\(\int \frac {-1+x^2}{-2 x+x^3} \, dx\) [433]

Optimal result

Integrand size = 15, antiderivative size = 19 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {\log (x)}{2}+\frac {1}{4} \log \left (2-x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1607, 457, 78} \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {1}{4} \log \left (2-x^2\right )+\frac {\log (x)}{2} \]

[In]

Int[(-1 + x^2)/(-2*x + x^3),x]

[Out]

Log[x]/2 + Log[2 - x^2]/4

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {\log (x)}{2}+\frac {1}{4} \log \left (2-x^2\right ) \]

[In]

Integrate[(-1 + x^2)/(-2*x + x^3),x]

[Out]

Log[x]/2 + Log[2 - x^2]/4

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x^{2}-2\right )}{4}\)	\(14\)
norman	\(\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x^{2}-2\right )}{4}\)	\(14\)
risch	\(\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x^{2}-2\right )}{4}\)	\(14\)
parallelrisch	\(\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x^{2}-2\right )}{4}\)	\(14\)
meijerg	\(\frac {\ln \left (1-\frac {x^{2}}{2}\right )}{4}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2\right )}{4}+\frac {i \pi }{4}\)	\(24\)

[In]

int((x^2-1)/(x^3-2*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {1}{4} \, \log \left (x^{2} - 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate((x^2-1)/(x^3-2*x),x, algorithm="fricas")

[Out]

1/4*log(x^2 - 2) + 1/2*log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {\log {\left (x \right )}}{2} + \frac {\log {\left (x^{2} - 2 \right )}}{4} \]

[In]

integrate((x**2-1)/(x**3-2*x),x)

[Out]

log(x)/2 + log(x**2 - 2)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {1}{4} \, \log \left (x^{2} - 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate((x^2-1)/(x^3-2*x),x, algorithm="maxima")

[Out]

1/4*log(x^2 - 2) + 1/2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{4} \, \log \left ({\left | x^{2} - 2 \right |}\right ) \]

[In]

integrate((x^2-1)/(x^3-2*x),x, algorithm="giac")

[Out]

1/4*log(x^2) + 1/4*log(abs(x^2 - 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-1+x^2}{-2 x+x^3} \, dx=\frac {\ln \left (x^2-2\right )}{4}+\frac {\ln \left (x\right )}{2} \]

[In]

int(-(x^2 - 1)/(2*x - x^3),x)

[Out]

log(x^2 - 2)/4 + log(x)/2