\(\int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log (\frac {1}{4} \log (3 x))}{45 \log (3 x)-30 \log (3 x) \log (\frac {1}{4} \log (3 x))+5 \log (3 x) \log ^2(\frac {1}{4} \log (3 x))} \, dx\) [594]

Optimal result

Integrand size = 69, antiderivative size = 20 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=\frac {x^2}{5 \left (-3+\log \left (\frac {1}{4} \log (3 x)\right )\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=\int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx \]

[In]

Int[(-x - 6*x*Log[3*x] + 2*x*Log[3*x]*Log[Log[3*x]/4])/(45*Log[3*x] - 30*Log[3*x]*Log[Log[3*x]/4] + 5*Log[3*x]
*Log[Log[3*x]/4]^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=\frac {x^2}{5 \left (-3+\log \left (\frac {1}{4} \log (3 x)\right )\right )} \]

[In]

Integrate[(-x - 6*x*Log[3*x] + 2*x*Log[3*x]*Log[Log[3*x]/4])/(45*Log[3*x] - 30*Log[3*x]*Log[Log[3*x]/4] + 5*Lo
g[3*x]*Log[Log[3*x]/4]^2),x]

[Out]

x^2/(5*(-3 + Log[Log[3*x]/4]))

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
norman	\(\frac {x^{2}}{5 \ln \left (\frac {\ln \left (3 x \right )}{4}\right )-15}\)	\(17\)
risch	\(\frac {x^{2}}{5 \ln \left (\frac {\ln \left (3 x \right )}{4}\right )-15}\)	\(17\)
parallelrisch	\(\frac {x^{2}}{5 \ln \left (\frac {\ln \left (3 x \right )}{4}\right )-15}\)	\(17\)

[In]

int((2*x*ln(3*x)*ln(1/4*ln(3*x))-6*x*ln(3*x)-x)/(5*ln(3*x)*ln(1/4*ln(3*x))^2-30*ln(3*x)*ln(1/4*ln(3*x))+45*ln(
3*x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=\frac {x^{2}}{5 \, {\left (\log \left (\frac {1}{4} \, \log \left (3 \, x\right )\right ) - 3\right )}} \]

[In]

integrate((2*x*log(3*x)*log(1/4*log(3*x))-6*x*log(3*x)-x)/(5*log(3*x)*log(1/4*log(3*x))^2-30*log(3*x)*log(1/4*
log(3*x))+45*log(3*x)),x, algorithm="fricas")

[Out]

1/5*x^2/(log(1/4*log(3*x)) - 3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=\frac {x^{2}}{5 \log {\left (\frac {\log {\left (3 x \right )}}{4} \right )} - 15} \]

[In]

integrate((2*x*ln(3*x)*ln(1/4*ln(3*x))-6*x*ln(3*x)-x)/(5*ln(3*x)*ln(1/4*ln(3*x))**2-30*ln(3*x)*ln(1/4*ln(3*x))
+45*ln(3*x)),x)

[Out]

x**2/(5*log(log(3*x)/4) - 15)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=-\frac {x^{2}}{5 \, {\left (2 \, \log \left (2\right ) - \log \left (\log \left (3\right ) + \log \left (x\right )\right ) + 3\right )}} \]

[In]

integrate((2*x*log(3*x)*log(1/4*log(3*x))-6*x*log(3*x)-x)/(5*log(3*x)*log(1/4*log(3*x))^2-30*log(3*x)*log(1/4*
log(3*x))+45*log(3*x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=-\frac {x^{2}}{5 \, {\left (2 \, \log \left (2\right ) - \log \left (\log \left (3 \, x\right )\right ) + 3\right )}} \]

[In]

integrate((2*x*log(3*x)*log(1/4*log(3*x))-6*x*log(3*x)-x)/(5*log(3*x)*log(1/4*log(3*x))^2-30*log(3*x)*log(1/4*
log(3*x))+45*log(3*x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-x-6 x \log (3 x)+2 x \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )}{45 \log (3 x)-30 \log (3 x) \log \left (\frac {1}{4} \log (3 x)\right )+5 \log (3 x) \log ^2\left (\frac {1}{4} \log (3 x)\right )} \, dx=\frac {x^2}{5\,\left (\ln \left (\frac {\ln \left (3\,x\right )}{4}\right )-3\right )} \]

[In]

int(-(x + 6*x*log(3*x) - 2*x*log(log(3*x)/4)*log(3*x))/(45*log(3*x) + 5*log(log(3*x)/4)^2*log(3*x) - 30*log(lo
g(3*x)/4)*log(3*x)),x)

[Out]

x^2/(5*(log(log(3*x)/4) - 3))