Integrand size = 124, antiderivative size = 25 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=\log \left (4 x^2 (1+x)^2 \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )\right ) \]
Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \log (x)+2 \log (1+x)+\log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right ) \]
Integrate[(-4*x + 8*x^2 + 20*x^3 + (-3*x - 5*x^2)*Log[Log[2]] + (-8*x - 16 *x^2 + (2 + 4*x)*Log[Log[2]])*Log[(4*x - Log[Log[2]])/4])/(4*x^3 + 4*x^4 + (-x^2 - x^3)*Log[Log[2]] + (-4*x^2 - 4*x^3 + (x + x^2)*Log[Log[2]])*Log[( 4*x - Log[Log[2]])/4]),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.
\(\displaystyle \int \frac {20 x^3+8 x^2+\left (-16 x^2-8 x+(4 x+2) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )+\left (-5 x^2-3 x\right ) \log (\log (2))-4 x}{4 x^4+4 x^3+\left (-4 x^3-4 x^2+\left (x^2+x\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )+\left (-x^3-x^2\right ) \log (\log (2))} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {20 x^3+8 x^2+\left (-16 x^2-8 x+(4 x+2) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )+\left (-5 x^2-3 x\right ) \log (\log (2))-4 x}{x (x+1) (4 x-\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-20 x^3+16 x^2 \log \left (x-\frac {1}{4} \log (\log (2))\right )-8 x^2 \left (1-\frac {5}{8} \log (\log (2))\right )+8 x \left (1-\frac {1}{2} \log (\log (2))\right ) \log \left (x-\frac {1}{4} \log (\log (2))\right )+4 x \left (1+\frac {3}{4} \log (\log (2))\right )-2 \log (\log (2)) \log \left (x-\frac {1}{4} \log (\log (2))\right )}{x \log (\log (2)) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}+\frac {20 x^3-16 x^2 \log \left (x-\frac {1}{4} \log (\log (2))\right )+8 x^2 \left (1-\frac {5}{8} \log (\log (2))\right )-8 x \left (1-\frac {1}{2} \log (\log (2))\right ) \log \left (x-\frac {1}{4} \log (\log (2))\right )-4 x \left (1+\frac {3}{4} \log (\log (2))\right )+2 \log (\log (2)) \log \left (x-\frac {1}{4} \log (\log (2))\right )}{(x+1) (4+\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}+\frac {16 \left (20 x^3-16 x^2 \log \left (x-\frac {1}{4} \log (\log (2))\right )+8 x^2 \left (1-\frac {5}{8} \log (\log (2))\right )-8 x \left (1-\frac {1}{2} \log (\log (2))\right ) \log \left (x-\frac {1}{4} \log (\log (2))\right )-4 x \left (1+\frac {3}{4} \log (\log (2))\right )+2 \log (\log (2)) \log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}{\log (\log (2)) (4+\log (\log (2))) (4 x-\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {16 \int \frac {x^2}{x-\log \left (x-\frac {1}{4} \log (\log (2))\right )}dx}{\log (\log (2)) (4+\log (\log (2)))}+\frac {4 \int \frac {x^2}{x-\log \left (x-\frac {1}{4} \log (\log (2))\right )}dx}{4+\log (\log (2))}-\frac {4 \int \frac {x^2}{x-\log \left (x-\frac {1}{4} \log (\log (2))\right )}dx}{\log (\log (2))}+\frac {(4+\log (\log (2))) \int \frac {1}{x-\log \left (x-\frac {1}{4} \log (\log (2))\right )}dx}{\log (\log (2))}-\frac {4 \int \frac {1}{x-\log \left (x-\frac {1}{4} \log (\log (2))\right )}dx}{\log (\log (2))}-4 \int \frac {1}{(4 x-\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}dx-\frac {8 x^2}{\log (\log (2))}-4 x-\frac {4 x (2-\log (\log (2)))}{\log (\log (2))}+2 \log (x)+2 \log (x+1)+\frac {2 (2 x+1)^2}{4+\log (\log (2))}+\frac {8 (2 x+1)^2}{\log (\log (2)) (4+\log (\log (2)))}\) |
Int[(-4*x + 8*x^2 + 20*x^3 + (-3*x - 5*x^2)*Log[Log[2]] + (-8*x - 16*x^2 + (2 + 4*x)*Log[Log[2]])*Log[(4*x - Log[Log[2]])/4])/(4*x^3 + 4*x^4 + (-x^2 - x^3)*Log[Log[2]] + (-4*x^2 - 4*x^3 + (x + x^2)*Log[Log[2]])*Log[(4*x - Log[Log[2]])/4]),x]
3.16.76.3.1 Defintions of rubi rules used
Time = 0.62 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(2 \ln \left (x^{2}+x \right )+\ln \left (-x +\ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )\right )\) | \(23\) |
norman | \(2 \ln \left (x \right )+2 \ln \left (1+x \right )+\ln \left (x -\ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )\right )\) | \(25\) |
parallelrisch | \(2 \ln \left (x \right )+2 \ln \left (1+x \right )+\ln \left (x -\ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )\right )\) | \(25\) |
derivativedivides | \(2 \ln \left (4 x \right )+\ln \left (-4 \ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )+4 x \right )+2 \ln \left (4+4 x \right )\) | \(31\) |
default | \(2 \ln \left (4 x \right )+\ln \left (-4 \ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )+4 x \right )+2 \ln \left (4+4 x \right )\) | \(31\) |
int((((4*x+2)*ln(ln(2))-16*x^2-8*x)*ln(-1/4*ln(ln(2))+x)+(-5*x^2-3*x)*ln(l n(2))+20*x^3+8*x^2-4*x)/(((x^2+x)*ln(ln(2))-4*x^3-4*x^2)*ln(-1/4*ln(ln(2)) +x)+(-x^3-x^2)*ln(ln(2))+4*x^4+4*x^3),x,method=_RETURNVERBOSE)
Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \, \log \left (x^{2} + x\right ) + \log \left (-x + \log \left (x - \frac {1}{4} \, \log \left (\log \left (2\right )\right )\right )\right ) \]
integrate((((4*x+2)*log(log(2))-16*x^2-8*x)*log(-1/4*log(log(2))+x)+(-5*x^ 2-3*x)*log(log(2))+20*x^3+8*x^2-4*x)/(((x^2+x)*log(log(2))-4*x^3-4*x^2)*lo g(-1/4*log(log(2))+x)+(-x^3-x^2)*log(log(2))+4*x^4+4*x^3),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=\log {\left (- x + \log {\left (x - \frac {\log {\left (\log {\left (2 \right )} \right )}}{4} \right )} \right )} + 2 \log {\left (x^{2} + x \right )} \]
integrate((((4*x+2)*ln(ln(2))-16*x**2-8*x)*ln(-1/4*ln(ln(2))+x)+(-5*x**2-3 *x)*ln(ln(2))+20*x**3+8*x**2-4*x)/(((x**2+x)*ln(ln(2))-4*x**3-4*x**2)*ln(- 1/4*ln(ln(2))+x)+(-x**3-x**2)*ln(ln(2))+4*x**4+4*x**3),x)
Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \, \log \left (x + 1\right ) + 2 \, \log \left (x\right ) + \log \left (-x - 2 \, \log \left (2\right ) + \log \left (4 \, x - \log \left (\log \left (2\right )\right )\right )\right ) \]
integrate((((4*x+2)*log(log(2))-16*x^2-8*x)*log(-1/4*log(log(2))+x)+(-5*x^ 2-3*x)*log(log(2))+20*x^3+8*x^2-4*x)/(((x^2+x)*log(log(2))-4*x^3-4*x^2)*lo g(-1/4*log(log(2))+x)+(-x^3-x^2)*log(log(2))+4*x^4+4*x^3),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \, \log \left (x + 1\right ) + 2 \, \log \left (x\right ) + \log \left (-x - 2 \, \log \left (2\right ) + \log \left (4 \, x - \log \left (\log \left (2\right )\right )\right )\right ) \]
integrate((((4*x+2)*log(log(2))-16*x^2-8*x)*log(-1/4*log(log(2))+x)+(-5*x^ 2-3*x)*log(log(2))+20*x^3+8*x^2-4*x)/(((x^2+x)*log(log(2))-4*x^3-4*x^2)*lo g(-1/4*log(log(2))+x)+(-x^3-x^2)*log(log(2))+4*x^4+4*x^3),x, algorithm=\
Time = 13.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2\,\ln \left (x\,\left (x+1\right )\right )+\ln \left (\ln \left (x-\frac {\ln \left (\ln \left (2\right )\right )}{4}\right )-x\right ) \]