Integrand size = 71, antiderivative size = 27 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=4+x+\frac {5}{2} x^2 \left (-4-x-\frac {x}{\log \left (\log \left (\frac {x}{3}\right )\right )}\right ) \] Output:
5/2*x^2*(-4-x/ln(ln(1/3*x))-x)+4+x
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=x-10 x^2-\frac {5 x^3}{2}-\frac {5 x^3}{2 \log \left (\log \left (\frac {x}{3}\right )\right )} \] Input:
Integrate[(5*x^2 - 15*x^2*Log[x/3]*Log[Log[x/3]] + (2 - 40*x - 15*x^2)*Log [x/3]*Log[Log[x/3]]^2)/(2*Log[x/3]*Log[Log[x/3]]^2),x]
Output:
x - 10*x^2 - (5*x^3)/2 - (5*x^3)/(2*Log[Log[x/3]])
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.
\(\displaystyle \int \frac {5 x^2+\left (-15 x^2-40 x+2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {-15 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right ) x^2+5 x^2+\left (-15 x^2-40 x+2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{\log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {15 x^2}{\log \left (\log \left (\frac {x}{3}\right )\right )}+\frac {5 x^2}{\log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}-15 x^2-40 x+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (5 \int \frac {x^2}{\log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}dx-15 \int \frac {x^2}{\log \left (\log \left (\frac {x}{3}\right )\right )}dx-5 x^3-20 x^2+2 x\right )\) |
Input:
Int[(5*x^2 - 15*x^2*Log[x/3]*Log[Log[x/3]] + (2 - 40*x - 15*x^2)*Log[x/3]* Log[Log[x/3]]^2)/(2*Log[x/3]*Log[Log[x/3]]^2),x]
Output:
$Aborted
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-\frac {5 x^{3}}{2}-10 x^{2}+x -\frac {5 x^{3}}{2 \ln \left (\ln \left (\frac {x}{3}\right )\right )}\) | \(25\) |
parallelrisch | \(\frac {-5 \ln \left (\ln \left (\frac {x}{3}\right )\right ) x^{3}-5 x^{3}-20 x^{2} \ln \left (\ln \left (\frac {x}{3}\right )\right )+2 x \ln \left (\ln \left (\frac {x}{3}\right )\right )}{2 \ln \left (\ln \left (\frac {x}{3}\right )\right )}\) | \(44\) |
Input:
int(1/2*((-15*x^2-40*x+2)*ln(1/3*x)*ln(ln(1/3*x))^2-15*x^2*ln(1/3*x)*ln(ln (1/3*x))+5*x^2)/ln(1/3*x)/ln(ln(1/3*x))^2,x,method=_RETURNVERBOSE)
Output:
-5/2*x^3-10*x^2+x-5/2*x^3/ln(ln(1/3*x))
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=-\frac {5 \, x^{3} + {\left (5 \, x^{3} + 20 \, x^{2} - 2 \, x\right )} \log \left (\log \left (\frac {1}{3} \, x\right )\right )}{2 \, \log \left (\log \left (\frac {1}{3} \, x\right )\right )} \] Input:
integrate(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/ 3*x)*log(log(1/3*x))+5*x^2)/log(1/3*x)/log(log(1/3*x))^2,x, algorithm="fri cas")
Output:
-1/2*(5*x^3 + (5*x^3 + 20*x^2 - 2*x)*log(log(1/3*x)))/log(log(1/3*x))
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=- \frac {5 x^{3}}{2} - \frac {5 x^{3}}{2 \log {\left (\log {\left (\frac {x}{3} \right )} \right )}} - 10 x^{2} + x \] Input:
integrate(1/2*((-15*x**2-40*x+2)*ln(1/3*x)*ln(ln(1/3*x))**2-15*x**2*ln(1/3 *x)*ln(ln(1/3*x))+5*x**2)/ln(1/3*x)/ln(ln(1/3*x))**2,x)
Output:
-5*x**3/2 - 5*x**3/(2*log(log(x/3))) - 10*x**2 + x
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=-\frac {5}{2} \, x^{3} - 10 \, x^{2} - \frac {5 \, x^{3}}{2 \, \log \left (-\log \left (3\right ) + \log \left (x\right )\right )} + x \] Input:
integrate(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/ 3*x)*log(log(1/3*x))+5*x^2)/log(1/3*x)/log(log(1/3*x))^2,x, algorithm="max ima")
Output:
-5/2*x^3 - 10*x^2 - 5/2*x^3/log(-log(3) + log(x)) + x
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=-\frac {5}{2} \, x^{3} - 10 \, x^{2} - \frac {5 \, x^{3}}{2 \, \log \left (\log \left (\frac {1}{3} \, x\right )\right )} + x \] Input:
integrate(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/ 3*x)*log(log(1/3*x))+5*x^2)/log(1/3*x)/log(log(1/3*x))^2,x, algorithm="gia c")
Output:
-5/2*x^3 - 10*x^2 - 5/2*x^3/log(log(1/3*x)) + x
Time = 2.97 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=x-\frac {5\,x^3}{2\,\ln \left (\ln \left (x\right )-\ln \left (3\right )\right )}-10\,x^2-\frac {5\,x^3}{2} \] Input:
int(-((log(x/3)*log(log(x/3))^2*(40*x + 15*x^2 - 2))/2 - (5*x^2)/2 + (15*x ^2*log(x/3)*log(log(x/3)))/2)/(log(x/3)*log(log(x/3))^2),x)
Output:
x - (5*x^3)/(2*log(log(x) - log(3))) - 10*x^2 - (5*x^3)/2
Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=\frac {x \left (-5 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{3}\right )\right ) x^{2}-20 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{3}\right )\right ) x +2 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{3}\right )\right )-5 x^{2}\right )}{2 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x}{3}\right )\right )} \] Input:
int(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/3*x)*l og(log(1/3*x))+5*x^2)/log(1/3*x)/log(log(1/3*x))^2,x)
Output:
(x*( - 5*log(log(x/3))*x**2 - 20*log(log(x/3))*x + 2*log(log(x/3)) - 5*x** 2))/(2*log(log(x/3)))