Integrand size = 53, antiderivative size = 19 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=\log \left (e^x-\frac {12 \log \left (\frac {\log (2)}{3}\right )}{\log (\log (x))}\right ) \] Output:
ln(exp(x)-12*ln(1/3*ln(2))/ln(ln(x)))
Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=-\log (\log (\log (x)))+\log \left (-12 \log \left (\frac {\log (2)}{3}\right )+e^x \log (\log (x))\right ) \] Input:
Integrate[(12*Log[Log[2]/3] + E^x*x*Log[x]*Log[Log[x]]^2)/(-12*x*Log[x]*Lo g[Log[2]/3]*Log[Log[x]] + E^x*x*Log[x]*Log[Log[x]]^2),x]
Output:
-Log[Log[Log[x]]] + Log[-12*Log[Log[2]/3] + E^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.
\(\displaystyle \int \frac {e^x x \log (x) \log ^2(\log (x))+12 \log \left (\frac {\log (2)}{3}\right )}{e^x x \log (x) \log ^2(\log (x))-12 x \log \left (\frac {\log (2)}{3}\right ) \log (x) \log (\log (x))} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^x x \log (x) \log ^2(\log (x))-12 \log \left (\frac {\log (2)}{3}\right )}{x \log (x) \log (\log (x)) \left (12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (1-\frac {12 \log \left (\frac {\log (2)}{3}\right ) (x \log (x) \log (\log (x))+1)}{x \log (x) \log (\log (x)) \left (12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -12 \log \left (\frac {\log (2)}{3}\right ) \int \frac {1}{12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))}dx+12 \log \left (\frac {\log (2)}{3}\right ) \int \frac {1}{x \log (x) \log (\log (x)) \left (e^x \log (\log (x))-12 \log \left (\frac {\log (2)}{3}\right )\right )}dx+x\) |
Input:
Int[(12*Log[Log[2]/3] + E^x*x*Log[x]*Log[Log[x]]^2)/(-12*x*Log[x]*Log[Log[ 2]/3]*Log[Log[x]] + E^x*x*Log[x]*Log[Log[x]]^2),x]
Output:
$Aborted
Time = 4.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (\ln \left (x \right )\right )\right )+\ln \left ({\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )-12 \ln \left (\frac {\ln \left (2\right )}{3}\right )\right )\) | \(23\) |
risch | \(x -\ln \left (\ln \left (\ln \left (x \right )\right )\right )+\ln \left (\ln \left (\ln \left (x \right )\right )-12 \left (-\ln \left (3\right )+\ln \left (\ln \left (2\right )\right )\right ) {\mathrm e}^{-x}\right )\) | \(28\) |
Input:
int((x*exp(x)*ln(x)*ln(ln(x))^2+12*ln(1/3*ln(2)))/(x*exp(x)*ln(x)*ln(ln(x) )^2-12*x*ln(x)*ln(1/3*ln(2))*ln(ln(x))),x,method=_RETURNVERBOSE)
Output:
-ln(ln(ln(x)))+ln(exp(x)*ln(ln(x))-12*ln(1/3*ln(2)))
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=x + \log \left ({\left (e^{x} \log \left (\log \left (x\right )\right ) - 12 \, \log \left (\frac {1}{3} \, \log \left (2\right )\right )\right )} e^{\left (-x\right )}\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:
integrate((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log (x)*log(log(x))^2-12*x*log(x)*log(1/3*log(2))*log(log(x))),x, algorithm="f ricas")
Output:
x + log((e^x*log(log(x)) - 12*log(1/3*log(2)))*e^(-x)) - log(log(log(x)))
Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=\log {\left (e^{x} + \frac {- 12 \log {\left (\log {\left (2 \right )} \right )} + 12 \log {\left (3 \right )}}{\log {\left (\log {\left (x \right )} \right )}} \right )} \] Input:
integrate((x*exp(x)*ln(x)*ln(ln(x))**2+12*ln(1/3*ln(2)))/(x*exp(x)*ln(x)*l n(ln(x))**2-12*x*ln(x)*ln(1/3*ln(2))*ln(ln(x))),x)
Output:
log(exp(x) + (-12*log(log(2)) + 12*log(3))/log(log(x)))
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=x + \log \left ({\left (e^{x} \log \left (\log \left (x\right )\right ) + 12 \, \log \left (3\right ) - 12 \, \log \left (\log \left (2\right )\right )\right )} e^{\left (-x\right )}\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:
integrate((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log (x)*log(log(x))^2-12*x*log(x)*log(1/3*log(2))*log(log(x))),x, algorithm="m axima")
Output:
x + log((e^x*log(log(x)) + 12*log(3) - 12*log(log(2)))*e^(-x)) - log(log(l og(x)))
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=\log \left (e^{x} \log \left (\log \left (x\right )\right ) + 12 \, \log \left (3\right ) - 12 \, \log \left (\log \left (2\right )\right )\right ) - \log \left (\log \left (\log \left (x\right )\right )\right ) \] Input:
integrate((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log (x)*log(log(x))^2-12*x*log(x)*log(1/3*log(2))*log(log(x))),x, algorithm="g iac")
Output:
log(e^x*log(log(x)) + 12*log(3) - 12*log(log(2))) - log(log(log(x)))
Timed out. \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=\int -\frac {x\,{\mathrm {e}}^x\,\ln \left (x\right )\,{\ln \left (\ln \left (x\right )\right )}^2+12\,\ln \left (\frac {\ln \left (2\right )}{3}\right )}{12\,x\,\ln \left (\frac {\ln \left (2\right )}{3}\right )\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )-x\,{\ln \left (\ln \left (x\right )\right )}^2\,{\mathrm {e}}^x\,\ln \left (x\right )} \,d x \] Input:
int(-(12*log(log(2)/3) + x*log(log(x))^2*exp(x)*log(x))/(12*x*log(log(2)/3 )*log(log(x))*log(x) - x*log(log(x))^2*exp(x)*log(x)),x)
Output:
int(-(12*log(log(2)/3) + x*log(log(x))^2*exp(x)*log(x))/(12*x*log(log(2)/3 )*log(log(x))*log(x) - x*log(log(x))^2*exp(x)*log(x)), x)
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx=-\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )\right )+\mathrm {log}\left (e^{x} \mathrm {log}\left (\mathrm {log}\left (x \right )\right )-12 \,\mathrm {log}\left (\frac {\mathrm {log}\left (2\right )}{3}\right )\right ) \] Input:
int((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log(x)*lo g(log(x))^2-12*x*log(x)*log(1/3*log(2))*log(log(x))),x)
Output:
- log(log(log(x))) + log(e**x*log(log(x)) - 12*log(log(2)/3))