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

Optimal. Leaf size=18 \[ \frac{1}{2} \log \left (1-b x^2\right )-\log (x) \]

[Out]

-Log[x] + Log[1 - b*x^2]/2

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Rubi [A]  time = 0.0102575, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {1593, 266, 36, 29, 31} \[ \frac{1}{2} \log \left (1-b x^2\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Int[(-x + b*x^3)^(-1),x]

[Out]

-Log[x] + Log[1 - b*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 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]

Rubi steps

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

Mathematica [A]  time = 0.0034299, size = 18, normalized size = 1. \[ \frac{1}{2} \log \left (1-b x^2\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-x + b*x^3)^(-1),x]

[Out]

-Log[x] + Log[1 - b*x^2]/2

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Maple [A]  time = 0.003, size = 16, normalized size = 0.9 \begin{align*} -\ln \left ( x \right ) +{\frac{\ln \left ( b{x}^{2}-1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05053, size = 20, normalized size = 1.11 \begin{align*} \frac{1}{2} \, \log \left (b x^{2} - 1\right ) - \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*x^2 - 1) - log(x)

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Fricas [A]  time = 1.40847, size = 39, normalized size = 2.17 \begin{align*} \frac{1}{2} \, \log \left (b x^{2} - 1\right ) - \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*x^2 - 1) - log(x)

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Sympy [A]  time = 0.122316, size = 12, normalized size = 0.67 \begin{align*} - \log{\left (x \right )} + \frac{\log{\left (x^{2} - \frac{1}{b} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x**2 - 1/b)/2

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Giac [A]  time = 1.23484, size = 24, normalized size = 1.33 \begin{align*} -\frac{1}{2} \, \log \left (x^{2}\right ) + \frac{1}{2} \, \log \left ({\left | b x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(x^2) + 1/2*log(abs(b*x^2 - 1))