\(\int \frac {4+2 \log (x^2)+2 \log (x^2) \log (x \log (x^2))-2 \log (x^2) \log (x \log (x^2)) \log (3 \log (x \log (x^2)))}{(-2 x-3 x^2) \log (x^2) \log (x \log (x^2))+2 x \log (x^2) \log (x \log (x^2)) \log (3 \log (x \log (x^2)))} \, dx\) [7024]

Optimal result

Integrand size = 93, antiderivative size = 23 \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=\log \left (-3-\frac {2}{x}+\frac {2 \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{x}\right ) \]

[Out]

ln(-3+2*ln(3*ln(x*ln(x^2)))/x-2/x)

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6820, 6874, 6816} \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=\log \left (-2 \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )+3 x+2\right )-\log (x) \]

[In]

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

[Out]

-Log[x] + Log[2 + 3*x - 2*Log[3*Log[x*Log[x^2]]]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=-\log (x)+\log \left (2+3 x-2 \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )\right ) \]

[In]

Integrate[(4 + 2*Log[x^2] + 2*Log[x^2]*Log[x*Log[x^2]] - 2*Log[x^2]*Log[x*Log[x^2]]*Log[3*Log[x*Log[x^2]]])/((
-2*x - 3*x^2)*Log[x^2]*Log[x*Log[x^2]] + 2*x*Log[x^2]*Log[x*Log[x^2]]*Log[3*Log[x*Log[x^2]]]),x]

[Out]

-Log[x] + Log[2 + 3*x - 2*Log[3*Log[x*Log[x^2]]]]

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(-\ln \left (x \right )+\ln \left (-\frac {2 \ln \left (3 \ln \left (x \ln \left (x^{2}\right )\right )\right )}{3}+x +\frac {2}{3}\right )\)	\(22\)

[In]

int((-2*ln(x^2)*ln(x*ln(x^2))*ln(3*ln(x*ln(x^2)))+2*ln(x^2)*ln(x*ln(x^2))+2*ln(x^2)+4)/(2*x*ln(x^2)*ln(x*ln(x^
2))*ln(3*ln(x*ln(x^2)))+(-3*x^2-2*x)*ln(x^2)*ln(x*ln(x^2))),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(-2/3*ln(3*ln(x*ln(x^2)))+x+2/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=-\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left (-3 \, x + 2 \, \log \left (3 \, \log \left (x \log \left (x^{2}\right )\right )\right ) - 2\right ) \]

[In]

integrate((-2*log(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2)))+2*log(x^2)*log(x*log(x^2))+2*log(x^2)+4)/(2*x*lo
g(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2)))+(-3*x^2-2*x)*log(x^2)*log(x*log(x^2))),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=- \log {\left (x \right )} + \log {\left (- \frac {3 x}{2} + \log {\left (3 \log {\left (x \log {\left (x^{2} \right )} \right )} \right )} - 1 \right )} \]

[In]

integrate((-2*ln(x**2)*ln(x*ln(x**2))*ln(3*ln(x*ln(x**2)))+2*ln(x**2)*ln(x*ln(x**2))+2*ln(x**2)+4)/(2*x*ln(x**
2)*ln(x*ln(x**2))*ln(3*ln(x*ln(x**2)))+(-3*x**2-2*x)*ln(x**2)*ln(x*ln(x**2))),x)

[Out]

-log(x) + log(-3*x/2 + log(3*log(x*log(x**2))) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=-\log \left (x\right ) + \log \left (-\frac {3}{2} \, x + \log \left (3\right ) + \log \left (\log \left (2\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right )\right ) - 1\right ) \]

[In]

integrate((-2*log(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2)))+2*log(x^2)*log(x*log(x^2))+2*log(x^2)+4)/(2*x*lo
g(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2)))+(-3*x^2-2*x)*log(x^2)*log(x*log(x^2))),x, algorithm="maxima")

[Out]

-log(x) + log(-3/2*x + log(3) + log(log(2) + log(x) + log(log(x))) - 1)

Giac [F]

\[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=\int { -\frac {2 \, {\left (\log \left (x^{2}\right ) \log \left (x \log \left (x^{2}\right )\right ) \log \left (3 \, \log \left (x \log \left (x^{2}\right )\right )\right ) - \log \left (x^{2}\right ) \log \left (x \log \left (x^{2}\right )\right ) - \log \left (x^{2}\right ) - 2\right )}}{2 \, x \log \left (x^{2}\right ) \log \left (x \log \left (x^{2}\right )\right ) \log \left (3 \, \log \left (x \log \left (x^{2}\right )\right )\right ) - {\left (3 \, x^{2} + 2 \, x\right )} \log \left (x^{2}\right ) \log \left (x \log \left (x^{2}\right )\right )} \,d x } \]

[In]

integrate((-2*log(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2)))+2*log(x^2)*log(x*log(x^2))+2*log(x^2)+4)/(2*x*lo
g(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2)))+(-3*x^2-2*x)*log(x^2)*log(x*log(x^2))),x, algorithm="giac")

[Out]

integrate(-2*(log(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2))) - log(x^2)*log(x*log(x^2)) - log(x^2) - 2)/(2*x*
log(x^2)*log(x*log(x^2))*log(3*log(x*log(x^2))) - (3*x^2 + 2*x)*log(x^2)*log(x*log(x^2))), x)

Mupad [B] (verification not implemented)

Time = 12.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4+2 \log \left (x^2\right )+2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )-2 \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )}{\left (-2 x-3 x^2\right ) \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right )+2 x \log \left (x^2\right ) \log \left (x \log \left (x^2\right )\right ) \log \left (3 \log \left (x \log \left (x^2\right )\right )\right )} \, dx=\ln \left (\ln \left (3\right )-\frac {3\,x}{2}+\ln \left (\ln \left (x\,\ln \left (x^2\right )\right )\right )-1\right )-\ln \left (x\right ) \]

[In]

int(-(2*log(x^2) + 2*log(x^2)*log(x*log(x^2)) - 2*log(x^2)*log(3*log(x*log(x^2)))*log(x*log(x^2)) + 4)/(log(x^
2)*log(x*log(x^2))*(2*x + 3*x^2) - 2*x*log(x^2)*log(3*log(x*log(x^2)))*log(x*log(x^2))),x)

[Out]

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