Integrand size = 99, antiderivative size = 20 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=2 x+\frac {12 x}{-9+x+\log \left (-4+\frac {x}{3}\right )} \] Output:
2*x+12*x/(ln(1/3*x-4)-9+x)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=2 \left (x+\frac {6 x}{-9+x+\log \left (-4+\frac {x}{3}\right )}\right ) \] Input:
Integrate[(-648 + 474*x - 60*x^2 + 2*x^3 + (288 - 72*x + 4*x^2)*Log[(-12 + x)/3] + (-24 + 2*x)*Log[(-12 + x)/3]^2)/(-972 + 297*x - 30*x^2 + x^3 + (2 16 - 42*x + 2*x^2)*Log[(-12 + x)/3] + (-12 + x)*Log[(-12 + x)/3]^2),x]
Output:
2*(x + (6*x)/(-9 + x + Log[-4 + x/3]))
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.
\(\displaystyle \int \frac {2 x^3-60 x^2+\left (4 x^2-72 x+288\right ) \log \left (\frac {x-12}{3}\right )+474 x+(2 x-24) \log ^2\left (\frac {x-12}{3}\right )-648}{x^3-30 x^2+\left (2 x^2-42 x+216\right ) \log \left (\frac {x-12}{3}\right )+297 x+(x-12) \log ^2\left (\frac {x-12}{3}\right )-972} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-2 x^3+60 x^2-\left (4 x^2-72 x+288\right ) \log \left (\frac {x-12}{3}\right )-474 x-(2 x-24) \log ^2\left (\frac {x-12}{3}\right )+648}{(12-x) \left (-x-\log \left (\frac {x}{3}-4\right )+9\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^3}{(x-12) \left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}-\frac {60 x^2}{(x-12) \left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}+\frac {2 \log ^2\left (\frac {x}{3}-4\right )}{\left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}+\frac {474 x}{(x-12) \left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}+\frac {4 (x-6) \log \left (\frac {x}{3}-4\right )}{\left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}-\frac {648}{(x-12) \left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -12 \int \frac {1}{\left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}dx-144 \int \frac {1}{(x-12) \left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}dx-12 \int \frac {x}{\left (x+\log \left (\frac {x}{3}-4\right )-9\right )^2}dx+12 \int \frac {1}{x+\log \left (\frac {x}{3}-4\right )-9}dx+2 x\) |
Input:
Int[(-648 + 474*x - 60*x^2 + 2*x^3 + (288 - 72*x + 4*x^2)*Log[(-12 + x)/3] + (-24 + 2*x)*Log[(-12 + x)/3]^2)/(-972 + 297*x - 30*x^2 + x^3 + (216 - 4 2*x + 2*x^2)*Log[(-12 + x)/3] + (-12 + x)*Log[(-12 + x)/3]^2),x]
Output:
$Aborted
Time = 0.50 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
risch | \(2 x +\frac {12 x}{\ln \left (\frac {x}{3}-4\right )-9+x}\) | \(19\) |
norman | \(\frac {6 \ln \left (\frac {x}{3}-4\right )+2 \ln \left (\frac {x}{3}-4\right ) x +2 x^{2}-54}{\ln \left (\frac {x}{3}-4\right )-9+x}\) | \(37\) |
derivativedivides | \(\frac {6 \ln \left (\frac {x}{3}-4\right ) \left (\frac {x}{3}-4\right )+18 x -72+18 \left (\frac {x}{3}-4\right )^{2}}{\ln \left (\frac {x}{3}-4\right )-9+x}\) | \(40\) |
default | \(\frac {6 \ln \left (\frac {x}{3}-4\right ) \left (\frac {x}{3}-4\right )+18 x -72+18 \left (\frac {x}{3}-4\right )^{2}}{\ln \left (\frac {x}{3}-4\right )-9+x}\) | \(40\) |
parallelrisch | \(\frac {-432+2 x^{2}+2 \ln \left (\frac {x}{3}-4\right ) x +42 x +48 \ln \left (\frac {x}{3}-4\right )}{\ln \left (\frac {x}{3}-4\right )-9+x}\) | \(40\) |
Input:
int(((2*x-24)*ln(1/3*x-4)^2+(4*x^2-72*x+288)*ln(1/3*x-4)+2*x^3-60*x^2+474* x-648)/((x-12)*ln(1/3*x-4)^2+(2*x^2-42*x+216)*ln(1/3*x-4)+x^3-30*x^2+297*x -972),x,method=_RETURNVERBOSE)
Output:
2*x+12*x/(ln(1/3*x-4)-9+x)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=\frac {2 \, {\left (x^{2} + x \log \left (\frac {1}{3} \, x - 4\right ) - 3 \, x\right )}}{x + \log \left (\frac {1}{3} \, x - 4\right ) - 9} \] Input:
integrate(((2*x-24)*log(1/3*x-4)^2+(4*x^2-72*x+288)*log(1/3*x-4)+2*x^3-60* x^2+474*x-648)/((x-12)*log(1/3*x-4)^2+(2*x^2-42*x+216)*log(1/3*x-4)+x^3-30 *x^2+297*x-972),x, algorithm="fricas")
Output:
2*(x^2 + x*log(1/3*x - 4) - 3*x)/(x + log(1/3*x - 4) - 9)
Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=2 x + \frac {12 x}{x + \log {\left (\frac {x}{3} - 4 \right )} - 9} \] Input:
integrate(((2*x-24)*ln(1/3*x-4)**2+(4*x**2-72*x+288)*ln(1/3*x-4)+2*x**3-60 *x**2+474*x-648)/((x-12)*ln(1/3*x-4)**2+(2*x**2-42*x+216)*ln(1/3*x-4)+x**3 -30*x**2+297*x-972),x)
Output:
2*x + 12*x/(x + log(x/3 - 4) - 9)
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=\frac {2 \, {\left (x^{2} - x {\left (\log \left (3\right ) + 3\right )} + x \log \left (x - 12\right )\right )}}{x - \log \left (3\right ) + \log \left (x - 12\right ) - 9} \] Input:
integrate(((2*x-24)*log(1/3*x-4)^2+(4*x^2-72*x+288)*log(1/3*x-4)+2*x^3-60* x^2+474*x-648)/((x-12)*log(1/3*x-4)^2+(2*x^2-42*x+216)*log(1/3*x-4)+x^3-30 *x^2+297*x-972),x, algorithm="maxima")
Output:
2*(x^2 - x*(log(3) + 3) + x*log(x - 12))/(x - log(3) + log(x - 12) - 9)
Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=2 \, x + \frac {12 \, x}{x + \log \left (\frac {1}{3} \, x - 4\right ) - 9} \] Input:
integrate(((2*x-24)*log(1/3*x-4)^2+(4*x^2-72*x+288)*log(1/3*x-4)+2*x^3-60* x^2+474*x-648)/((x-12)*log(1/3*x-4)^2+(2*x^2-42*x+216)*log(1/3*x-4)+x^3-30 *x^2+297*x-972),x, algorithm="giac")
Output:
2*x + 12*x/(x + log(1/3*x - 4) - 9)
Time = 3.82 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=\frac {2\,\left (x\,\ln \left (\frac {x}{3}-4\right )-6\,\ln \left (\frac {x}{3}-4\right )-9\,x+x^2+54\right )}{x+\ln \left (\frac {x}{3}-4\right )-9} \] Input:
int((474*x + log(x/3 - 4)*(4*x^2 - 72*x + 288) + log(x/3 - 4)^2*(2*x - 24) - 60*x^2 + 2*x^3 - 648)/(297*x + log(x/3 - 4)^2*(x - 12) + log(x/3 - 4)*( 2*x^2 - 42*x + 216) - 30*x^2 + x^3 - 972),x)
Output:
(2*(x*log(x/3 - 4) - 6*log(x/3 - 4) - 9*x + x^2 + 54))/(x + log(x/3 - 4) - 9)
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.95 \[ \int \frac {-648+474 x-60 x^2+2 x^3+\left (288-72 x+4 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-24+2 x) \log ^2\left (\frac {1}{3} (-12+x)\right )}{-972+297 x-30 x^2+x^3+\left (216-42 x+2 x^2\right ) \log \left (\frac {1}{3} (-12+x)\right )+(-12+x) \log ^2\left (\frac {1}{3} (-12+x)\right )} \, dx=\frac {2 \,\mathrm {log}\left (x -12\right ) \mathrm {log}\left (\frac {x}{3}-4\right )+2 \,\mathrm {log}\left (x -12\right ) x -18 \,\mathrm {log}\left (x -12\right )-2 \mathrm {log}\left (\frac {x}{3}-4\right )^{2}+24 \,\mathrm {log}\left (\frac {x}{3}-4\right )+2 x^{2}-54}{\mathrm {log}\left (\frac {x}{3}-4\right )+x -9} \] Input:
int(((2*x-24)*log(1/3*x-4)^2+(4*x^2-72*x+288)*log(1/3*x-4)+2*x^3-60*x^2+47 4*x-648)/((x-12)*log(1/3*x-4)^2+(2*x^2-42*x+216)*log(1/3*x-4)+x^3-30*x^2+2 97*x-972),x)
Output:
(2*(log(x - 12)*log((x - 12)/3) + log(x - 12)*x - 9*log(x - 12) - log((x - 12)/3)**2 + 12*log((x - 12)/3) + x**2 - 27))/(log((x - 12)/3) + x - 9)