\(\int \frac {4 x+4 x^3-4 \log (3)+(4 x+12 x^3) \log (x)}{(x^3+2 x^5+x^7+(-2 x^2-2 x^4) \log (3)+x \log ^2(3)) \log ^2(x)} \, dx\) [4296]

Optimal result

Integrand size = 64, antiderivative size = 22 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=2-\frac {4}{\left (x \left (1+x^2\right )-\log (3)\right ) \log (x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=\int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx \]

[In]

Int[(4*x + 4*x^3 - 4*Log[3] + (4*x + 12*x^3)*Log[x])/((x^3 + 2*x^5 + x^7 + (-2*x^2 - 2*x^4)*Log[3] + x*Log[3]^
2)*Log[x]^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=\frac {4}{\left (-x-x^3+\log (3)\right ) \log (x)} \]

[In]

Integrate[(4*x + 4*x^3 - 4*Log[3] + (4*x + 12*x^3)*Log[x])/((x^3 + 2*x^5 + x^7 + (-2*x^2 - 2*x^4)*Log[3] + x*L
og[3]^2)*Log[x]^2),x]

[Out]

4/((-x - x^3 + Log[3])*Log[x])

Maple [A] (verified)

Time = 6.99 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {4}{\left (-x^{3}+\ln \left (3\right )-x \right ) \ln \left (x \right )}\)	\(20\)
norman	\(\frac {4}{\left (-x^{3}+\ln \left (3\right )-x \right ) \ln \left (x \right )}\)	\(20\)
risch	\(\frac {4}{\left (-x^{3}+\ln \left (3\right )-x \right ) \ln \left (x \right )}\)	\(20\)
parallelrisch	\(\frac {4}{\left (-x^{3}+\ln \left (3\right )-x \right ) \ln \left (x \right )}\)	\(20\)

[In]

int(((12*x^3+4*x)*ln(x)-4*ln(3)+4*x^3+4*x)/(x*ln(3)^2+(-2*x^4-2*x^2)*ln(3)+x^7+2*x^5+x^3)/ln(x)^2,x,method=_RE
TURNVERBOSE)

[Out]

4/(-x^3+ln(3)-x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=-\frac {4}{{\left (x^{3} + x - \log \left (3\right )\right )} \log \left (x\right )} \]

[In]

integrate(((12*x^3+4*x)*log(x)-4*log(3)+4*x^3+4*x)/(x*log(3)^2+(-2*x^4-2*x^2)*log(3)+x^7+2*x^5+x^3)/log(x)^2,x
, algorithm="fricas")

[Out]

-4/((x^3 + x - log(3))*log(x))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=- \frac {4}{\left (x^{3} + x - \log {\left (3 \right )}\right ) \log {\left (x \right )}} \]

[In]

integrate(((12*x**3+4*x)*ln(x)-4*ln(3)+4*x**3+4*x)/(x*ln(3)**2+(-2*x**4-2*x**2)*ln(3)+x**7+2*x**5+x**3)/ln(x)*
*2,x)

[Out]

-4/((x**3 + x - log(3))*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=-\frac {4}{{\left (x^{3} + x - \log \left (3\right )\right )} \log \left (x\right )} \]

[In]

integrate(((12*x^3+4*x)*log(x)-4*log(3)+4*x^3+4*x)/(x*log(3)^2+(-2*x^4-2*x^2)*log(3)+x^7+2*x^5+x^3)/log(x)^2,x
, algorithm="maxima")

[Out]

-4/((x^3 + x - log(3))*log(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=-\frac {4}{x^{3} \log \left (x\right ) + x \log \left (x\right ) - \log \left (3\right ) \log \left (x\right )} \]

[In]

integrate(((12*x^3+4*x)*log(x)-4*log(3)+4*x^3+4*x)/(x*log(3)^2+(-2*x^4-2*x^2)*log(3)+x^7+2*x^5+x^3)/log(x)^2,x
, algorithm="giac")

[Out]

-4/(x^3*log(x) + x*log(x) - log(3)*log(x))

Mupad [B] (verification not implemented)

Time = 14.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {4 x+4 x^3-4 \log (3)+\left (4 x+12 x^3\right ) \log (x)}{\left (x^3+2 x^5+x^7+\left (-2 x^2-2 x^4\right ) \log (3)+x \log ^2(3)\right ) \log ^2(x)} \, dx=-\frac {4\,x^3+4\,x-\ln \left (81\right )}{\ln \left (x\right )\,{\left (x^3+x-\ln \left (3\right )\right )}^2} \]

[In]

int((4*x - 4*log(3) + log(x)*(4*x + 12*x^3) + 4*x^3)/(log(x)^2*(x*log(3)^2 - log(3)*(2*x^2 + 2*x^4) + x^3 + 2*
x^5 + x^7)),x)

[Out]

-(4*x - log(81) + 4*x^3)/(log(x)*(x - log(3) + x^3)^2)