Integrand size = 115, antiderivative size = 23 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\frac {3 x}{4+\frac {1}{4} \log \left (\frac {2}{(-1+x)^2}\right )-\log (x)} \] Output:
3/(4-ln(x)+1/4*ln(2/(-1+x)^2))*x
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\frac {12 x}{16+\log \left (\frac {2}{(-1+x)^2}\right )-4 \log (x)} \] Input:
Integrate[(-240 + 264*x + (48 - 48*x)*Log[x] + (-12 + 12*x)*Log[2/(1 - 2*x + x^2)])/(-256 + 256*x + (-16 + 16*x)*Log[x]^2 + (-32 + 32*x)*Log[2/(1 - 2*x + x^2)] + (-1 + x)*Log[2/(1 - 2*x + x^2)]^2 + Log[x]*(128 - 128*x + (8 - 8*x)*Log[2/(1 - 2*x + x^2)])),x]
Output:
(12*x)/(16 + Log[2/(-1 + x)^2] - 4*Log[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.
| \(\displaystyle \int \frac {(12 x-12) \log \left (\frac {2}{x^2-2 x+1}\right )+264 x+(48-48 x) \log (x)-240}{(x-1) \log ^2\left (\frac {2}{x^2-2 x+1}\right )+\left ((8-8 x) \log \left (\frac {2}{x^2-2 x+1}\right )-128 x+128\right ) \log (x)+(32 x-32) \log \left (\frac {2}{x^2-2 x+1}\right )+256 x+(16 x-16) \log ^2(x)-256} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {12 \left (-22 x-(x-1) \log \left (\frac {2}{(x-1)^2}\right )+4 (x-1) \log (x)+20\right )}{(1-x) \left (\log \left (\frac {2}{(x-1)^2}\right )-4 \log (x)+16\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 12 \int \frac {-22 x+(1-x) \log \left (\frac {2}{(1-x)^2}\right )-4 (1-x) \log (x)+20}{(1-x) \left (\log \left (\frac {2}{(1-x)^2}\right )-4 \log (x)+16\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 12 \int \left (\frac {2 (3 x-2)}{(x-1) \left (\log \left (\frac {2}{(x-1)^2}\right )-4 \log (x)+16\right )^2}+\frac {1}{\log \left (\frac {2}{(x-1)^2}\right )-4 \log (x)+16}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 12 \left (6 \int \frac {1}{\left (\log \left (\frac {2}{(x-1)^2}\right )-4 \log (x)+16\right )^2}dx+2 \int \frac {1}{(x-1) \left (\log \left (\frac {2}{(x-1)^2}\right )-4 \log (x)+16\right )^2}dx+\int \frac {1}{\log \left (\frac {2}{(x-1)^2}\right )-4 \log (x)+16}dx\right )\) |
Input:
Int[(-240 + 264*x + (48 - 48*x)*Log[x] + (-12 + 12*x)*Log[2/(1 - 2*x + x^2 )])/(-256 + 256*x + (-16 + 16*x)*Log[x]^2 + (-32 + 32*x)*Log[2/(1 - 2*x + x^2)] + (-1 + x)*Log[2/(1 - 2*x + x^2)]^2 + Log[x]*(128 - 128*x + (8 - 8*x )*Log[2/(1 - 2*x + x^2)])),x]
Output:
$Aborted
Time = 7.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17
| method | result | size |
| parallelrisch | \(-\frac {12 x}{4 \ln \left (x \right )-\ln \left (\frac {2}{x^{2}-2 x +1}\right )-16}\) | \(27\) |
| default | \(-\frac {24 i x}{\pi \operatorname {csgn}\left (i \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-1+x \right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (i \left (-1+x \right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \left (-1+x \right )^{2}\right )^{3}-2 i \ln \left (2\right )+8 i \ln \left (x \right )+4 i \ln \left (-1+x \right )-32 i}\) | \(81\) |
| risch | \(\frac {24 i x}{-\pi \operatorname {csgn}\left (i \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-1+x \right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (i \left (-1+x \right )^{2}\right )^{2}-\pi \operatorname {csgn}\left (i \left (-1+x \right )^{2}\right )^{3}+2 i \ln \left (2\right )-8 i \ln \left (x \right )-4 i \ln \left (-1+x \right )+32 i}\) | \(83\) |
Input:
int(((-48*x+48)*ln(x)+(12*x-12)*ln(2/(x^2-2*x+1))+264*x-240)/((16*x-16)*ln (x)^2+((-8*x+8)*ln(2/(x^2-2*x+1))-128*x+128)*ln(x)+(-1+x)*ln(2/(x^2-2*x+1) )^2+(32*x-32)*ln(2/(x^2-2*x+1))+256*x-256),x,method=_RETURNVERBOSE)
Output:
-12*x/(4*ln(x)-ln(2/(x^2-2*x+1))-16)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=-\frac {12 \, x}{4 \, \log \left (x\right ) - \log \left (\frac {2}{x^{2} - 2 \, x + 1}\right ) - 16} \] Input:
integrate(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16* x-16)*log(x)^2+((-8*x+8)*log(2/(x^2-2*x+1))-128*x+128)*log(x)+(-1+x)*log(2 /(x^2-2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x, algorithm="fric as")
Output:
-12*x/(4*log(x) - log(2/(x^2 - 2*x + 1)) - 16)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\frac {12 x}{- 4 \log {\left (x \right )} + \log {\left (\frac {2}{x^{2} - 2 x + 1} \right )} + 16} \] Input:
integrate(((-48*x+48)*ln(x)+(12*x-12)*ln(2/(x**2-2*x+1))+264*x-240)/((16*x -16)*ln(x)**2+((-8*x+8)*ln(2/(x**2-2*x+1))-128*x+128)*ln(x)+(-1+x)*ln(2/(x **2-2*x+1))**2+(32*x-32)*ln(2/(x**2-2*x+1))+256*x-256),x)
Output:
12*x/(-4*log(x) + log(2/(x**2 - 2*x + 1)) + 16)
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\frac {12 \, x}{\log \left (2\right ) - 2 \, \log \left (x - 1\right ) - 4 \, \log \left (x\right ) + 16} \] Input:
integrate(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16* x-16)*log(x)^2+((-8*x+8)*log(2/(x^2-2*x+1))-128*x+128)*log(x)+(-1+x)*log(2 /(x^2-2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x, algorithm="maxi ma")
Output:
12*x/(log(2) - 2*log(x - 1) - 4*log(x) + 16)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\frac {12 \, x}{\log \left (2\right ) - \log \left (x^{2} - 2 \, x + 1\right ) - 4 \, \log \left (x\right ) + 16} \] Input:
integrate(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16* x-16)*log(x)^2+((-8*x+8)*log(2/(x^2-2*x+1))-128*x+128)*log(x)+(-1+x)*log(2 /(x^2-2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x, algorithm="giac ")
Output:
12*x/(log(2) - log(x^2 - 2*x + 1) - 4*log(x) + 16)
Timed out. \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\int \frac {264\,x-\ln \left (x\right )\,\left (48\,x-48\right )+\ln \left (\frac {2}{x^2-2\,x+1}\right )\,\left (12\,x-12\right )-240}{256\,x-\ln \left (x\right )\,\left (128\,x+\ln \left (\frac {2}{x^2-2\,x+1}\right )\,\left (8\,x-8\right )-128\right )+\ln \left (\frac {2}{x^2-2\,x+1}\right )\,\left (32\,x-32\right )+{\ln \left (\frac {2}{x^2-2\,x+1}\right )}^2\,\left (x-1\right )+{\ln \left (x\right )}^2\,\left (16\,x-16\right )-256} \,d x \] Input:
int((264*x - log(x)*(48*x - 48) + log(2/(x^2 - 2*x + 1))*(12*x - 12) - 240 )/(256*x - log(x)*(128*x + log(2/(x^2 - 2*x + 1))*(8*x - 8) - 128) + log(2 /(x^2 - 2*x + 1))*(32*x - 32) + log(2/(x^2 - 2*x + 1))^2*(x - 1) + log(x)^ 2*(16*x - 16) - 256),x)
Output:
int((264*x - log(x)*(48*x - 48) + log(2/(x^2 - 2*x + 1))*(12*x - 12) - 240 )/(256*x - log(x)*(128*x + log(2/(x^2 - 2*x + 1))*(8*x - 8) - 128) + log(2 /(x^2 - 2*x + 1))*(32*x - 32) + log(2/(x^2 - 2*x + 1))^2*(x - 1) + log(x)^ 2*(16*x - 16) - 256), x)
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-240+264 x+(48-48 x) \log (x)+(-12+12 x) \log \left (\frac {2}{1-2 x+x^2}\right )}{-256+256 x+(-16+16 x) \log ^2(x)+(-32+32 x) \log \left (\frac {2}{1-2 x+x^2}\right )+(-1+x) \log ^2\left (\frac {2}{1-2 x+x^2}\right )+\log (x) \left (128-128 x+(8-8 x) \log \left (\frac {2}{1-2 x+x^2}\right )\right )} \, dx=\frac {12 x}{\mathrm {log}\left (\frac {2}{x^{2}-2 x +1}\right )-4 \,\mathrm {log}\left (x \right )+16} \] Input:
int(((-48*x+48)*log(x)+(12*x-12)*log(2/(x^2-2*x+1))+264*x-240)/((16*x-16)* log(x)^2+((-8*x+8)*log(2/(x^2-2*x+1))-128*x+128)*log(x)+(-1+x)*log(2/(x^2- 2*x+1))^2+(32*x-32)*log(2/(x^2-2*x+1))+256*x-256),x)
Output:
(12*x)/(log(2/(x**2 - 2*x + 1)) - 4*log(x) + 16)