\(\int \frac {-2+(-1+3 \log (x)) \log (x^2)+\log (x^2) \log (\log (x^2))}{(2 x+3 x \log (x)) \log (x^2)+x \log (x^2) \log (\log (x^2))} \, dx\) [4452]

Optimal result

Integrand size = 52, antiderivative size = 31 \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=3+\log \left (\frac {x}{5+e^{e^5}}\right )-\log \left (\log (x)+\frac {1}{3} \left (2+\log \left (\log \left (x^2\right )\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6873, 6874, 6816} \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\log (x)-\log \left (\log \left (\log \left (x^2\right )\right )+3 \log (x)+2\right ) \]

[In]

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

[Out]

Log[x] - Log[2 + 3*Log[x] + Log[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 6873

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

Rule 6874

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\log (x)-\log \left (4+6 \left (\log (x)-\frac {\log \left (x^2\right )}{2}\right )+3 \log \left (x^2\right )+2 \log \left (\log \left (x^2\right )\right )\right ) \]

[In]

Integrate[(-2 + (-1 + 3*Log[x])*Log[x^2] + Log[x^2]*Log[Log[x^2]])/((2*x + 3*x*Log[x])*Log[x^2] + x*Log[x^2]*L
og[Log[x^2]]),x]

[Out]

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

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\ln \left (x \right )-\ln \left (\frac {\ln \left (\ln \left (x^{2}\right )\right )}{3}+\frac {2}{3}+\ln \left (x \right )\right )\)	\(18\)
risch	\(\ln \left (x \right )-\ln \left (3 \ln \left (x \right )+\ln \left (2 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}\right )+2\right )\)	\(47\)

[In]

int((ln(x^2)*ln(ln(x^2))+(3*ln(x)-1)*ln(x^2)-2)/(x*ln(x^2)*ln(ln(x^2))+(3*x*ln(x)+2*x)*ln(x^2)),x,method=_RETU
RNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\log \left (x\right ) - \log \left (3 \, \log \left (x\right ) + \log \left (2 \, \log \left (x\right )\right ) + 2\right ) \]

[In]

integrate((log(x^2)*log(log(x^2))+(3*log(x)-1)*log(x^2)-2)/(x*log(x^2)*log(log(x^2))+(3*x*log(x)+2*x)*log(x^2)
),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\log {\left (x \right )} - \log {\left (3 \log {\left (x \right )} + \log {\left (2 \log {\left (x \right )} \right )} + 2 \right )} \]

[In]

integrate((ln(x**2)*ln(ln(x**2))+(3*ln(x)-1)*ln(x**2)-2)/(x*ln(x**2)*ln(ln(x**2))+(3*x*ln(x)+2*x)*ln(x**2)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\log \left (x\right ) - \log \left (\log \left (2\right ) + 3 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) + 2\right ) \]

[In]

integrate((log(x^2)*log(log(x^2))+(3*log(x)-1)*log(x^2)-2)/(x*log(x^2)*log(log(x^2))+(3*x*log(x)+2*x)*log(x^2)
),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\int { \frac {{\left (3 \, \log \left (x\right ) - 1\right )} \log \left (x^{2}\right ) + \log \left (x^{2}\right ) \log \left (\log \left (x^{2}\right )\right ) - 2}{x \log \left (x^{2}\right ) \log \left (\log \left (x^{2}\right )\right ) + {\left (3 \, x \log \left (x\right ) + 2 \, x\right )} \log \left (x^{2}\right )} \,d x } \]

[In]

integrate((log(x^2)*log(log(x^2))+(3*log(x)-1)*log(x^2)-2)/(x*log(x^2)*log(log(x^2))+(3*x*log(x)+2*x)*log(x^2)
),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.63 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-2+(-1+3 \log (x)) \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )}{(2 x+3 x \log (x)) \log \left (x^2\right )+x \log \left (x^2\right ) \log \left (\log \left (x^2\right )\right )} \, dx=\ln \left (x\right )-\ln \left (\ln \left (\ln \left (x^2\right )\right )+3\,\ln \left (x\right )+2\right ) \]

[In]

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

[Out]

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