\(\int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+(10 x-4 x^2-2 x^3) \log (6)+x^2 \log ^2(6)+(-30+12 x+6 x^2-6 x \log (6)) \log (x)+9 \log ^2(x)} \, dx\) [553]

Optimal result

Integrand size = 81, antiderivative size = 23 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=\frac {2}{-2+\frac {5}{x}-x+\log (6)-\frac {3 \log (x)}{x}} \]

[Out]

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

Rubi [F]

\[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=\int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx \]

[In]

Int[(16 + 2*x^2 - 6*Log[x])/(25 - 20*x - 6*x^2 + 4*x^3 + x^4 + (10*x - 4*x^2 - 2*x^3)*Log[6] + x^2*Log[6]^2 +
(-30 + 12*x + 6*x^2 - 6*x*Log[6])*Log[x] + 9*Log[x]^2),x]

[Out]

6*Defer[Int][(5 - x^2 - 2*x*(1 - Log[6]/2) - 3*Log[x])^(-2), x] + 2*(2 - Log[6])*Defer[Int][x/(5 - x^2 - 2*x*(
1 - Log[6]/2) - 3*Log[x])^2, x] + 4*Defer[Int][x^2/(5 - x^2 - 2*x*(1 - Log[6]/2) - 3*Log[x])^2, x] + 2*Defer[I
nt][(5 - x^2 - 2*x*(1 - Log[6]/2) - 3*Log[x])^(-1), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {16+2 x^2-6 \log (x)}{25-20 x+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \left (-6+\log ^2(6)\right )+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx \\ & = \int \frac {2 \left (8+x^2-3 \log (x)\right )}{\left (5-x^2+x (-2+\log (6))-3 \log (x)\right )^2} \, dx \\ & = 2 \int \frac {8+x^2-3 \log (x)}{\left (5-x^2+x (-2+\log (6))-3 \log (x)\right )^2} \, dx \\ & = 2 \int \left (\frac {3+2 x^2+x (2-\log (6))}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2}+\frac {1}{5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)}\right ) \, dx \\ & = 2 \int \frac {3+2 x^2+x (2-\log (6))}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2} \, dx+2 \int \frac {1}{5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)} \, dx \\ & = 2 \int \left (\frac {3}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2}+\frac {2 x^2}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2}+\frac {x (2-\log (6))}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2}\right ) \, dx+2 \int \frac {1}{5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)} \, dx \\ & = 2 \int \frac {1}{5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)} \, dx+4 \int \frac {x^2}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2} \, dx+6 \int \frac {1}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2} \, dx+(2 (2-\log (6))) \int \frac {x}{\left (5-x^2-2 x \left (1-\frac {\log (6)}{2}\right )-3 \log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 x}{-5+2 x+x^2-x \log (6)+3 \log (x)} \]

[In]

Integrate[(16 + 2*x^2 - 6*Log[x])/(25 - 20*x - 6*x^2 + 4*x^3 + x^4 + (10*x - 4*x^2 - 2*x^3)*Log[6] + x^2*Log[6
]^2 + (-30 + 12*x + 6*x^2 - 6*x*Log[6])*Log[x] + 9*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {2 x}{x \ln \left (6\right )-x^{2}-3 \ln \left (x \right )-2 x +5}\)	\(24\)
norman	\(\frac {2 x}{x \ln \left (6\right )-x^{2}-3 \ln \left (x \right )-2 x +5}\)	\(24\)
parallelrisch	\(\frac {2 x}{x \ln \left (6\right )-x^{2}-3 \ln \left (x \right )-2 x +5}\)	\(24\)
risch	\(\frac {2 x}{x \ln \left (3\right )+x \ln \left (2\right )-x^{2}-2 x -3 \ln \left (x \right )+5}\)	\(28\)

[In]

int((-6*ln(x)+2*x^2+16)/(9*ln(x)^2+(-6*x*ln(6)+6*x^2+12*x-30)*ln(x)+x^2*ln(6)^2+(-2*x^3-4*x^2+10*x)*ln(6)+x^4+
4*x^3-6*x^2-20*x+25),x,method=_RETURNVERBOSE)

[Out]

2*x/(x*ln(6)-x^2-3*ln(x)-2*x+5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 \, x}{x^{2} - x \log \left (6\right ) + 2 \, x + 3 \, \log \left (x\right ) - 5} \]

[In]

integrate((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log(x)+x^2*log(6)^2+(-2*x^3-4*x^2+10*x)
*log(6)+x^4+4*x^3-6*x^2-20*x+25),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=- \frac {2 x}{x^{2} - x \log {\left (6 \right )} + 2 x + 3 \log {\left (x \right )} - 5} \]

[In]

integrate((-6*ln(x)+2*x**2+16)/(9*ln(x)**2+(-6*x*ln(6)+6*x**2+12*x-30)*ln(x)+x**2*ln(6)**2+(-2*x**3-4*x**2+10*
x)*ln(6)+x**4+4*x**3-6*x**2-20*x+25),x)

[Out]

-2*x/(x**2 - x*log(6) + 2*x + 3*log(x) - 5)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 \, x}{x^{2} - x {\left (\log \left (3\right ) + \log \left (2\right ) - 2\right )} + 3 \, \log \left (x\right ) - 5} \]

[In]

integrate((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log(x)+x^2*log(6)^2+(-2*x^3-4*x^2+10*x)
*log(6)+x^4+4*x^3-6*x^2-20*x+25),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2 \, x}{x^{2} - x \log \left (6\right ) + 2 \, x + 3 \, \log \left (x\right ) - 5} \]

[In]

integrate((-6*log(x)+2*x^2+16)/(9*log(x)^2+(-6*x*log(6)+6*x^2+12*x-30)*log(x)+x^2*log(6)^2+(-2*x^3-4*x^2+10*x)
*log(6)+x^4+4*x^3-6*x^2-20*x+25),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {16+2 x^2-6 \log (x)}{25-20 x-6 x^2+4 x^3+x^4+\left (10 x-4 x^2-2 x^3\right ) \log (6)+x^2 \log ^2(6)+\left (-30+12 x+6 x^2-6 x \log (6)\right ) \log (x)+9 \log ^2(x)} \, dx=-\frac {2\,x}{3\,\ln \left (x\right )-x\,\left (\ln \left (6\right )-2\right )+x^2-5} \]

[In]

int((2*x^2 - 6*log(x) + 16)/(x^2*log(6)^2 - 20*x + 9*log(x)^2 - log(6)*(4*x^2 - 10*x + 2*x^3) - 6*x^2 + 4*x^3
+ x^4 + log(x)*(12*x - 6*x*log(6) + 6*x^2 - 30) + 25),x)

[Out]

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