\(\int \frac {-8+2 x^4-4 x^4 \log (x^2) \log (\log (x^2))+5 x \log (x^2) \log ^2(\log (x^2))}{(4 x-x^5) \log (x^2) \log (\log (x^2))+5 x^2 \log (x^2) \log ^2(\log (x^2))} \, dx\) [7213]

Optimal result

Integrand size = 74, antiderivative size = 21 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\log \left (-x+\frac {-4+x^4}{5 \log \left (\log \left (x^2\right )\right )}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx \]

[In]

Int[(-8 + 2*x^4 - 4*x^4*Log[x^2]*Log[Log[x^2]] + 5*x*Log[x^2]*Log[Log[x^2]]^2)/((4*x - x^5)*Log[x^2]*Log[Log[x
^2]] + 5*x^2*Log[x^2]*Log[Log[x^2]]^2),x]

[Out]

Log[x] - Log[Log[Log[x^2]]] + 4*Defer[Int][1/(x*(-4 + x^4 - 5*x*Log[Log[x^2]])), x] + 3*Defer[Int][x^3/(-4 + x
^4 - 5*x*Log[Log[x^2]]), x] - 10*Defer[Int][1/(Log[x^2]*(-4 + x^4 - 5*x*Log[Log[x^2]])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right ) \left (4-x^4+5 x \log \left (\log \left (x^2\right )\right )\right )} \, dx \\ & = \int \left (\frac {1}{x}-\frac {2}{x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}+\frac {-10 x+4 \log \left (x^2\right )+3 x^4 \log \left (x^2\right )}{x \log \left (x^2\right ) \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )}\right ) \, dx \\ & = \log (x)-2 \int \frac {1}{x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx+\int \frac {-10 x+4 \log \left (x^2\right )+3 x^4 \log \left (x^2\right )}{x \log \left (x^2\right ) \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )} \, dx \\ & = \log (x)+\int \left (\frac {4}{x \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )}+\frac {3 x^3}{-4+x^4-5 x \log \left (\log \left (x^2\right )\right )}-\frac {10}{\log \left (x^2\right ) \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\log \left (x^2\right )\right ) \\ & = \log (x)+3 \int \frac {x^3}{-4+x^4-5 x \log \left (\log \left (x^2\right )\right )} \, dx+4 \int \frac {1}{x \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )} \, dx-10 \int \frac {1}{\log \left (x^2\right ) \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )} \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\log \left (x^2\right )\right )\right ) \\ & = \log (x)-\log \left (\log \left (\log \left (x^2\right )\right )\right )+3 \int \frac {x^3}{-4+x^4-5 x \log \left (\log \left (x^2\right )\right )} \, dx+4 \int \frac {1}{x \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )} \, dx-10 \int \frac {1}{\log \left (x^2\right ) \left (-4+x^4-5 x \log \left (\log \left (x^2\right )\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=-\log \left (\log \left (\log \left (x^2\right )\right )\right )+\log \left (4-x^4+5 x \log \left (\log \left (x^2\right )\right )\right ) \]

[In]

Integrate[(-8 + 2*x^4 - 4*x^4*Log[x^2]*Log[Log[x^2]] + 5*x*Log[x^2]*Log[Log[x^2]]^2)/((4*x - x^5)*Log[x^2]*Log
[Log[x^2]] + 5*x^2*Log[x^2]*Log[Log[x^2]]^2),x]

[Out]

-Log[Log[Log[x^2]]] + Log[4 - x^4 + 5*x*Log[Log[x^2]]]

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14

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

[In]

int((5*x*ln(x^2)*ln(ln(x^2))^2-4*x^4*ln(x^2)*ln(ln(x^2))+2*x^4-8)/(5*x^2*ln(x^2)*ln(ln(x^2))^2+(-x^5+4*x)*ln(x
^2)*ln(ln(x^2))),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(ln(x^2)))+ln(x^4-5*x*ln(ln(x^2))-4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left (-\frac {x^{4} - 5 \, x \log \left (\log \left (x^{2}\right )\right ) - 4}{x}\right ) - \log \left (\log \left (\log \left (x^{2}\right )\right )\right ) \]

[In]

integrate((5*x*log(x^2)*log(log(x^2))^2-4*x^4*log(x^2)*log(log(x^2))+2*x^4-8)/(5*x^2*log(x^2)*log(log(x^2))^2+
(-x^5+4*x)*log(x^2)*log(log(x^2))),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (\log {\left (x^{2} \right )} \right )} + \frac {8 - 2 x^{4}}{10 x} \right )} - \log {\left (\log {\left (\log {\left (x^{2} \right )} \right )} \right )} \]

[In]

integrate((5*x*ln(x**2)*ln(ln(x**2))**2-4*x**4*ln(x**2)*ln(ln(x**2))+2*x**4-8)/(5*x**2*ln(x**2)*ln(ln(x**2))**
2+(-x**5+4*x)*ln(x**2)*ln(ln(x**2))),x)

[Out]

log(x) + log(log(log(x**2)) + (8 - 2*x**4)/(10*x)) - log(log(log(x**2)))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\log \left (x\right ) + \log \left (-\frac {x^{4} - 5 \, x \log \left (2\right ) - 5 \, x \log \left (\log \left (x\right )\right ) - 4}{5 \, x}\right ) - \log \left (\log \left (2\right ) + \log \left (\log \left (x\right )\right )\right ) \]

[In]

integrate((5*x*log(x^2)*log(log(x^2))^2-4*x^4*log(x^2)*log(log(x^2))+2*x^4-8)/(5*x^2*log(x^2)*log(log(x^2))^2+
(-x^5+4*x)*log(x^2)*log(log(x^2))),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\log \left (-x^{4} + 5 \, x \log \left (\log \left (x^{2}\right )\right ) + 4\right ) - \log \left (\log \left (\log \left (x^{2}\right )\right )\right ) \]

[In]

integrate((5*x*log(x^2)*log(log(x^2))^2-4*x^4*log(x^2)*log(log(x^2))+2*x^4-8)/(5*x^2*log(x^2)*log(log(x^2))^2+
(-x^5+4*x)*log(x^2)*log(log(x^2))),x, algorithm="giac")

[Out]

log(-x^4 + 5*x*log(log(x^2)) + 4) - log(log(log(x^2)))

Mupad [B] (verification not implemented)

Time = 12.63 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.71 \[ \int \frac {-8+2 x^4-4 x^4 \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )}{\left (4 x-x^5\right ) \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )+5 x^2 \log \left (x^2\right ) \log ^2\left (\log \left (x^2\right )\right )} \, dx=\ln \left (\frac {20\,x\,\ln \left (\ln \left (x^2\right )\right )-4\,x^4+16}{\ln \left (x^2\right )}\right )-\ln \left (\frac {\ln \left (\ln \left (x^2\right )\right )\,\left (4\,\ln \left (x^2\right )-10\,x+3\,x^4\,\ln \left (x^2\right )\right )}{\ln \left (x^2\right )}\right )+\ln \left (4\,\ln \left (x^2\right )-10\,x+3\,x^4\,\ln \left (x^2\right )\right ) \]

[In]

int((2*x^4 + 5*x*log(x^2)*log(log(x^2))^2 - 4*x^4*log(x^2)*log(log(x^2)) - 8)/(5*x^2*log(x^2)*log(log(x^2))^2
+ log(x^2)*log(log(x^2))*(4*x - x^5)),x)

[Out]

log((20*x*log(log(x^2)) - 4*x^4 + 16)/log(x^2)) - log((log(log(x^2))*(4*log(x^2) - 10*x + 3*x^4*log(x^2)))/log
(x^2)) + log(4*log(x^2) - 10*x + 3*x^4*log(x^2))