Integrand size = 67, antiderivative size = 29 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=4+\frac {-\frac {x^2}{2+\log (4)}+\log \left (\frac {1}{\log (x)}\right )}{-1+\frac {2}{x}} \] Output:
(ln(1/ln(x))-x^2/(2+2*ln(2)))/(2/x-1)+4
Time = 0.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\frac {8-4 x+x^3-2 (2+\log (4)) \log \left (\frac {1}{\log (x)}\right )+(-2+x) (2+\log (4)) \log (\log (x))}{(-2+x) (2+\log (4))} \] Input:
Integrate[(-4 + 2*x + (-2 + x)*Log[4] + (-6*x^2 + 2*x^3)*Log[x] + (4 + 2*L og[4])*Log[x]*Log[Log[x]^(-1)])/((8 - 8*x + 2*x^2 + (4 - 4*x + x^2)*Log[4] )*Log[x]),x]
Output:
(8 - 4*x + x^3 - 2*(2 + Log[4])*Log[Log[x]^(-1)] + (-2 + x)*(2 + Log[4])*L og[Log[x]])/((-2 + x)*(2 + Log[4]))
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.
\(\displaystyle \int \frac {\left (2 x^3-6 x^2\right ) \log (x)+2 x+(x-2) \log (4)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )-4}{\left (2 x^2+\left (x^2-4 x+4\right ) \log (4)-8 x+8\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (\frac {(x-3) x^2}{2+\log (4)}+\log \left (\frac {1}{\log (x)}\right )\right )+\frac {x-2}{\log (x)}}{(2-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^3 \log (x)-6 x^2 \log (x)+2 x (1+\log (2))-4 (1+\log (2))}{(2-x)^2 (2+\log (4)) \log (x)}+\frac {2 \log \left (\frac {1}{\log (x)}\right )}{(x-2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (1+\log (2)) \int \frac {1}{(x-2) \log (x)}dx}{2+\log (4)}+2 \int \frac {\log \left (\frac {1}{\log (x)}\right )}{(x-2)^2}dx-\frac {x^3}{(2-x) (2+\log (4))}\) |
Input:
Int[(-4 + 2*x + (-2 + x)*Log[4] + (-6*x^2 + 2*x^3)*Log[x] + (4 + 2*Log[4]) *Log[x]*Log[Log[x]^(-1)])/((8 - 8*x + 2*x^2 + (4 - 4*x + x^2)*Log[4])*Log[ x]),x]
Output:
$Aborted
Time = 1.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(\frac {-2 \ln \left (2\right ) x \ln \left (\frac {1}{\ln \left (x \right )}\right )+x^{3}-2 x \ln \left (\frac {1}{\ln \left (x \right )}\right )}{2 \left (1+\ln \left (2\right )\right ) \left (-2+x \right )}\) | \(36\) |
risch | \(\frac {2 \ln \left (\ln \left (x \right )\right )}{-2+x}+\frac {2 \ln \left (2\right ) \ln \left (\ln \left (x \right )\right ) x +x^{3}-4 \ln \left (2\right ) \ln \left (\ln \left (x \right )\right )+2 x \ln \left (\ln \left (x \right )\right )-4 \ln \left (\ln \left (x \right )\right )-4 x +8}{2 x \ln \left (2\right )-4 \ln \left (2\right )+2 x -4}\) | \(61\) |
Input:
int(((4*ln(2)+4)*ln(x)*ln(1/ln(x))+(2*x^3-6*x^2)*ln(x)+2*(-2+x)*ln(2)+2*x- 4)/(2*(x^2-4*x+4)*ln(2)+2*x^2-8*x+8)/ln(x),x,method=_RETURNVERBOSE)
Output:
1/2*(-2*ln(2)*x*ln(1/ln(x))+x^3-2*x*ln(1/ln(x)))/(1+ln(2))/(-2+x)
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\frac {x^{3} - 2 \, {\left (x \log \left (2\right ) + x\right )} \log \left (\frac {1}{\log \left (x\right )}\right ) - 4 \, x + 8}{2 \, {\left ({\left (x - 2\right )} \log \left (2\right ) + x - 2\right )}} \] Input:
integrate(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(-2+x) *log(2)+2*x-4)/(2*(x^2-4*x+4)*log(2)+2*x^2-8*x+8)/log(x),x, algorithm="fri cas")
Output:
1/2*(x^3 - 2*(x*log(2) + x)*log(1/log(x)) - 4*x + 8)/((x - 2)*log(2) + x - 2)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\frac {x^{2}}{2 \log {\left (2 \right )} + 2} + \frac {x}{\log {\left (2 \right )} + 1} + \log {\left (\log {\left (x \right )} \right )} + \frac {4}{x \left (\log {\left (2 \right )} + 1\right ) - 2 - 2 \log {\left (2 \right )}} - \frac {2 \log {\left (\frac {1}{\log {\left (x \right )}} \right )}}{x - 2} \] Input:
integrate(((4*ln(2)+4)*ln(x)*ln(1/ln(x))+(2*x**3-6*x**2)*ln(x)+2*(-2+x)*ln (2)+2*x-4)/(2*(x**2-4*x+4)*ln(2)+2*x**2-8*x+8)/ln(x),x)
Output:
x**2/(2*log(2) + 2) + x/(log(2) + 1) + log(log(x)) + 4/(x*(log(2) + 1) - 2 - 2*log(2)) - 2*log(1/log(x))/(x - 2)
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\frac {x^{3} + 2 \, x {\left (\log \left (2\right ) + 1\right )} \log \left (\log \left (x\right )\right ) - 4 \, x + 8}{2 \, {\left (x {\left (\log \left (2\right ) + 1\right )} - 2 \, \log \left (2\right ) - 2\right )}} \] Input:
integrate(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(-2+x) *log(2)+2*x-4)/(2*(x^2-4*x+4)*log(2)+2*x^2-8*x+8)/log(x),x, algorithm="max ima")
Output:
1/2*(x^3 + 2*x*(log(2) + 1)*log(log(x)) - 4*x + 8)/(x*(log(2) + 1) - 2*log (2) - 2)
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\frac {x^{2}}{2 \, {\left (\log \left (2\right ) + 1\right )}} + \frac {x}{\log \left (2\right ) + 1} + \frac {2 \, \log \left (\log \left (x\right )\right )}{x - 2} + \frac {4}{x \log \left (2\right ) + x - 2 \, \log \left (2\right ) - 2} + \log \left (\log \left (x\right )\right ) \] Input:
integrate(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(-2+x) *log(2)+2*x-4)/(2*(x^2-4*x+4)*log(2)+2*x^2-8*x+8)/log(x),x, algorithm="gia c")
Output:
1/2*x^2/(log(2) + 1) + x/(log(2) + 1) + 2*log(log(x))/(x - 2) + 4/(x*log(2 ) + x - 2*log(2) - 2) + log(log(x))
Timed out. \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\int \frac {2\,x-\ln \left (x\right )\,\left (6\,x^2-2\,x^3\right )+2\,\ln \left (2\right )\,\left (x-2\right )+\ln \left (\frac {1}{\ln \left (x\right )}\right )\,\ln \left (x\right )\,\left (4\,\ln \left (2\right )+4\right )-4}{\ln \left (x\right )\,\left (2\,x^2-8\,x+2\,\ln \left (2\right )\,\left (x^2-4\,x+4\right )+8\right )} \,d x \] Input:
int((2*x - log(x)*(6*x^2 - 2*x^3) + 2*log(2)*(x - 2) + log(1/log(x))*log(x )*(4*log(2) + 4) - 4)/(log(x)*(2*x^2 - 8*x + 2*log(2)*(x^2 - 4*x + 4) + 8) ),x)
Output:
int((2*x - log(x)*(6*x^2 - 2*x^3) + 2*log(2)*(x - 2) + log(1/log(x))*log(x )*(4*log(2) + 4) - 4)/(log(x)*(2*x^2 - 8*x + 2*log(2)*(x^2 - 4*x + 4) + 8) ), x)
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-4+2 x+(-2+x) \log (4)+\left (-6 x^2+2 x^3\right ) \log (x)+(4+2 \log (4)) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (8-8 x+2 x^2+\left (4-4 x+x^2\right ) \log (4)\right ) \log (x)} \, dx=\frac {x \left (2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (2\right )+2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+x^{2}\right )}{2 \,\mathrm {log}\left (2\right ) x -4 \,\mathrm {log}\left (2\right )+2 x -4} \] Input:
int(((4*log(2)+4)*log(x)*log(1/log(x))+(2*x^3-6*x^2)*log(x)+2*(-2+x)*log(2 )+2*x-4)/(2*(x^2-4*x+4)*log(2)+2*x^2-8*x+8)/log(x),x)
Output:
(x*(2*log(log(x))*log(2) + 2*log(log(x)) + x**2))/(2*(log(2)*x - 2*log(2) + x - 2))