Integrand size = 154, antiderivative size = 29 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x-\log \left (\log \left (3-\frac {e^{2+x}}{3}-x+\frac {x}{5-10 x}\right )\right ) \] Output:
x-ln(ln(3-x+x/(-10*x+5)-1/3*exp(2+x)))
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x-\log \left (\log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )\right ) \] Input:
Integrate[(-12 + 60*x - 60*x^2 + E^(2 + x)*(-5 + 20*x - 20*x^2) + (-45 + 1 92*x - 234*x^2 + 60*x^3 + E^(2 + x)*(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)])/((-45 + 192*x - 234*x^2 + 60*x^3 + E^(2 + x)*(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]),x]
Output:
x - Log[Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*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.
| \(\displaystyle \int \frac {-60 x^2+e^{x+2} \left (-20 x^2+20 x-5\right )+\left (60 x^3-234 x^2+e^{x+2} \left (20 x^2-20 x+5\right )+192 x-45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )+60 x-12}{\left (60 x^3-234 x^2+e^{x+2} \left (20 x^2-20 x+5\right )+192 x-45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {60 x^2-e^{x+2} \left (-20 x^2+20 x-5\right )-\left (60 x^3-234 x^2+e^{x+2} \left (20 x^2-20 x+5\right )+192 x-45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )-60 x+12}{(1-2 x) \left (30 x^2+10 e^{x+2} x-102 x-5 e^{x+2}+45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {\log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )-1}{\log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}+\frac {3 \left (20 x^3-98 x^2+84 x-19\right )}{(2 x-1) \left (30 x^2+10 e^{x+2} x-102 x-5 e^{x+2}+45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \frac {1}{\log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}dx+60 \int \frac {1}{\left (30 x^2+10 e^{x+2} x-102 x-5 e^{x+2}+45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}dx-132 \int \frac {x}{\left (30 x^2+10 e^{x+2} x-102 x-5 e^{x+2}+45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}dx+30 \int \frac {x^2}{\left (30 x^2+10 e^{x+2} x-102 x-5 e^{x+2}+45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}dx+3 \int \frac {1}{(2 x-1) \left (30 x^2+10 e^{x+2} x-102 x-5 e^{x+2}+45\right ) \log \left (\frac {-30 x^2+102 x+e^{x+2} (5-10 x)-45}{30 x-15}\right )}dx+x\) |
Input:
Int[(-12 + 60*x - 60*x^2 + E^(2 + x)*(-5 + 20*x - 20*x^2) + (-45 + 192*x - 234*x^2 + 60*x^3 + E^(2 + x)*(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)])/((-45 + 192*x - 234*x^2 + 60*x^3 + E^(2 + x)*(5 - 20*x + 20*x^2))*Log[(-45 + E^(2 + x)*(5 - 10*x) + 102*x - 30*x^2)/(-15 + 30*x)]),x]
Output:
$Aborted
Time = 0.86 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21
| method | result | size |
| norman | \(x -\ln \left (\ln \left (\frac {\left (-10 x +5\right ) {\mathrm e}^{2+x}-30 x^{2}+102 x -45}{30 x -15}\right )\right )\) | \(35\) |
| parallelrisch | \(2-\ln \left (\ln \left (\frac {\left (-10 x +5\right ) {\mathrm e}^{2+x}-30 x^{2}+102 x -45}{30 x -15}\right )\right )+x\) | \(37\) |
| risch | \(x -\ln \left (\ln \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )+\frac {i \left (-2 \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{-\frac {1}{2}+x}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{-\frac {1}{2}+x}\right ) \operatorname {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{-\frac {1}{2}+x}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{-\frac {1}{2}+x}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{-\frac {1}{2}+x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{-\frac {1}{2}+x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+\left (\frac {{\mathrm e}^{2+x}}{3}-\frac {17}{5}\right ) x -\frac {{\mathrm e}^{2+x}}{6}+\frac {3}{2}\right )}{-\frac {1}{2}+x}\right )}^{3}+2 \pi +2 i \ln \left (-\frac {1}{2}+x \right )\right )}{2}\right )\) | \(281\) |
Input:
int((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*ln(((-10*x+5)*exp( 2+x)-30*x^2+102*x-45)/(30*x-15))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+60*x-12) /((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/ln(((-10*x+5)*exp(2+x) -30*x^2+102*x-45)/(30*x-15)),x,method=_RETURNVERBOSE)
Output:
x-ln(ln(((-10*x+5)*exp(2+x)-30*x^2+102*x-45)/(30*x-15)))
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x - \log \left (\log \left (-\frac {30 \, x^{2} + 5 \, {\left (2 \, x - 1\right )} e^{\left (x + 2\right )} - 102 \, x + 45}{15 \, {\left (2 \, x - 1\right )}}\right )\right ) \] Input:
integrate((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+ 5)*exp(2+x)-30*x^2+102*x-45)/(30*x-15))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+6 0*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5)* exp(2+x)-30*x^2+102*x-45)/(30*x-15)),x, algorithm="fricas")
Output:
x - log(log(-1/15*(30*x^2 + 5*(2*x - 1)*e^(x + 2) - 102*x + 45)/(2*x - 1)) )
Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x - \log {\left (\log {\left (\frac {- 30 x^{2} + 102 x + \left (5 - 10 x\right ) e^{x + 2} - 45}{30 x - 15} \right )} \right )} \] Input:
integrate((((20*x**2-20*x+5)*exp(2+x)+60*x**3-234*x**2+192*x-45)*ln(((-10* x+5)*exp(2+x)-30*x**2+102*x-45)/(30*x-15))+(-20*x**2+20*x-5)*exp(2+x)-60*x **2+60*x-12)/((20*x**2-20*x+5)*exp(2+x)+60*x**3-234*x**2+192*x-45)/ln(((-1 0*x+5)*exp(2+x)-30*x**2+102*x-45)/(30*x-15)),x)
Output:
x - log(log((-30*x**2 + 102*x + (5 - 10*x)*exp(x + 2) - 45)/(30*x - 15)))
Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x - \log \left (-\log \left (5\right ) - \log \left (3\right ) + \log \left (-30 \, x^{2} - 5 \, {\left (2 \, x e^{2} - e^{2}\right )} e^{x} + 102 \, x - 45\right ) - \log \left (2 \, x - 1\right )\right ) \] Input:
integrate((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+ 5)*exp(2+x)-30*x^2+102*x-45)/(30*x-15))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+6 0*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5)* exp(2+x)-30*x^2+102*x-45)/(30*x-15)),x, algorithm="maxima")
Output:
x - log(-log(5) - log(3) + log(-30*x^2 - 5*(2*x*e^2 - e^2)*e^x + 102*x - 4 5) - log(2*x - 1))
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x - \log \left (\log \left (-\frac {30 \, x^{2} + 10 \, x e^{\left (x + 2\right )} - 102 \, x - 5 \, e^{\left (x + 2\right )} + 45}{15 \, {\left (2 \, x - 1\right )}}\right )\right ) \] Input:
integrate((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+ 5)*exp(2+x)-30*x^2+102*x-45)/(30*x-15))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+6 0*x-12)/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5)* exp(2+x)-30*x^2+102*x-45)/(30*x-15)),x, algorithm="giac")
Output:
x - log(log(-1/15*(30*x^2 + 10*x*e^(x + 2) - 102*x - 5*e^(x + 2) + 45)/(2* x - 1)))
Time = 4.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=x-\ln \left (\ln \left (-\frac {30\,x^2-102\,x+{\mathrm {e}}^2\,{\mathrm {e}}^x\,\left (10\,x-5\right )+45}{30\,x-15}\right )\right ) \] Input:
int(-(exp(x + 2)*(20*x^2 - 20*x + 5) - 60*x - log(-(exp(x + 2)*(10*x - 5) - 102*x + 30*x^2 + 45)/(30*x - 15))*(192*x + exp(x + 2)*(20*x^2 - 20*x + 5 ) - 234*x^2 + 60*x^3 - 45) + 60*x^2 + 12)/(log(-(exp(x + 2)*(10*x - 5) - 1 02*x + 30*x^2 + 45)/(30*x - 15))*(192*x + exp(x + 2)*(20*x^2 - 20*x + 5) - 234*x^2 + 60*x^3 - 45)),x)
Output:
x - log(log(-(30*x^2 - 102*x + exp(2)*exp(x)*(10*x - 5) + 45)/(30*x - 15)) )
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {-12+60 x-60 x^2+e^{2+x} \left (-5+20 x-20 x^2\right )+\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )}{\left (-45+192 x-234 x^2+60 x^3+e^{2+x} \left (5-20 x+20 x^2\right )\right ) \log \left (\frac {-45+e^{2+x} (5-10 x)+102 x-30 x^2}{-15+30 x}\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (\frac {-10 e^{x} e^{2} x +5 e^{x} e^{2}-30 x^{2}+102 x -45}{30 x -15}\right )\right )+x \] Input:
int((((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)*log(((-10*x+5)*exp (2+x)-30*x^2+102*x-45)/(30*x-15))+(-20*x^2+20*x-5)*exp(2+x)-60*x^2+60*x-12 )/((20*x^2-20*x+5)*exp(2+x)+60*x^3-234*x^2+192*x-45)/log(((-10*x+5)*exp(2+ x)-30*x^2+102*x-45)/(30*x-15)),x)
Output:
- log(log(( - 10*e**x*e**2*x + 5*e**x*e**2 - 30*x**2 + 102*x - 45)/(30*x - 15))) + x