 \,{\mathrm e}^{x} \ln \left (\frac {3}{x}\right )-x \right )\) | \(37\) |
parallelrisch | \(-\ln \left (-1+x \right )+\ln \left (-x \,{\mathrm e}^{x} \ln \left (\frac {3}{x}\right )+x^{2}+4 \,{\mathrm e}^{x} \ln \left (\frac {3}{x}\right )-x \right )\) | \(37\) |
risch | \(x +\ln \left (x -4\right )-\ln \left (-1+x \right )+\ln \left (\frac {-x \ln \left (3\right )+4 \ln \left (3\right )+x^{2} {\mathrm e}^{-x}-x \,{\mathrm e}^{-x}+x \ln \left (x \right )-4 \ln \left (x \right )}{x -4}\right )\) | \(53\) |
default | \(-\ln \left (-1+x \right )+\ln \left (x \,{\mathrm e}^{x} \ln \left (x \right )-{\mathrm e}^{x} x \left (\ln \left (\frac {3}{x}\right )+\ln \left (x \right )\right )+x^{2}-4 \,{\mathrm e}^{x} \ln \left (x \right )+4 \,{\mathrm e}^{x} \left (\ln \left (\frac {3}{x}\right )+\ln \left (x \right )\right )-x \right )\) | \(55\) |
int(((x^3-5*x^2+7*x)*exp(x)*ln(3/x)+(-x^2+5*x-4)*exp(x)-x^3+2*x^2-x)/((x^3 -5*x^2+4*x)*exp(x)*ln(3/x)-x^4+2*x^3-x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=x - \log \left (x - 1\right ) + \log \left (x - 4\right ) + \log \left (\frac {{\left ({\left (x - 4\right )} e^{x} \log \left (\frac {3}{x}\right ) - x^{2} + x\right )} e^{\left (-x\right )}}{x - 4}\right ) \]
integrate(((x^3-5*x^2+7*x)*exp(x)*log(3/x)+(-x^2+5*x-4)*exp(x)-x^3+2*x^2-x )/((x^3-5*x^2+4*x)*exp(x)*log(3/x)-x^4+2*x^3-x^2),x, algorithm=\
Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=\log {\left (x - 4 \right )} - \log {\left (x - 1 \right )} + \log {\left (\frac {- x^{2} + x}{x \log {\left (\frac {3}{x} \right )} - 4 \log {\left (\frac {3}{x} \right )}} + e^{x} \right )} + \log {\left (\log {\left (\frac {3}{x} \right )} \right )} \]
integrate(((x**3-5*x**2+7*x)*exp(x)*ln(3/x)+(-x**2+5*x-4)*exp(x)-x**3+2*x* *2-x)/((x**3-5*x**2+4*x)*exp(x)*ln(3/x)-x**4+2*x**3-x**2),x)
log(x - 4) - log(x - 1) + log((-x**2 + x)/(x*log(3/x) - 4*log(3/x)) + exp( x)) + log(log(3/x))
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (26) = 52\).
Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=-\log \left (x - 1\right ) + \log \left (x - 4\right ) + \log \left (-\frac {x^{2} - {\left (x \log \left (3\right ) - {\left (x - 4\right )} \log \left (x\right ) - 4 \, \log \left (3\right )\right )} e^{x} - x}{x \log \left (3\right ) - {\left (x - 4\right )} \log \left (x\right ) - 4 \, \log \left (3\right )}\right ) + \log \left (-\log \left (3\right ) + \log \left (x\right )\right ) \]
integrate(((x^3-5*x^2+7*x)*exp(x)*log(3/x)+(-x^2+5*x-4)*exp(x)-x^3+2*x^2-x )/((x^3-5*x^2+4*x)*exp(x)*log(3/x)-x^4+2*x^3-x^2),x, algorithm=\
-log(x - 1) + log(x - 4) + log(-(x^2 - (x*log(3) - (x - 4)*log(x) - 4*log( 3))*e^x - x)/(x*log(3) - (x - 4)*log(x) - 4*log(3))) + log(-log(3) + log(x ))
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=\log \left (x e^{x} \log \left (\frac {3}{x}\right ) - x^{2} - 4 \, e^{x} \log \left (\frac {3}{x}\right ) + x\right ) - \log \left (x - 1\right ) \]
integrate(((x^3-5*x^2+7*x)*exp(x)*log(3/x)+(-x^2+5*x-4)*exp(x)-x^3+2*x^2-x )/((x^3-5*x^2+4*x)*exp(x)*log(3/x)-x^4+2*x^3-x^2),x, algorithm=\
Timed out. \[ \int \frac {-x+2 x^2-x^3+e^x \left (-4+5 x-x^2\right )+e^x \left (7 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )}{-x^2+2 x^3-x^4+e^x \left (4 x-5 x^2+x^3\right ) \log \left (\frac {3}{x}\right )} \, dx=\int \frac {x+{\mathrm {e}}^x\,\left (x^2-5\,x+4\right )-2\,x^2+x^3-{\mathrm {e}}^x\,\ln \left (\frac {3}{x}\right )\,\left (x^3-5\,x^2+7\,x\right )}{x^2-2\,x^3+x^4-{\mathrm {e}}^x\,\ln \left (\frac {3}{x}\right )\,\left (x^3-5\,x^2+4\,x\right )} \,d x \]
int((x + exp(x)*(x^2 - 5*x + 4) - 2*x^2 + x^3 - exp(x)*log(3/x)*(7*x - 5*x ^2 + x^3))/(x^2 - 2*x^3 + x^4 - exp(x)*log(3/x)*(4*x - 5*x^2 + x^3)),x)