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

Optimal. Leaf size=30 \[ \frac{1}{2 \left (1-b x^2\right )}-\frac{1}{2} \log \left (1-b x^2\right )+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.0194258, antiderivative size = 30, 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} \[ \frac{1}{2 \left (1-b x^2\right )}-\frac{1}{2} \log \left (1-b x^2\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(-1 + b*x^2)^2),x]

[Out]

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

Mathematica [A]  time = 0.0121764, size = 26, normalized size = 0.87 \[ \frac{1}{2-2 b x^2}-\frac{1}{2} \log \left (1-b x^2\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(-1 + b*x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.01, size = 25, normalized size = 0.8 \begin{align*} \ln \left ( x \right ) -{\frac{\ln \left ( b{x}^{2}-1 \right ) }{2}}-{\frac{1}{2\,b{x}^{2}-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.48238, size = 38, normalized size = 1.27 \begin{align*} -\frac{1}{2 \,{\left (b x^{2} - 1\right )}} - \frac{1}{2} \, \log \left (b x^{2} - 1\right ) + \frac{1}{2} \, \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.40382, size = 100, normalized size = 3.33 \begin{align*} -\frac{{\left (b x^{2} - 1\right )} \log \left (b x^{2} - 1\right ) - 2 \,{\left (b x^{2} - 1\right )} \log \left (x\right ) + 1}{2 \,{\left (b x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.354297, size = 22, normalized size = 0.73 \begin{align*} \log{\left (x \right )} - \frac{\log{\left (x^{2} - \frac{1}{b} \right )}}{2} - \frac{1}{2 b x^{2} - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.77783, size = 49, normalized size = 1.63 \begin{align*} \frac{b x^{2} - 2}{2 \,{\left (b x^{2} - 1\right )}} + \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/x/(b*x^2-1)^2,x, algorithm="giac")

[Out]

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