Integrand size = 73, antiderivative size = 23 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x} \] Output:
ln(1+9*ln(x^2)/x-exp(exp(2)))/x
Time = 0.44 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\log \left (1-e^{e^2}+\frac {9 \log \left (x^2\right )}{x}\right )}{x} \] Input:
Integrate[(18 - 9*Log[x^2] + (-x + E^E^2*x - 9*Log[x^2])*Log[(x - E^E^2*x + 9*Log[x^2])/x])/(x^3 - E^E^2*x^3 + 9*x^2*Log[x^2]),x]
Output:
Log[1 - E^E^2 + (9*Log[x^2])/x]/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.
\(\displaystyle \int \frac {-9 \log \left (x^2\right )+\left (-9 \log \left (x^2\right )+e^{e^2} x-x\right ) \log \left (\frac {9 \log \left (x^2\right )-e^{e^2} x+x}{x}\right )+18}{-e^{e^2} x^3+x^3+9 x^2 \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-9 \log \left (x^2\right )+\left (-9 \log \left (x^2\right )+e^{e^2} x-x\right ) \log \left (\frac {9 \log \left (x^2\right )-e^{e^2} x+x}{x}\right )+18}{\left (1-e^{e^2}\right ) x^3+9 x^2 \log \left (x^2\right )}dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {-9 \log \left (x^2\right )+\left (-9 \log \left (x^2\right )+e^{e^2} x-x\right ) \log \left (\frac {9 \log \left (x^2\right )-e^{e^2} x+x}{x}\right )+18}{x^2 \left (9 \log \left (x^2\right )+\left (1-e^{e^2}\right ) x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {9 \left (2-\log \left (x^2\right )\right )}{x^2 \left (9 \log \left (x^2\right )+\left (1-e^{e^2}\right ) x\right )}-\frac {\log \left (\frac {9 \log \left (x^2\right )}{x}-e^{e^2}+1\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 18 \int \frac {1}{x^2 \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )}dx+\left (1-e^{e^2}\right ) \int \frac {1}{x \left (\left (1-e^{e^2}\right ) x+9 \log \left (x^2\right )\right )}dx-\int \frac {\log \left (\frac {9 \log \left (x^2\right )}{x}-e^{e^2}+1\right )}{x^2}dx+\frac {1}{x}\) |
Input:
Int[(18 - 9*Log[x^2] + (-x + E^E^2*x - 9*Log[x^2])*Log[(x - E^E^2*x + 9*Lo g[x^2])/x])/(x^3 - E^E^2*x^3 + 9*x^2*Log[x^2]),x]
Output:
$Aborted
Time = 1.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {9 \ln \left (x^{2}\right )-x \,{\mathrm e}^{{\mathrm e}^{2}}+x}{x}\right )}{x}\) | \(24\) |
Input:
int(((-9*ln(x^2)+x*exp(exp(2))-x)*ln((9*ln(x^2)-x*exp(exp(2))+x)/x)-9*ln(x ^2)+18)/(9*x^2*ln(x^2)-x^3*exp(exp(2))+x^3),x,method=_RETURNVERBOSE)
Output:
ln((9*ln(x^2)-x*exp(exp(2))+x)/x)/x
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\log \left (-\frac {x e^{\left (e^{2}\right )} - x - 9 \, \log \left (x^{2}\right )}{x}\right )}{x} \] Input:
integrate(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/ x)-9*log(x^2)+18)/(9*x^2*log(x^2)-x^3*exp(exp(2))+x^3),x, algorithm="frica s")
Output:
log(-(x*e^(e^2) - x - 9*log(x^2))/x)/x
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\log {\left (\frac {- x e^{e^{2}} + x + 9 \log {\left (x^{2} \right )}}{x} \right )}}{x} \] Input:
integrate(((-9*ln(x**2)+x*exp(exp(2))-x)*ln((9*ln(x**2)-x*exp(exp(2))+x)/x )-9*ln(x**2)+18)/(9*x**2*ln(x**2)-x**3*exp(exp(2))+x**3),x)
Output:
log((-x*exp(exp(2)) + x + 9*log(x**2))/x)/x
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\log \left (-x {\left (e^{\left (e^{2}\right )} - 1\right )} + 18 \, \log \left (x\right )\right ) - \log \left (x\right )}{x} \] Input:
integrate(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/ x)-9*log(x^2)+18)/(9*x^2*log(x^2)-x^3*exp(exp(2))+x^3),x, algorithm="maxim a")
Output:
(log(-x*(e^(e^2) - 1) + 18*log(x)) - log(x))/x
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\log \left (-x e^{\left (e^{2}\right )} + x + 9 \, \log \left (x^{2}\right )\right ) - \log \left (x\right )}{x} \] Input:
integrate(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/ x)-9*log(x^2)+18)/(9*x^2*log(x^2)-x^3*exp(exp(2))+x^3),x, algorithm="giac" )
Output:
(log(-x*e^(e^2) + x + 9*log(x^2)) - log(x))/x
Time = 2.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\ln \left (\frac {x+\ln \left (x^{18}\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^2}}{x}\right )}{x} \] Input:
int(-(9*log(x^2) + log((x + 9*log(x^2) - x*exp(exp(2)))/x)*(x + 9*log(x^2) - x*exp(exp(2))) - 18)/(9*x^2*log(x^2) - x^3*exp(exp(2)) + x^3),x)
Output:
log((x + log(x^18) - x*exp(exp(2)))/x)/x
Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {18-9 \log \left (x^2\right )+\left (-x+e^{e^2} x-9 \log \left (x^2\right )\right ) \log \left (\frac {x-e^{e^2} x+9 \log \left (x^2\right )}{x}\right )}{x^3-e^{e^2} x^3+9 x^2 \log \left (x^2\right )} \, dx=\frac {\mathrm {log}\left (\frac {-e^{e^{2}} x +9 \,\mathrm {log}\left (x^{2}\right )+x}{x}\right )}{x} \] Input:
int(((-9*log(x^2)+x*exp(exp(2))-x)*log((9*log(x^2)-x*exp(exp(2))+x)/x)-9*l og(x^2)+18)/(9*x^2*log(x^2)-x^3*exp(exp(2))+x^3),x)
Output:
log(( - e**(e**2)*x + 9*log(x**2) + x)/x)/x