\(\int \frac {6+x^2+24 x^3+(3-2 x-48 x^2) \log (\frac {1}{x^2})+(1+24 x) \log ^2(\frac {1}{x^2})}{3 x^2-6 x \log (\frac {1}{x^2})+3 \log ^2(\frac {1}{x^2})} \, dx\) [8387]

Optimal result

Integrand size = 61, antiderivative size = 25 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=-2+x-2 \left (\frac {1}{3}-2 x\right ) x+\frac {x}{-x+\log \left (\frac {1}{x^2}\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=\int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx \]

[In]

Int[(6 + x^2 + 24*x^3 + (3 - 2*x - 48*x^2)*Log[x^(-2)] + (1 + 24*x)*Log[x^(-2)]^2)/(3*x^2 - 6*x*Log[x^(-2)] +
3*Log[x^(-2)]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 \left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \, dx \\ & = \frac {1}{3} \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \, dx \\ & = \frac {1}{3} \int \left (1+24 x+\frac {3 (2+x)}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2}-\frac {3}{x-\log \left (\frac {1}{x^2}\right )}\right ) \, dx \\ & = \frac {x}{3}+4 x^2+\int \frac {2+x}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \, dx-\int \frac {1}{x-\log \left (\frac {1}{x^2}\right )} \, dx \\ & = \frac {x}{3}+4 x^2+\int \left (\frac {2}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2}+\frac {x}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2}\right ) \, dx-\int \frac {1}{x-\log \left (\frac {1}{x^2}\right )} \, dx \\ & = \frac {x}{3}+4 x^2+2 \int \frac {1}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \, dx+\int \frac {x}{\left (x-\log \left (\frac {1}{x^2}\right )\right )^2} \, dx-\int \frac {1}{x-\log \left (\frac {1}{x^2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=\frac {1}{3} \left (x+12 x^2+\frac {3 x}{-x+\log \left (\frac {1}{x^2}\right )}\right ) \]

[In]

Integrate[(6 + x^2 + 24*x^3 + (3 - 2*x - 48*x^2)*Log[x^(-2)] + (1 + 24*x)*Log[x^(-2)]^2)/(3*x^2 - 6*x*Log[x^(-
2)] + 3*Log[x^(-2)]^2),x]

[Out]

(x + 12*x^2 + (3*x)/(-x + Log[x^(-2)]))/3

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
risch	\(4 x^{2}+\frac {x}{3}-\frac {x}{x -\ln \left (\frac {1}{x^{2}}\right )}\)	\(23\)
parallelrisch	\(\frac {24 x^{3}-24 x^{2} \ln \left (\frac {1}{x^{2}}\right )+2 x^{2}-2 x \ln \left (\frac {1}{x^{2}}\right )-6 x}{6 x -6 \ln \left (\frac {1}{x^{2}}\right )}\)	\(43\)
norman	\(\frac {-\ln \left (\frac {1}{x^{2}}\right )+\frac {x^{2}}{3}+4 x^{3}-\frac {x \ln \left (\frac {1}{x^{2}}\right )}{3}-4 x^{2} \ln \left (\frac {1}{x^{2}}\right )}{x -\ln \left (\frac {1}{x^{2}}\right )}\)	\(45\)

[In]

int(((24*x+1)*ln(1/x^2)^2+(-48*x^2-2*x+3)*ln(1/x^2)+24*x^3+x^2+6)/(3*ln(1/x^2)^2-6*x*ln(1/x^2)+3*x^2),x,method
=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=\frac {12 \, x^{3} + x^{2} - {\left (12 \, x^{2} + x\right )} \log \left (\frac {1}{x^{2}}\right ) - 3 \, x}{3 \, {\left (x - \log \left (\frac {1}{x^{2}}\right )\right )}} \]

[In]

integrate(((24*x+1)*log(1/x^2)^2+(-48*x^2-2*x+3)*log(1/x^2)+24*x^3+x^2+6)/(3*log(1/x^2)^2-6*x*log(1/x^2)+3*x^2
),x, algorithm="fricas")

[Out]

1/3*(12*x^3 + x^2 - (12*x^2 + x)*log(x^(-2)) - 3*x)/(x - log(x^(-2)))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=4 x^{2} + \frac {x}{3} + \frac {x}{- x + \log {\left (\frac {1}{x^{2}} \right )}} \]

[In]

integrate(((24*x+1)*ln(1/x**2)**2+(-48*x**2-2*x+3)*ln(1/x**2)+24*x**3+x**2+6)/(3*ln(1/x**2)**2-6*x*ln(1/x**2)+
3*x**2),x)

[Out]

4*x**2 + x/3 + x/(-x + log(x**(-2)))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=\frac {12 \, x^{3} + x^{2} + 2 \, {\left (12 \, x^{2} + x\right )} \log \left (x\right ) - 3 \, x}{3 \, {\left (x + 2 \, \log \left (x\right )\right )}} \]

[In]

integrate(((24*x+1)*log(1/x^2)^2+(-48*x^2-2*x+3)*log(1/x^2)+24*x^3+x^2+6)/(3*log(1/x^2)^2-6*x*log(1/x^2)+3*x^2
),x, algorithm="maxima")

[Out]

1/3*(12*x^3 + x^2 + 2*(12*x^2 + x)*log(x) - 3*x)/(x + 2*log(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=4 \, x^{2} + \frac {1}{3} \, x - \frac {x}{x + \log \left (x^{2}\right )} \]

[In]

integrate(((24*x+1)*log(1/x^2)^2+(-48*x^2-2*x+3)*log(1/x^2)+24*x^3+x^2+6)/(3*log(1/x^2)^2-6*x*log(1/x^2)+3*x^2
),x, algorithm="giac")

[Out]

4*x^2 + 1/3*x - x/(x + log(x^2))

Mupad [B] (verification not implemented)

Time = 12.80 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {6+x^2+24 x^3+\left (3-2 x-48 x^2\right ) \log \left (\frac {1}{x^2}\right )+(1+24 x) \log ^2\left (\frac {1}{x^2}\right )}{3 x^2-6 x \log \left (\frac {1}{x^2}\right )+3 \log ^2\left (\frac {1}{x^2}\right )} \, dx=\frac {x}{3}-\frac {\ln \left (\frac {1}{x^2}\right )}{x-\ln \left (\frac {1}{x^2}\right )}+4\,x^2 \]

[In]

int((log(1/x^2)^2*(24*x + 1) - log(1/x^2)*(2*x + 48*x^2 - 3) + x^2 + 24*x^3 + 6)/(3*log(1/x^2)^2 - 6*x*log(1/x
^2) + 3*x^2),x)

[Out]

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