3.126 \(\int \frac {1}{-x+x^3} \, dx\)

Optimal. Leaf size=17 \[ \frac {1}{2} \log \left (1-x^2\right )-\log (x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {1593, 266, 36, 31, 29} \[ \frac {1}{2} \log \left (1-x^2\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Int[(-x + x^3)^(-1),x]

[Out]

-Log[x] + Log[1 - x^2]/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

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

Rule 1593

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*} \int \frac {1}{-x+x^3} \, dx &=\int \frac {1}{x \left (-1+x^2\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(-1+x) x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-1+x} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )\\ &=-\log (x)+\frac {1}{2} \log \left (1-x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.00 \[ \frac {1}{2} \log \left (1-x^2\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-x + x^3)^(-1),x]

[Out]

-Log[x] + Log[1 - x^2]/2

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fricas [A]  time = 0.38, size = 13, normalized size = 0.76 \[ \frac {1}{2} \, \log \left (x^{2} - 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x^2 - 1) - log(x)

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giac [A]  time = 0.01, size = 16, normalized size = 0.94 \[ -\frac {1}{2} \, \log \left (x^{2}\right ) + \frac {1}{2} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(x^2) + 1/2*log(abs(x^2 - 1))

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maple [A]  time = 0.01, size = 18, normalized size = 1.06 \[ -\ln \relax (x )+\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-x),x)

[Out]

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

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maxima [A]  time = 0.52, size = 17, normalized size = 1.00 \[ \frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x + 1) + 1/2*log(x - 1) - log(x)

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mupad [B]  time = 0.11, size = 13, normalized size = 0.76 \[ \frac {\ln \left (x^2-1\right )}{2}-\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x - x^3),x)

[Out]

log(x^2 - 1)/2 - log(x)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.59 \[ - \log {\relax (x )} + \frac {\log {\left (x^{2} - 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x**2 - 1)/2

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