Integrand size = 92, antiderivative size = 19 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{(\log (16)+\log (1+x)) \log (-3+2 x)} \] Output:
3/(ln(1+x)+4*ln(2))/ln(-3+2*x)
Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3 (\log (256)+2 \log (1+x))}{2 \log ^2(16 (1+x)) \log (-3+2 x)} \] Input:
Integrate[((-6 - 6*x)*Log[16] + (-6 - 6*x)*Log[1 + x] + (9 - 6*x)*Log[-3 + 2*x])/(((-3 - x + 2*x^2)*Log[16]^2 + (-6 - 2*x + 4*x^2)*Log[16]*Log[1 + x ] + (-3 - x + 2*x^2)*Log[1 + x]^2)*Log[-3 + 2*x]^2),x]
Output:
(3*(Log[256] + 2*Log[1 + x]))/(2*Log[16*(1 + x)]^2*Log[-3 + 2*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.
\(\displaystyle \int \frac {(-6 x-6) \log (x+1)+(-6 x-6) \log (16)+(9-6 x) \log (2 x-3)}{\left (\left (2 x^2-x-3\right ) \log ^2(x+1)+\left (2 x^2-x-3\right ) \log ^2(16)+\left (4 x^2-2 x-6\right ) \log (16) \log (x+1)\right ) \log ^2(2 x-3)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 (2 (x+1) \log (x+1)+(x+1) \log (256)+(2 x-3) \log (2 x-3))}{\left (-2 x^2+x+3\right ) \log ^2(2 x-3) \log ^2(16 x+16)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int \frac {2 \log (x+1) (x+1)+\log (256) (x+1)-(3-2 x) \log (2 x-3)}{\left (-2 x^2+x+3\right ) \log ^2(2 x-3) \log ^2(16 x+16)}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle 3 \int \left (\frac {2 (2 \log (x+1) x+2 \log (2 x-3) x+\log (256) x+2 \log (x+1)-3 \log (2 x-3)+\log (256))}{5 (3-2 x) \log ^2(2 x-3) \log ^2(16 x+16)}+\frac {2 \log (x+1) x+2 \log (2 x-3) x+\log (256) x+2 \log (x+1)-3 \log (2 x-3)+\log (256)}{5 (x+1) \log ^2(2 x-3) \log ^2(16 x+16)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (\log (256) \int \frac {1}{(3-2 x) \log ^2(2 x-3) \log ^2(16 x+16)}dx+\frac {1}{5} \log (256) \int \frac {1}{(-x-1) \log ^2(2 x-3) \log ^2(16 x+16)}dx+\frac {1}{5} \log (256) \int \frac {1}{(x+1) \log ^2(2 x-3) \log ^2(16 x+16)}dx+2 \int \frac {\log (x+1)}{(3-2 x) \log ^2(2 x-3) \log ^2(16 x+16)}dx+\frac {2}{5} \int \frac {\log (x+1)}{(-x-1) \log ^2(2 x-3) \log ^2(16 x+16)}dx+\frac {2}{5} \int \frac {\log (x+1)}{(x+1) \log ^2(2 x-3) \log ^2(16 x+16)}dx+\frac {6}{5} \int \frac {1}{(3-2 x) \log (2 x-3) \log ^2(16 x+16)}dx+\int \frac {1}{(-x-1) \log (2 x-3) \log ^2(16 x+16)}dx+\frac {6}{5} \int \frac {1}{(2 x-3) \log (2 x-3) \log ^2(16 x+16)}dx\right )\) |
Input:
Int[((-6 - 6*x)*Log[16] + (-6 - 6*x)*Log[1 + x] + (9 - 6*x)*Log[-3 + 2*x]) /(((-3 - x + 2*x^2)*Log[16]^2 + (-6 - 2*x + 4*x^2)*Log[16]*Log[1 + x] + (- 3 - x + 2*x^2)*Log[1 + x]^2)*Log[-3 + 2*x]^2),x]
Output:
$Aborted
Time = 2.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {3}{\left (\ln \left (1+x \right )+4 \ln \left (2\right )\right ) \ln \left (-3+2 x \right )}\) | \(22\) |
parallelrisch | \(\frac {3}{\left (\ln \left (1+x \right )+4 \ln \left (2\right )\right ) \ln \left (-3+2 x \right )}\) | \(22\) |
Input:
int(((-6*x+9)*ln(-3+2*x)+(-6*x-6)*ln(1+x)+4*(-6*x-6)*ln(2))/((2*x^2-x-3)*l n(1+x)^2+4*(4*x^2-2*x-6)*ln(2)*ln(1+x)+16*(2*x^2-x-3)*ln(2)^2)/ln(-3+2*x)^ 2,x,method=_RETURNVERBOSE)
Output:
3/(ln(1+x)+4*ln(2))/ln(-3+2*x)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{{\left (4 \, \log \left (2\right ) + \log \left (x + 1\right )\right )} \log \left (2 \, x - 3\right )} \] Input:
integrate(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x ^2-x-3)*log(1+x)^2+4*(4*x^2-2*x-6)*log(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2 )/log(-3+2*x)^2,x, algorithm="fricas")
Output:
3/((4*log(2) + log(x + 1))*log(2*x - 3))
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{\left (\log {\left (x + 1 \right )} + 4 \log {\left (2 \right )}\right ) \log {\left (2 x - 3 \right )}} \] Input:
integrate(((-6*x+9)*ln(-3+2*x)+(-6*x-6)*ln(1+x)+4*(-6*x-6)*ln(2))/((2*x**2 -x-3)*ln(1+x)**2+4*(4*x**2-2*x-6)*ln(2)*ln(1+x)+16*(2*x**2-x-3)*ln(2)**2)/ ln(-3+2*x)**2,x)
Output:
3/((log(x + 1) + 4*log(2))*log(2*x - 3))
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{{\left (4 \, \log \left (2\right ) + \log \left (x + 1\right )\right )} \log \left (2 \, x - 3\right )} \] Input:
integrate(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x ^2-x-3)*log(1+x)^2+4*(4*x^2-2*x-6)*log(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2 )/log(-3+2*x)^2,x, algorithm="maxima")
Output:
3/((4*log(2) + log(x + 1))*log(2*x - 3))
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{4 \, \log \left (2\right ) \log \left (2 \, x - 3\right ) + \log \left (2 \, x - 3\right ) \log \left (x + 1\right )} \] Input:
integrate(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x ^2-x-3)*log(1+x)^2+4*(4*x^2-2*x-6)*log(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2 )/log(-3+2*x)^2,x, algorithm="giac")
Output:
3/(4*log(2)*log(2*x - 3) + log(2*x - 3)*log(x + 1))
Time = 3.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 10.21 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {15}{4\,\left (x+1\right )}-\frac {\frac {15\,\left (4\,\ln \left (2\right )+1\right )}{4\,\left (x+1\right )}+\frac {15\,\ln \left (x+1\right )}{4\,\left (x+1\right )}}{\ln \left (x+1\right )+4\,\ln \left (2\right )}+\frac {\frac {3}{\ln \left (x+1\right )+4\,\ln \left (2\right )}+\frac {3\,\ln \left (2\,x-3\right )\,\left (2\,x-3\right )}{2\,\left (8\,\ln \left (x+1\right )\,\ln \left (2\right )+x\,{\ln \left (x+1\right )}^2+16\,x\,{\ln \left (2\right )}^2+{\ln \left (x+1\right )}^2+16\,{\ln \left (2\right )}^2+8\,x\,\ln \left (x+1\right )\,\ln \left (2\right )\right )}}{\ln \left (2\,x-3\right )}+\frac {\frac {3\,\left (10\,\ln \left (2\right )-2\,x+3\right )}{2\,\left (x+1\right )}+\frac {15\,\ln \left (x+1\right )}{4\,\left (x+1\right )}}{{\ln \left (x+1\right )}^2+8\,\ln \left (2\right )\,\ln \left (x+1\right )+16\,{\ln \left (2\right )}^2} \] Input:
int((4*log(2)*(6*x + 6) + log(2*x - 3)*(6*x - 9) + log(x + 1)*(6*x + 6))/( log(2*x - 3)^2*(log(x + 1)^2*(x - 2*x^2 + 3) + 16*log(2)^2*(x - 2*x^2 + 3) + 4*log(x + 1)*log(2)*(2*x - 4*x^2 + 6))),x)
Output:
15/(4*(x + 1)) - ((15*(4*log(2) + 1))/(4*(x + 1)) + (15*log(x + 1))/(4*(x + 1)))/(log(x + 1) + 4*log(2)) + (3/(log(x + 1) + 4*log(2)) + (3*log(2*x - 3)*(2*x - 3))/(2*(8*log(x + 1)*log(2) + x*log(x + 1)^2 + 16*x*log(2)^2 + log(x + 1)^2 + 16*log(2)^2 + 8*x*log(x + 1)*log(2))))/log(2*x - 3) + ((3*( 10*log(2) - 2*x + 3))/(2*(x + 1)) + (15*log(x + 1))/(4*(x + 1)))/(8*log(x + 1)*log(2) + log(x + 1)^2 + 16*log(2)^2)
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {(-6-6 x) \log (16)+(-6-6 x) \log (1+x)+(9-6 x) \log (-3+2 x)}{\left (\left (-3-x+2 x^2\right ) \log ^2(16)+\left (-6-2 x+4 x^2\right ) \log (16) \log (1+x)+\left (-3-x+2 x^2\right ) \log ^2(1+x)\right ) \log ^2(-3+2 x)} \, dx=\frac {3}{\mathrm {log}\left (2 x -3\right ) \left (\mathrm {log}\left (x +1\right )+4 \,\mathrm {log}\left (2\right )\right )} \] Input:
int(((-6*x+9)*log(-3+2*x)+(-6*x-6)*log(1+x)+4*(-6*x-6)*log(2))/((2*x^2-x-3 )*log(1+x)^2+4*(4*x^2-2*x-6)*log(2)*log(1+x)+16*(2*x^2-x-3)*log(2)^2)/log( -3+2*x)^2,x)
Output:
3/(log(2*x - 3)*(log(x + 1) + 4*log(2)))