Integrand size = 76, antiderivative size = 27 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\left (-\frac {5}{3}+x-x^2\right ) \left (4+\frac {x^3}{2 \log \left (\log \left (x^2\right )\right )}\right ) \]
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=4 x-4 x^2-\frac {x^3 \left (5-3 x+3 x^2\right )}{6 \log \left (\log \left (x^2\right )\right )} \]
Integrate[(10*x^2 - 6*x^3 + 6*x^4 + (-15*x^2 + 12*x^3 - 15*x^4)*Log[x^2]*L og[Log[x^2]] + (24 - 48*x)*Log[x^2]*Log[Log[x^2]]^2)/(6*Log[x^2]*Log[Log[x ^2]]^2),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.
\(\displaystyle \int \frac {6 x^4-6 x^3+10 x^2+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )+\left (-15 x^4+12 x^3-15 x^2\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {6 x^4-6 x^3+10 x^2+24 (1-2 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )-3 \left (5 x^4-4 x^3+5 x^2\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{6} \int \left (-\frac {3 \left (5 x^2-4 x+5\right ) x^2}{\log \left (\log \left (x^2\right )\right )}+\frac {2 \left (3 x^2-3 x+5\right ) x^2}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}-24 (2 x-1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (-3 \text {Subst}\left (\int \frac {x}{\log (x) \log ^2(\log (x))}dx,x,x^2\right )+6 \text {Subst}\left (\int \frac {x}{\log (\log (x))}dx,x,x^2\right )+10 \int \frac {x^2}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}dx-15 \int \frac {x^2}{\log \left (\log \left (x^2\right )\right )}dx+6 \int \frac {x^4}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}dx-15 \int \frac {x^4}{\log \left (\log \left (x^2\right )\right )}dx-6 (1-2 x)^2\right )\) |
Int[(10*x^2 - 6*x^3 + 6*x^4 + (-15*x^2 + 12*x^3 - 15*x^4)*Log[x^2]*Log[Log [x^2]] + (24 - 48*x)*Log[x^2]*Log[Log[x^2]]^2)/(6*Log[x^2]*Log[Log[x^2]]^2 ),x]
3.7.20.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.59 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63
method | result | size |
parallelrisch | \(-\frac {3 x^{5}-3 x^{4}+5 x^{3}+24 x^{2} \ln \left (\ln \left (x^{2}\right )\right )-24 x \ln \left (\ln \left (x^{2}\right )\right )}{6 \ln \left (\ln \left (x^{2}\right )\right )}\) | \(44\) |
int(1/6*((-48*x+24)*ln(x^2)*ln(ln(x^2))^2+(-15*x^4+12*x^3-15*x^2)*ln(x^2)* ln(ln(x^2))+6*x^4-6*x^3+10*x^2)/ln(x^2)/ln(ln(x^2))^2,x,method=_RETURNVERB OSE)
Time = 0.52 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-\frac {3 \, x^{5} - 3 \, x^{4} + 5 \, x^{3} + 24 \, {\left (x^{2} - x\right )} \log \left (\log \left (x^{2}\right )\right )}{6 \, \log \left (\log \left (x^{2}\right )\right )} \]
integrate(1/6*((-48*x+24)*log(x^2)*log(log(x^2))^2+(-15*x^4+12*x^3-15*x^2) *log(x^2)*log(log(x^2))+6*x^4-6*x^3+10*x^2)/log(x^2)/log(log(x^2))^2,x, al gorithm=\
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=- 4 x^{2} + 4 x + \frac {- 3 x^{5} + 3 x^{4} - 5 x^{3}}{6 \log {\left (\log {\left (x^{2} \right )} \right )}} \]
integrate(1/6*((-48*x+24)*ln(x**2)*ln(ln(x**2))**2+(-15*x**4+12*x**3-15*x* *2)*ln(x**2)*ln(ln(x**2))+6*x**4-6*x**3+10*x**2)/ln(x**2)/ln(ln(x**2))**2, x)
Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-4 \, x^{2} + 4 \, x - \frac {3 \, x^{5} - 3 \, x^{4} + 5 \, x^{3}}{6 \, {\left (\log \left (2\right ) + \log \left (\log \left (x\right )\right )\right )}} \]
integrate(1/6*((-48*x+24)*log(x^2)*log(log(x^2))^2+(-15*x^4+12*x^3-15*x^2) *log(x^2)*log(log(x^2))+6*x^4-6*x^3+10*x^2)/log(x^2)/log(log(x^2))^2,x, al gorithm=\
Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-4 \, x^{2} + 4 \, x - \frac {3 \, x^{5} - 3 \, x^{4} + 5 \, x^{3}}{6 \, \log \left (\log \left (x^{2}\right )\right )} \]
integrate(1/6*((-48*x+24)*log(x^2)*log(log(x^2))^2+(-15*x^4+12*x^3-15*x^2) *log(x^2)*log(log(x^2))+6*x^4-6*x^3+10*x^2)/log(x^2)/log(log(x^2))^2,x, al gorithm=\
Time = 11.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.96 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=4\,x-\frac {\frac {x^3\,\left (3\,x^2-3\,x+5\right )}{6}-\frac {x^3\,\ln \left (x^2\right )\,\ln \left (\ln \left (x^2\right )\right )\,\left (5\,x^2-4\,x+5\right )}{4}}{\ln \left (\ln \left (x^2\right )\right )}-\ln \left (x^2\right )\,\left (\frac {5\,x^5}{4}-x^4+\frac {5\,x^3}{4}\right )-4\,x^2 \]
int(-(x^3 - (5*x^2)/3 - x^4 + (log(x^2)*log(log(x^2))*(15*x^2 - 12*x^3 + 1 5*x^4))/6 + (log(x^2)*log(log(x^2))^2*(48*x - 24))/6)/(log(x^2)*log(log(x^ 2))^2),x)