\(\int \frac {-2-8 x-6 x^8+2 \log (x^2)}{-8 x^2+4 x^3+x^9+x \log (x^2)} \, dx\) [66]

Optimal result

Integrand size = 39, antiderivative size = 26 \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=\log \left (\frac {x}{-3+x+\frac {x+\frac {1}{4} \left (x^8+\log \left (x^2\right )\right )}{x}}\right ) \]

[Out]

ln(x/((x+1/4*ln(x^2)+1/4*x^8)/x-3+x))

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6873, 12, 6874, 6816} \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=2 \log (x)-\log \left (-x^8-4 x^2-\log \left (x^2\right )+8 x\right ) \]

[In]

Int[(-2 - 8*x - 6*x^8 + 2*Log[x^2])/(-8*x^2 + 4*x^3 + x^9 + x*Log[x^2]),x]

[Out]

2*Log[x] - Log[8*x - 4*x^2 - x^8 - Log[x^2]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=2 \log (x)-\log \left (-8 x+4 x^2+x^8+\log \left (x^2\right )\right ) \]

[In]

Integrate[(-2 - 8*x - 6*x^8 + 2*Log[x^2])/(-8*x^2 + 4*x^3 + x^9 + x*Log[x^2]),x]

[Out]

2*Log[x] - Log[-8*x + 4*x^2 + x^8 + Log[x^2]]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
norman	\(\ln \left (x^{2}\right )-\ln \left (x^{8}+4 x^{2}+\ln \left (x^{2}\right )-8 x \right )\)	\(25\)
risch	\(2 \ln \left (x \right )-\ln \left (x^{8}+4 x^{2}+\ln \left (x^{2}\right )-8 x \right )\)	\(25\)
parallelrisch	\(\ln \left (x^{2}\right )-\ln \left (x^{8}+4 x^{2}+\ln \left (x^{2}\right )-8 x \right )\)	\(25\)

[In]

int((2*ln(x^2)-6*x^8-8*x-2)/(x*ln(x^2)+x^9+4*x^3-8*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x^2)-ln(x^8+4*x^2+ln(x^2)-8*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=-\log \left (x^{8} + 4 \, x^{2} - 8 \, x + \log \left (x^{2}\right )\right ) + \log \left (x^{2}\right ) \]

[In]

integrate((2*log(x^2)-6*x^8-8*x-2)/(x*log(x^2)+x^9+4*x^3-8*x^2),x, algorithm="fricas")

[Out]

-log(x^8 + 4*x^2 - 8*x + log(x^2)) + log(x^2)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=2 \log {\left (x \right )} - \log {\left (x^{8} + 4 x^{2} - 8 x + \log {\left (x^{2} \right )} \right )} \]

[In]

integrate((2*ln(x**2)-6*x**8-8*x-2)/(x*ln(x**2)+x**9+4*x**3-8*x**2),x)

[Out]

2*log(x) - log(x**8 + 4*x**2 - 8*x + log(x**2))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=-\log \left (\frac {1}{2} \, x^{8} + 2 \, x^{2} - 4 \, x + \log \left (x\right )\right ) + 2 \, \log \left (x\right ) \]

[In]

integrate((2*log(x^2)-6*x^8-8*x-2)/(x*log(x^2)+x^9+4*x^3-8*x^2),x, algorithm="maxima")

[Out]

-log(1/2*x^8 + 2*x^2 - 4*x + log(x)) + 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=-\log \left (x^{8} + 4 \, x^{2} - 8 \, x + \log \left (x^{2}\right )\right ) + 2 \, \log \left (x\right ) \]

[In]

integrate((2*log(x^2)-6*x^8-8*x-2)/(x*log(x^2)+x^9+4*x^3-8*x^2),x, algorithm="giac")

[Out]

-log(x^8 + 4*x^2 - 8*x + log(x^2)) + 2*log(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-2-8 x-6 x^8+2 \log \left (x^2\right )}{-8 x^2+4 x^3+x^9+x \log \left (x^2\right )} \, dx=\int -\frac {8\,x-2\,\ln \left (x^2\right )+6\,x^8+2}{x\,\ln \left (x^2\right )-8\,x^2+4\,x^3+x^9} \,d x \]

[In]

int(-(8*x - 2*log(x^2) + 6*x^8 + 2)/(x*log(x^2) - 8*x^2 + 4*x^3 + x^9),x)

[Out]

int(-(8*x - 2*log(x^2) + 6*x^8 + 2)/(x*log(x^2) - 8*x^2 + 4*x^3 + x^9), x)