Integrand size = 90, antiderivative size = 26 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (\frac {(-5+x) x}{4-\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )\right ) \]
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )\right ) \]
Integrate[(35 - 16*x + x^2 + (-5 + 2*x)*Log[x^3/(100*E^x)])/((20*x - 4*x^2 + (-5*x + x^2)*Log[x^3/(100*E^x)])*Log[(5*x - x^2)/(-4 + Log[x^3/(100*E^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.
\(\displaystyle \int \frac {(2 x-5) \log \left (\frac {1}{100} e^{-x} x^3\right )+x^2-16 x+35}{\left (-4 x^2+\left (x^2-5 x\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )+20 x\right ) \log \left (\frac {5 x-x^2}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(2 x-5) \log \left (\frac {1}{100} e^{-x} x^3\right )+x^2-16 x+35}{(5-x) x \left (4-\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {(5-x) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-2 x \log \left (\frac {1}{100} e^{-x} x^3\right )+5 \log \left (\frac {1}{100} e^{-x} x^3\right )-x^2+16 x-35}{5 x \left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}+\frac {2 x \log \left (\frac {1}{100} e^{-x} x^3\right )-5 \log \left (\frac {1}{100} e^{-x} x^3\right )+x^2-16 x+35}{5 (x-5) \left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{\left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}dx-4 \int \frac {1}{(x-5) \left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}dx-7 \int \frac {1}{x \left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}dx+\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{(x-5) \left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}dx+\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (\log \left (\frac {1}{100} e^{-x} x^3\right )-4\right ) \log \left (-\frac {(x-5) x}{\log \left (\frac {1}{100} e^{-x} x^3\right )-4}\right )}dx\) |
Int[(35 - 16*x + x^2 + (-5 + 2*x)*Log[x^3/(100*E^x)])/((20*x - 4*x^2 + (-5 *x + x^2)*Log[x^3/(100*E^x)])*Log[(5*x - x^2)/(-4 + Log[x^3/(100*E^x)])]), x]
3.20.55.3.1 Defintions of rubi rules used
Time = 1.81 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
default | \(\ln \left (\ln \left (\frac {\left (-5+x \right ) x}{4-\ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )}\right )\right )\) | \(24\) |
parallelrisch | \(\ln \left (\ln \left (\frac {-x^{2}+5 x}{\ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )-4}\right )\right )\) | \(27\) |
int(((-5+2*x)*ln(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*ln(1/100*x^3/ex p(x))-4*x^2+20*x)/ln((-x^2+5*x)/(ln(1/100*x^3/exp(x))-4)),x,method=_RETURN VERBOSE)
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {x^{2} - 5 \, x}{\log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 4}\right )\right ) \]
integrate(((-5+2*x)*log(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*log(1/10 0*x^3/exp(x))-4*x^2+20*x)/log((-x^2+5*x)/(log(1/100*x^3/exp(x))-4)),x, alg orithm=\
Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log {\left (\log {\left (\frac {- x^{2} + 5 x}{\log {\left (\frac {x^{3} e^{- x}}{100} \right )} - 4} \right )} \right )} \]
integrate(((-5+2*x)*ln(1/100*x**3/exp(x))+x**2-16*x+35)/((x**2-5*x)*ln(1/1 00*x**3/exp(x))-4*x**2+20*x)/ln((-x**2+5*x)/(ln(1/100*x**3/exp(x))-4)),x)
Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (x + 2 \, \log \left (5\right ) + 2 \, \log \left (2\right ) - 3 \, \log \left (x\right ) + 4\right ) - \log \left (x - 5\right ) - \log \left (x\right )\right ) \]
integrate(((-5+2*x)*log(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*log(1/10 0*x^3/exp(x))-4*x^2+20*x)/log((-x^2+5*x)/(log(1/100*x^3/exp(x))-4)),x, alg orithm=\
\[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\int { -\frac {x^{2} + {\left (2 \, x - 5\right )} \log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 16 \, x + 35}{{\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} \log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 20 \, x\right )} \log \left (-\frac {x^{2} - 5 \, x}{\log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 4}\right )} \,d x } \]
integrate(((-5+2*x)*log(1/100*x^3/exp(x))+x^2-16*x+35)/((x^2-5*x)*log(1/10 0*x^3/exp(x))-4*x^2+20*x)/log((-x^2+5*x)/(log(1/100*x^3/exp(x))-4)),x, alg orithm=\
integrate(-(x^2 + (2*x - 5)*log(1/100*x^3*e^(-x)) - 16*x + 35)/((4*x^2 - ( x^2 - 5*x)*log(1/100*x^3*e^(-x)) - 20*x)*log(-(x^2 - 5*x)/(log(1/100*x^3*e ^(-x)) - 4))), x)
Time = 14.67 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\ln \left (\ln \left (-\frac {5\,x-x^2}{x-\ln \left (x^3\right )+2\,\ln \left (10\right )+4}\right )\right ) \]