\(\int \frac {10 x^2-6 x^3+6 x^4+(-15 x^2+12 x^3-15 x^4) \log (x^2) \log (\log (x^2))+(24-48 x) \log (x^2) \log ^2(\log (x^2))}{6 \log (x^2) \log ^2(\log (x^2))} \, dx\) [6620]

Optimal result

Integrand size = 76, antiderivative size = 27 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\left (-\frac {5}{3}+x-x^2\right ) \left (4+\frac {x^3}{2 \log \left (\log \left (x^2\right )\right )}\right ) \]

[Out]

4*(1+1/8*x^3/ln(ln(x^2)))*(x-x^2-5/3)

Rubi [F]

\[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx \]

[In]

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[Lo
g[x^2]]^2)/(6*Log[x^2]*Log[Log[x^2]]^2),x]

[Out]

-(1 - 2*x)^2 + (5*Defer[Int][x^2/(Log[x^2]*Log[Log[x^2]]^2), x])/3 + Defer[Int][x^4/(Log[x^2]*Log[Log[x^2]]^2)
, x] - (5*Defer[Int][x^2/Log[Log[x^2]], x])/2 - (5*Defer[Int][x^4/Log[Log[x^2]], x])/2 - Defer[Subst][Defer[In
t][x/(Log[x]*Log[Log[x]]^2), x], x, x^2]/2 + Defer[Subst][Defer[Int][x/Log[Log[x]], x], x, x^2]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \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 )}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx \\ & = \frac {1}{6} \int \left (-24 (-1+2 x)+\frac {2 x^2 \left (5-3 x+3 x^2\right )}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}-\frac {3 x^2 \left (5-4 x+5 x^2\right )}{\log \left (\log \left (x^2\right )\right )}\right ) \, dx \\ & = -(1-2 x)^2+\frac {1}{3} \int \frac {x^2 \left (5-3 x+3 x^2\right )}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx-\frac {1}{2} \int \frac {x^2 \left (5-4 x+5 x^2\right )}{\log \left (\log \left (x^2\right )\right )} \, dx \\ & = -(1-2 x)^2+\frac {1}{3} \int \left (\frac {5 x^2}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}-\frac {3 x^3}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}+\frac {3 x^4}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}\right ) \, dx-\frac {1}{2} \int \left (\frac {5 x^2}{\log \left (\log \left (x^2\right )\right )}-\frac {4 x^3}{\log \left (\log \left (x^2\right )\right )}+\frac {5 x^4}{\log \left (\log \left (x^2\right )\right )}\right ) \, dx \\ & = -(1-2 x)^2+\frac {5}{3} \int \frac {x^2}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx+2 \int \frac {x^3}{\log \left (\log \left (x^2\right )\right )} \, dx-\frac {5}{2} \int \frac {x^2}{\log \left (\log \left (x^2\right )\right )} \, dx-\frac {5}{2} \int \frac {x^4}{\log \left (\log \left (x^2\right )\right )} \, dx-\int \frac {x^3}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx+\int \frac {x^4}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx \\ & = -(1-2 x)^2-\frac {1}{2} \text {Subst}\left (\int \frac {x}{\log (x) \log ^2(\log (x))} \, dx,x,x^2\right )+\frac {5}{3} \int \frac {x^2}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx-\frac {5}{2} \int \frac {x^2}{\log \left (\log \left (x^2\right )\right )} \, dx-\frac {5}{2} \int \frac {x^4}{\log \left (\log \left (x^2\right )\right )} \, dx+\int \frac {x^4}{\log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx+\text {Subst}\left (\int \frac {x}{\log (\log (x))} \, dx,x,x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=4 x-4 x^2-\frac {x^3 \left (5-3 x+3 x^2\right )}{6 \log \left (\log \left (x^2\right )\right )} \]

[In]

Integrate[(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]

[Out]

4*x - 4*x^2 - (x^3*(5 - 3*x + 3*x^2))/(6*Log[Log[x^2]])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63

method	result	size
parallelrisch	\(-\frac {3 x^{5}-3 x^{4}+5 x^{3}+24 x^{2} \ln \left (\ln \left (x^{2}\right )\right )-24 x \ln \left (\ln \left (x^{2}\right )\right )}{6 \ln \left (\ln \left (x^{2}\right )\right )}\)	\(44\)

[In]

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=_RETURNVERBOSE)

[Out]

-1/6*(3*x^5-3*x^4+5*x^3+24*x^2*ln(ln(x^2))-24*x*ln(ln(x^2)))/ln(ln(x^2))

Fricas [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-\frac {3 \, x^{5} - 3 \, x^{4} + 5 \, x^{3} + 24 \, {\left (x^{2} - x\right )} \log \left (\log \left (x^{2}\right )\right )}{6 \, \log \left (\log \left (x^{2}\right )\right )} \]

[In]

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, algorithm="fricas")

[Out]

-1/6*(3*x^5 - 3*x^4 + 5*x^3 + 24*(x^2 - x)*log(log(x^2)))/log(log(x^2))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=- 4 x^{2} + 4 x + \frac {- 3 x^{5} + 3 x^{4} - 5 x^{3}}{6 \log {\left (\log {\left (x^{2} \right )} \right )}} \]

[In]

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)

[Out]

-4*x**2 + 4*x + (-3*x**5 + 3*x**4 - 5*x**3)/(6*log(log(x**2)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-4 \, x^{2} + 4 \, x - \frac {3 \, x^{5} - 3 \, x^{4} + 5 \, x^{3}}{6 \, {\left (\log \left (2\right ) + \log \left (\log \left (x\right )\right )\right )}} \]

[In]

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, algorithm="maxima")

[Out]

-4*x^2 + 4*x - 1/6*(3*x^5 - 3*x^4 + 5*x^3)/(log(2) + log(log(x)))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-4 \, x^{2} + 4 \, x - \frac {3 \, x^{5} - 3 \, x^{4} + 5 \, x^{3}}{6 \, \log \left (\log \left (x^{2}\right )\right )} \]

[In]

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, algorithm="giac")

[Out]

-4*x^2 + 4*x - 1/6*(3*x^5 - 3*x^4 + 5*x^3)/log(log(x^2))

Mupad [B] (verification not implemented)

Time = 11.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.96 \[ \int \frac {10 x^2-6 x^3+6 x^4+\left (-15 x^2+12 x^3-15 x^4\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+(24-48 x) \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{6 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=4\,x-\frac {\frac {x^3\,\left (3\,x^2-3\,x+5\right )}{6}-\frac {x^3\,\ln \left (x^2\right )\,\ln \left (\ln \left (x^2\right )\right )\,\left (5\,x^2-4\,x+5\right )}{4}}{\ln \left (\ln \left (x^2\right )\right )}-\ln \left (x^2\right )\,\left (\frac {5\,x^5}{4}-x^4+\frac {5\,x^3}{4}\right )-4\,x^2 \]

[In]

int(-(x^3 - (5*x^2)/3 - x^4 + (log(x^2)*log(log(x^2))*(15*x^2 - 12*x^3 + 15*x^4))/6 + (log(x^2)*log(log(x^2))^
2*(48*x - 24))/6)/(log(x^2)*log(log(x^2))^2),x)

[Out]

4*x - ((x^3*(3*x^2 - 3*x + 5))/6 - (x^3*log(x^2)*log(log(x^2))*(5*x^2 - 4*x + 5))/4)/log(log(x^2)) - log(x^2)*
((5*x^3)/4 - x^4 + (5*x^5)/4) - 4*x^2