Integrand size = 46, antiderivative size = 22 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=\frac {3 (x-\log (\log (4)))}{1+\frac {3 x}{\log \left (\frac {1}{x}\right )}} \] Output:
3*(x-ln(2*ln(2)))/(3*x/ln(1/x)+1)
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=3 x-\frac {9 x (x-\log (\log (4)))}{3 x+\log \left (\frac {1}{x}\right )} \] Input:
Integrate[(-9*x + 3*Log[x^(-1)]^2 + (9 + 9*Log[x^(-1)])*Log[Log[4]])/(9*x^ 2 + 6*x*Log[x^(-1)] + Log[x^(-1)]^2),x]
Output:
3*x - (9*x*(x - Log[Log[4]]))/(3*x + Log[x^(-1)])
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.
\(\displaystyle \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\log (\log (4)) \left (9 \log \left (\frac {1}{x}\right )+9\right )}{9 x^2+\log ^2\left (\frac {1}{x}\right )+6 x \log \left (\frac {1}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\log (\log (4)) \left (9 \log \left (\frac {1}{x}\right )+9\right )}{\left (3 x+\log \left (\frac {1}{x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {9 (3 x-1) (x-\log (\log (4)))}{\left (3 x+\log \left (\frac {1}{x}\right )\right )^2}-\frac {9 (2 x-\log (\log (4)))}{3 x+\log \left (\frac {1}{x}\right )}+3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 27 \int \frac {x^2}{\left (3 x+\log \left (\frac {1}{x}\right )\right )^2}dx+9 \log (\log (4)) \int \frac {1}{\left (3 x+\log \left (\frac {1}{x}\right )\right )^2}dx-9 (1+3 \log (\log (4))) \int \frac {x}{\left (3 x+\log \left (\frac {1}{x}\right )\right )^2}dx+9 \log (\log (4)) \int \frac {1}{3 x+\log \left (\frac {1}{x}\right )}dx-18 \int \frac {x}{3 x+\log \left (\frac {1}{x}\right )}dx+3 x\) |
Input:
Int[(-9*x + 3*Log[x^(-1)]^2 + (9 + 9*Log[x^(-1)])*Log[Log[4]])/(9*x^2 + 6* x*Log[x^(-1)] + Log[x^(-1)]^2),x]
Output:
$Aborted
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23
method | result | size |
risch | \(3 x +\frac {9 \left (-x +\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x}{\ln \left (\frac {1}{x}\right )+3 x}\) | \(27\) |
parallelrisch | \(\frac {9 \ln \left (2 \ln \left (2\right )\right ) x +3 x \ln \left (\frac {1}{x}\right )}{\ln \left (\frac {1}{x}\right )+3 x}\) | \(28\) |
norman | \(\frac {\left (-3 \ln \left (2\right )-3 \ln \left (\ln \left (2\right )\right )\right ) \ln \left (\frac {1}{x}\right )+3 x \ln \left (\frac {1}{x}\right )}{\ln \left (\frac {1}{x}\right )+3 x}\) | \(35\) |
derivativedivides | \(\frac {3 \ln \left (\frac {1}{x}\right )}{\frac {\ln \left (\frac {1}{x}\right )}{x}+3}+\frac {9 \ln \left (2\right )}{\frac {\ln \left (\frac {1}{x}\right )}{x}+3}+\frac {9 \ln \left (\ln \left (2\right )\right )}{\frac {\ln \left (\frac {1}{x}\right )}{x}+3}\) | \(53\) |
default | \(\frac {3 \ln \left (\frac {1}{x}\right )}{\frac {\ln \left (\frac {1}{x}\right )}{x}+3}+\frac {9 \ln \left (2\right )}{\frac {\ln \left (\frac {1}{x}\right )}{x}+3}+\frac {9 \ln \left (\ln \left (2\right )\right )}{\frac {\ln \left (\frac {1}{x}\right )}{x}+3}\) | \(53\) |
Input:
int(((9*ln(1/x)+9)*ln(2*ln(2))+3*ln(1/x)^2-9*x)/(ln(1/x)^2+6*x*ln(1/x)+9*x ^2),x,method=_RETURNVERBOSE)
Output:
3*x+9*(-x+ln(2)+ln(ln(2)))*x/(ln(1/x)+3*x)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=\frac {3 \, {\left (3 \, x \log \left (2 \, \log \left (2\right )\right ) + x \log \left (\frac {1}{x}\right )\right )}}{3 \, x + \log \left (\frac {1}{x}\right )} \] Input:
integrate(((9*log(1/x)+9)*log(2*log(2))+3*log(1/x)^2-9*x)/(log(1/x)^2+6*x* log(1/x)+9*x^2),x, algorithm="fricas")
Output:
3*(3*x*log(2*log(2)) + x*log(1/x))/(3*x + log(1/x))
Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=3 x + \frac {- 9 x^{2} + 9 x \log {\left (\log {\left (2 \right )} \right )} + 9 x \log {\left (2 \right )}}{3 x + \log {\left (\frac {1}{x} \right )}} \] Input:
integrate(((9*ln(1/x)+9)*ln(2*ln(2))+3*ln(1/x)**2-9*x)/(ln(1/x)**2+6*x*ln( 1/x)+9*x**2),x)
Output:
3*x + (-9*x**2 + 9*x*log(log(2)) + 9*x*log(2))/(3*x + log(1/x))
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=\frac {3 \, {\left (3 \, x {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} - x \log \left (x\right )\right )}}{3 \, x - \log \left (x\right )} \] Input:
integrate(((9*log(1/x)+9)*log(2*log(2))+3*log(1/x)^2-9*x)/(log(1/x)^2+6*x* log(1/x)+9*x^2),x, algorithm="maxima")
Output:
3*(3*x*(log(2) + log(log(2))) - x*log(x))/(3*x - log(x))
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=3 \, x - \frac {9 \, {\left (x^{2} - x \log \left (2\right ) - x \log \left (\log \left (2\right )\right )\right )}}{3 \, x - \log \left (x\right )} \] Input:
integrate(((9*log(1/x)+9)*log(2*log(2))+3*log(1/x)^2-9*x)/(log(1/x)^2+6*x* log(1/x)+9*x^2),x, algorithm="giac")
Output:
3*x - 9*(x^2 - x*log(2) - x*log(log(2)))/(3*x - log(x))
Time = 3.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=3\,x+\frac {x\,\ln \left ({\ln \left (4\right )}^9\right )-9\,x^2}{3\,x+\ln \left (\frac {1}{x}\right )} \] Input:
int((log(2*log(2))*(9*log(1/x) + 9) - 9*x + 3*log(1/x)^2)/(6*x*log(1/x) + log(1/x)^2 + 9*x^2),x)
Output:
3*x + (x*log(log(4)^9) - 9*x^2)/(3*x + log(1/x))
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-9 x+3 \log ^2\left (\frac {1}{x}\right )+\left (9+9 \log \left (\frac {1}{x}\right )\right ) \log (\log (4))}{9 x^2+6 x \log \left (\frac {1}{x}\right )+\log ^2\left (\frac {1}{x}\right )} \, dx=\frac {3 \,\mathrm {log}\left (x \right ) \left (-\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )+x \right )}{\mathrm {log}\left (x \right )-3 x} \] Input:
int(((9*log(1/x)+9)*log(2*log(2))+3*log(1/x)^2-9*x)/(log(1/x)^2+6*x*log(1/ x)+9*x^2),x)
Output:
(3*log(x)*( - log(2*log(2)) + x))/(log(x) - 3*x)