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

Optimal. Leaf size=15 \[ \log (x)-\frac{1}{2} \log \left (1-x^2\right ) \]

[Out]

Log[x] - Log[1 - x^2]/2

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Rubi [A]  time = 0.0082264, antiderivative size = 15, normalized size of antiderivative = 1., 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} \[ \log (x)-\frac{1}{2} \log \left (1-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x - x^3)^(-1),x]

[Out]

Log[x] - Log[1 - x^2]/2

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]

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 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 31

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

Rule 29

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

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*}

Mathematica [A]  time = 0.0024746, size = 15, normalized size = 1. \[ \log (x)-\frac{1}{2} \log \left (1-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x - x^3)^(-1),x]

[Out]

Log[x] - Log[1 - x^2]/2

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Maple [A]  time = 0.006, size = 16, normalized size = 1.1 \begin{align*} \ln \left ( x \right ) -{\frac{\ln \left ( 1+x \right ) }{2}}-{\frac{\ln \left ( -1+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02813, size = 20, normalized size = 1.33 \begin{align*} -\frac{1}{2} \, \log \left (x + 1\right ) - \frac{1}{2} \, \log \left (x - 1\right ) + \log \left (x\right ) \end{align*}

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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Fricas [A]  time = 1.44473, size = 38, normalized size = 2.53 \begin{align*} -\frac{1}{2} \, \log \left (x^{2} - 1\right ) + \log \left (x\right ) \end{align*}

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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Sympy [A]  time = 0.087841, size = 10, normalized size = 0.67 \begin{align*} \log{\left (x \right )} - \frac{\log{\left (x^{2} - 1 \right )}}{2} \end{align*}

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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Giac [A]  time = 1.22125, size = 22, normalized size = 1.47 \begin{align*} \frac{1}{2} \, \log \left (x^{2}\right ) - \frac{1}{2} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

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))