\(\int \frac {3-3 x-5 x^2+x^3+(-6 x+6 x^3) \log (x)}{3 x-x^3+(-3 x^2+x^4) \log (x)} \, dx\) [5690]

Optimal result

Integrand size = 49, antiderivative size = 20 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=9+\log \left (x \left (3-x^2\right )^2 (-1+x \log (x))\right ) \]

[Out]

ln((-x^2+3)^2*(x*ln(x)-1)*x)+9

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=\log (x)+2 \log \left (3-x^2\right )+\log (1-x \log (x)) \]

[In]

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

[Out]

Log[x] + 2*Log[3 - x^2] + Log[1 - x*Log[x]]

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
default	\(\ln \left (x \right )+2 \ln \left (x^{2}-3\right )+\ln \left (x \ln \left (x \right )-1\right )\)	\(19\)
norman	\(\ln \left (x \right )+2 \ln \left (x^{2}-3\right )+\ln \left (x \ln \left (x \right )-1\right )\)	\(19\)
parallelrisch	\(\ln \left (x \right )+2 \ln \left (x^{2}-3\right )+\ln \left (x \ln \left (x \right )-1\right )\)	\(19\)
risch	\(2 \ln \left (x^{3}-3 x \right )+\ln \left (\ln \left (x \right )-\frac {1}{x}\right )\)	\(21\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=2 \, \log \left (x^{3} - 3 \, x\right ) + \log \left (\frac {x \log \left (x\right ) - 1}{x}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=2 \log {\left (x^{3} - 3 x \right )} + \log {\left (\log {\left (x \right )} - \frac {1}{x} \right )} \]

[In]

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

[Out]

2*log(x**3 - 3*x) + log(log(x) - 1/x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=2 \, \log \left (x^{2} - 3\right ) + 2 \, \log \left (x\right ) + \log \left (\frac {x \log \left (x\right ) - 1}{x}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=2 \, \log \left (x^{2} - 3\right ) + \log \left (-x \log \left (x\right ) + 1\right ) + \log \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {3-3 x-5 x^2+x^3+\left (-6 x+6 x^3\right ) \log (x)}{3 x-x^3+\left (-3 x^2+x^4\right ) \log (x)} \, dx=2\,\ln \left (x^3-3\,x\right )+\ln \left (\frac {x\,\ln \left (x\right )-1}{x}\right ) \]

[In]

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

[Out]

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