3.256 \(\int \frac{1}{x^3 (-1+b x^2)} \, dx\)

Optimal. Leaf size=27 \[ \frac{1}{2} b \log \left (1-b x^2\right )-b \log (x)+\frac{1}{2 x^2} \]

[Out]

1/(2*x^2) - b*Log[x] + (b*Log[1 - b*x^2])/2

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Rubi [A]  time = 0.017959, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 44} \[ \frac{1}{2} b \log \left (1-b x^2\right )-b \log (x)+\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-1 + b*x^2)),x]

[Out]

1/(2*x^2) - b*Log[x] + (b*Log[1 - b*x^2])/2

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.004213, size = 27, normalized size = 1. \[ \frac{1}{2} b \log \left (1-b x^2\right )-b \log (x)+\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(-1 + b*x^2)),x]

[Out]

1/(2*x^2) - b*Log[x] + (b*Log[1 - b*x^2])/2

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Maple [A]  time = 0.007, size = 23, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{x}^{2}}}-b\ln \left ( x \right ) +{\frac{b\ln \left ( b{x}^{2}-1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^2-1),x)

[Out]

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

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Maxima [A]  time = 2.2964, size = 32, normalized size = 1.19 \begin{align*} \frac{1}{2} \, b \log \left (b x^{2} - 1\right ) - \frac{1}{2} \, b \log \left (x^{2}\right ) + \frac{1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.24557, size = 72, normalized size = 2.67 \begin{align*} \frac{b x^{2} \log \left (b x^{2} - 1\right ) - 2 \, b x^{2} \log \left (x\right ) + 1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.193962, size = 22, normalized size = 0.81 \begin{align*} - b \log{\left (x \right )} + \frac{b \log{\left (x^{2} - \frac{1}{b} \right )}}{2} + \frac{1}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.85536, size = 43, normalized size = 1.59 \begin{align*} -\frac{1}{2} \, b \log \left (x^{2}\right ) + \frac{1}{2} \, b \log \left ({\left | b x^{2} - 1 \right |}\right ) + \frac{b x^{2} + 1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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