Integrand size = 80, antiderivative size = 19 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=61-x \left (-4+\frac {2 \log (2)}{1+x+\log (\log (x))}\right ) \] Output:
61-x*(ln(2)/(1/2+1/2*x+1/2*ln(ln(x)))-4)
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=2 \left (2 x-\frac {x \log (2)}{1+x+\log (\log (x))}\right ) \] Input:
Integrate[(2*Log[2] + (4 + 8*x + 4*x^2 - 2*Log[2])*Log[x] + (8 + 8*x - 2*L og[2])*Log[x]*Log[Log[x]] + 4*Log[x]*Log[Log[x]]^2)/((1 + 2*x + x^2)*Log[x ] + (2 + 2*x)*Log[x]*Log[Log[x]] + Log[x]*Log[Log[x]]^2),x]
Output:
2*(2*x - (x*Log[2])/(1 + x + Log[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.
\(\displaystyle \int \frac {\left (4 x^2+8 x+4-2 \log (2)\right ) \log (x)+4 \log (x) \log ^2(\log (x))+(8 x+8-2 \log (2)) \log (x) \log (\log (x))+2 \log (2)}{\left (x^2+2 x+1\right ) \log (x)+\log (x) \log ^2(\log (x))+(2 x+2) \log (x) \log (\log (x))} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (4 x^2+8 x+4-2 \log (2)\right ) \log (x)+4 \log (x) \log ^2(\log (x))+(8 x+8-2 \log (2)) \log (x) \log (\log (x))+2 \log (2)}{\log (x) (x+\log (\log (x))+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log (4) (x \log (x)+1)}{\log (x) (x+\log (\log (x))+1)^2}-\frac {\log (4)}{x+\log (\log (x))+1}+4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (4) \int \frac {x}{(x+\log (\log (x))+1)^2}dx+\log (4) \int \frac {1}{\log (x) (x+\log (\log (x))+1)^2}dx-\log (4) \int \frac {1}{x+\log (\log (x))+1}dx+4 x\) |
Input:
Int[(2*Log[2] + (4 + 8*x + 4*x^2 - 2*Log[2])*Log[x] + (8 + 8*x - 2*Log[2]) *Log[x]*Log[Log[x]] + 4*Log[x]*Log[Log[x]]^2)/((1 + 2*x + x^2)*Log[x] + (2 + 2*x)*Log[x]*Log[Log[x]] + Log[x]*Log[Log[x]]^2),x]
Output:
$Aborted
Time = 0.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
risch | \(4 x -\frac {2 \ln \left (2\right ) x}{1+\ln \left (\ln \left (x \right )\right )+x}\) | \(18\) |
parallelrisch | \(-\frac {2 x \ln \left (2\right )-4 x^{2}-4 x \ln \left (\ln \left (x \right )\right )-4 x}{1+\ln \left (\ln \left (x \right )\right )+x}\) | \(31\) |
default | \(4 x +2 \ln \left (2\right ) \left (-\frac {x}{\left (x \ln \left (x \right )+1\right ) \left (1+\ln \left (\ln \left (x \right )\right )+x \right )}-\frac {\ln \left (x \right ) x^{2}}{\left (x \ln \left (x \right )+1\right ) \left (1+\ln \left (\ln \left (x \right )\right )+x \right )}\right )\) | \(52\) |
Input:
int((4*ln(x)*ln(ln(x))^2+(-2*ln(2)+8*x+8)*ln(x)*ln(ln(x))+(-2*ln(2)+4*x^2+ 8*x+4)*ln(x)+2*ln(2))/(ln(x)*ln(ln(x))^2+(2+2*x)*ln(x)*ln(ln(x))+(x^2+2*x+ 1)*ln(x)),x,method=_RETURNVERBOSE)
Output:
4*x-2*ln(2)*x/(1+ln(ln(x))+x)
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=\frac {2 \, {\left (2 \, x^{2} - x \log \left (2\right ) + 2 \, x \log \left (\log \left (x\right )\right ) + 2 \, x\right )}}{x + \log \left (\log \left (x\right )\right ) + 1} \] Input:
integrate((4*log(x)*log(log(x))^2+(-2*log(2)+8*x+8)*log(x)*log(log(x))+(-2 *log(2)+4*x^2+8*x+4)*log(x)+2*log(2))/(log(x)*log(log(x))^2+(2+2*x)*log(x) *log(log(x))+(x^2+2*x+1)*log(x)),x, algorithm="fricas")
Output:
2*(2*x^2 - x*log(2) + 2*x*log(log(x)) + 2*x)/(x + log(log(x)) + 1)
Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=4 x - \frac {2 x \log {\left (2 \right )}}{x + \log {\left (\log {\left (x \right )} \right )} + 1} \] Input:
integrate((4*ln(x)*ln(ln(x))**2+(-2*ln(2)+8*x+8)*ln(x)*ln(ln(x))+(-2*ln(2) +4*x**2+8*x+4)*ln(x)+2*ln(2))/(ln(x)*ln(ln(x))**2+(2+2*x)*ln(x)*ln(ln(x))+ (x**2+2*x+1)*ln(x)),x)
Output:
4*x - 2*x*log(2)/(x + log(log(x)) + 1)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=\frac {2 \, {\left (2 \, x^{2} - x {\left (\log \left (2\right ) - 2\right )} + 2 \, x \log \left (\log \left (x\right )\right )\right )}}{x + \log \left (\log \left (x\right )\right ) + 1} \] Input:
integrate((4*log(x)*log(log(x))^2+(-2*log(2)+8*x+8)*log(x)*log(log(x))+(-2 *log(2)+4*x^2+8*x+4)*log(x)+2*log(2))/(log(x)*log(log(x))^2+(2+2*x)*log(x) *log(log(x))+(x^2+2*x+1)*log(x)),x, algorithm="maxima")
Output:
2*(2*x^2 - x*(log(2) - 2) + 2*x*log(log(x)))/(x + log(log(x)) + 1)
Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=4 \, x - \frac {2 \, x \log \left (2\right )}{x + \log \left (\log \left (x\right )\right ) + 1} \] Input:
integrate((4*log(x)*log(log(x))^2+(-2*log(2)+8*x+8)*log(x)*log(log(x))+(-2 *log(2)+4*x^2+8*x+4)*log(x)+2*log(2))/(log(x)*log(log(x))^2+(2+2*x)*log(x) *log(log(x))+(x^2+2*x+1)*log(x)),x, algorithm="giac")
Output:
4*x - 2*x*log(2)/(x + log(log(x)) + 1)
Time = 4.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=\frac {4\,x+\ln \left (4\right )+\ln \left (\ln \left (x\right )\right )\,\ln \left (4\right )+4\,x\,\ln \left (\ln \left (x\right )\right )+4\,x^2}{x+\ln \left (\ln \left (x\right )\right )+1} \] Input:
int((2*log(2) + log(x)*(8*x - 2*log(2) + 4*x^2 + 4) + 4*log(log(x))^2*log( x) + log(log(x))*log(x)*(8*x - 2*log(2) + 8))/(log(x)*(2*x + x^2 + 1) + lo g(log(x))^2*log(x) + log(log(x))*log(x)*(2*x + 2)),x)
Output:
(4*x + log(4) + log(log(x))*log(4) + 4*x*log(log(x)) + 4*x^2)/(x + log(log (x)) + 1)
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int \frac {2 \log (2)+\left (4+8 x+4 x^2-2 \log (2)\right ) \log (x)+(8+8 x-2 \log (2)) \log (x) \log (\log (x))+4 \log (x) \log ^2(\log (x))}{\left (1+2 x+x^2\right ) \log (x)+(2+2 x) \log (x) \log (\log (x))+\log (x) \log ^2(\log (x))} \, dx=\frac {2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (2\right )+4 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x -4 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+2 \,\mathrm {log}\left (2\right )+4 x^{2}-4}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+x +1} \] Input:
int((4*log(x)*log(log(x))^2+(-2*log(2)+8*x+8)*log(x)*log(log(x))+(-2*log(2 )+4*x^2+8*x+4)*log(x)+2*log(2))/(log(x)*log(log(x))^2+(2+2*x)*log(x)*log(l og(x))+(x^2+2*x+1)*log(x)),x)
Output:
(2*(log(log(x))*log(2) + 2*log(log(x))*x - 2*log(log(x)) + log(2) + 2*x**2 - 2))/(log(log(x)) + x + 1)