Integrand size = 102, antiderivative size = 30 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=\log \left (\frac {2 \left (x+4 \left (3+x^2 \left (e^{-2-x} x+\log (-3+x)\right )\right )\right )}{x}\right ) \] Output:
ln(2*(x+12+4*(ln(-3+x)+exp(ln(x)-x-2))*x^2)/x)
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=4 \left (-\frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{4} \log \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )\right ) \] Input:
Integrate[(36 - 12*x + 4*x^3 + E^(-2 - x)*x*(-24*x^2 + 20*x^3 - 4*x^4) + ( -12*x^2 + 4*x^3)*Log[-3 + x])/(-36*x + 9*x^2 + x^3 + E^(-2 - x)*x*(-12*x^3 + 4*x^4) + (-12*x^3 + 4*x^4)*Log[-3 + x]),x]
Output:
4*(-1/4*x - Log[x]/4 + Log[12*E^(2 + x) + E^(2 + x)*x + 4*x^3 + 4*E^(2 + x )*x^2*Log[-3 + x]]/4)
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.
\(\displaystyle \int \frac {4 x^3+\left (4 x^3-12 x^2\right ) \log (x-3)+e^{-x-2} \left (-4 x^4+20 x^3-24 x^2\right ) x-12 x+36}{x^3+9 x^2+e^{-x-2} \left (4 x^4-12 x^3\right ) x+\left (4 x^4-12 x^3\right ) \log (x-3)-36 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x+2} \left (-4 x^3-\left (4 x^3-12 x^2\right ) \log (x-3)-e^{-x-2} \left (-4 x^4+20 x^3-24 x^2\right ) x+12 x-36\right )}{(3-x) x \left (4 x^3+4 e^{x+2} x^2 \log (x-3)+e^{x+2} x+12 e^{x+2}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x+2} \left (4 x^4 \log (x-3)+5 x^3-16 x^3 \log (x-3)+7 x^2+12 x^2 \log (x-3)-66 x+108\right )}{(x-3) x \left (4 x^3+4 e^{x+2} x^2 \log (x-3)+e^{x+2} x+12 e^{x+2}\right )}-\frac {x-2}{x}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x+2} \left (4 x^4 \log (x-3)+5 x^3-16 x^3 \log (x-3)+7 x^2+12 x^2 \log (x-3)-66 x+108\right )}{(x-3) x \left (4 x^3+4 e^{x+2} x^2 \log (x-3)+e^{x+2} x+12 e^{x+2}\right )}+\frac {2-x}{x}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {e^{x+2} \left (4 x^4 \log (x-3)+5 x^3-16 x^3 \log (x-3)+7 x^2+12 x^2 \log (x-3)-66 x+108\right )}{(x-3) x \left (4 x^3+4 e^{x+2} x^2 \log (x-3)+e^{x+2} x+12 e^{x+2}\right )}+\frac {2-x}{x}\right )dx\) |
Input:
Int[(36 - 12*x + 4*x^3 + E^(-2 - x)*x*(-24*x^2 + 20*x^3 - 4*x^4) + (-12*x^ 2 + 4*x^3)*Log[-3 + x])/(-36*x + 9*x^2 + x^3 + E^(-2 - x)*x*(-12*x^3 + 4*x ^4) + (-12*x^3 + 4*x^4)*Log[-3 + x]),x]
Output:
$Aborted
Time = 5.76 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\ln \left (x \right )+2+\ln \left (x \,{\mathrm e}^{-2-x}+\frac {4 x^{2} \ln \left (-3+x \right )+x +12}{4 x^{2}}\right )\) | \(32\) |
parallelrisch | \(\ln \left (x^{2} \ln \left (-3+x \right )+x^{2} {\mathrm e}^{\ln \left (x \right )-x -2}+\frac {x}{4}+3\right )-\ln \left (x \right )\) | \(32\) |
Input:
int(((-4*x^4+20*x^3-24*x^2)*exp(ln(x)-x-2)+(4*x^3-12*x^2)*ln(-3+x)+4*x^3-1 2*x+36)/((4*x^4-12*x^3)*exp(ln(x)-x-2)+(4*x^4-12*x^3)*ln(-3+x)+x^3+9*x^2-3 6*x),x,method=_RETURNVERBOSE)
Output:
ln(x)+2+ln(x*exp(-2-x)+1/4*(4*x^2*ln(-3+x)+x+12)/x^2)
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=\log \left (x\right ) + \log \left (\frac {4 \, x^{2} e^{\left (-x + \log \left (x\right ) - 2\right )} + 4 \, x^{2} \log \left (x - 3\right ) + x + 12}{x^{2}}\right ) \] Input:
integrate(((-4*x^4+20*x^3-24*x^2)*exp(log(x)-x-2)+(4*x^3-12*x^2)*log(-3+x) +4*x^3-12*x+36)/((4*x^4-12*x^3)*exp(log(x)-x-2)+(4*x^4-12*x^3)*log(-3+x)+x ^3+9*x^2-36*x),x, algorithm="fricas")
Output:
log(x) + log((4*x^2*e^(-x + log(x) - 2) + 4*x^2*log(x - 3) + x + 12)/x^2)
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=2 \log {\left (x \right )} + \log {\left (e^{- x - 2} + \frac {4 x^{2} \log {\left (x - 3 \right )} + x + 12}{4 x^{3}} \right )} \] Input:
integrate(((-4*x**4+20*x**3-24*x**2)*exp(ln(x)-x-2)+(4*x**3-12*x**2)*ln(-3 +x)+4*x**3-12*x+36)/((4*x**4-12*x**3)*exp(ln(x)-x-2)+(4*x**4-12*x**3)*ln(- 3+x)+x**3+9*x**2-36*x),x)
Output:
2*log(x) + log(exp(-x - 2) + (4*x**2*log(x - 3) + x + 12)/(4*x**3))
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=\log \left (x\right ) + \log \left (\frac {{\left (4 \, x^{2} e^{\left (x + 2\right )} \log \left (x - 3\right ) + 4 \, x^{3} + {\left (x e^{2} + 12 \, e^{2}\right )} e^{x}\right )} e^{\left (-x - 2\right )}}{4 \, x^{2}}\right ) \] Input:
integrate(((-4*x^4+20*x^3-24*x^2)*exp(log(x)-x-2)+(4*x^3-12*x^2)*log(-3+x) +4*x^3-12*x+36)/((4*x^4-12*x^3)*exp(log(x)-x-2)+(4*x^4-12*x^3)*log(-3+x)+x ^3+9*x^2-36*x),x, algorithm="maxima")
Output:
log(x) + log(1/4*(4*x^2*e^(x + 2)*log(x - 3) + 4*x^3 + (x*e^2 + 12*e^2)*e^ x)*e^(-x - 2)/x^2)
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=-x + \log \left (4 \, x^{2} e^{\left (x + 2\right )} \log \left (x - 3\right ) + 4 \, x^{3} + x e^{\left (x + 2\right )} + 12 \, e^{\left (x + 2\right )}\right ) - \log \left (x\right ) \] Input:
integrate(((-4*x^4+20*x^3-24*x^2)*exp(log(x)-x-2)+(4*x^3-12*x^2)*log(-3+x) +4*x^3-12*x+36)/((4*x^4-12*x^3)*exp(log(x)-x-2)+(4*x^4-12*x^3)*log(-3+x)+x ^3+9*x^2-36*x),x, algorithm="giac")
Output:
-x + log(4*x^2*e^(x + 2)*log(x - 3) + 4*x^3 + x*e^(x + 2) + 12*e^(x + 2)) - log(x)
Time = 4.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=\ln \left (\frac {x+4\,x^2\,\ln \left (x-3\right )+4\,x^3\,{\mathrm {e}}^{-x-2}+12}{x^2}\right )+\ln \left (x\right ) \] Input:
int((12*x + log(x - 3)*(12*x^2 - 4*x^3) - 4*x^3 + exp(log(x) - x - 2)*(24* x^2 - 20*x^3 + 4*x^4) - 36)/(36*x + log(x - 3)*(12*x^3 - 4*x^4) + exp(log( x) - x - 2)*(12*x^3 - 4*x^4) - 9*x^2 - x^3),x)
Output:
log((x + 4*x^2*log(x - 3) + 4*x^3*exp(- x - 2) + 12)/x^2) + log(x)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, dx=\mathrm {log}\left (4 e^{x} \mathrm {log}\left (x -3\right ) e^{2} x^{2}+e^{x} e^{2} x +12 e^{x} e^{2}+4 x^{3}\right )-\mathrm {log}\left (x \right )-x \] Input:
int(((-4*x^4+20*x^3-24*x^2)*exp(log(x)-x-2)+(4*x^3-12*x^2)*log(-3+x)+4*x^3 -12*x+36)/((4*x^4-12*x^3)*exp(log(x)-x-2)+(4*x^4-12*x^3)*log(-3+x)+x^3+9*x ^2-36*x),x)
Output:
log(4*e**x*log(x - 3)*e**2*x**2 + e**x*e**2*x + 12*e**x*e**2 + 4*x**3) - l og(x) - x