3.236 \(\int \frac{1}{x (a-b x^2)^2} \, dx\)

Optimal. Leaf size=40 \[ -\frac{\log \left (a-b x^2\right )}{2 a^2}+\frac{\log (x)}{a^2}+\frac{1}{2 a \left (a-b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0279616, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {266, 44} \[ -\frac{\log \left (a-b x^2\right )}{2 a^2}+\frac{\log (x)}{a^2}+\frac{1}{2 a \left (a-b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a - b*x^2)^2),x]

[Out]

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

Mathematica [A]  time = 0.0160346, size = 35, normalized size = 0.88 \[ \frac{\frac{a}{a-b x^2}-\log \left (a-b x^2\right )+2 \log (x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a - b*x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.008, size = 39, normalized size = 1. \begin{align*}{\frac{\ln \left ( x \right ) }{{a}^{2}}}-{\frac{1}{2\,a \left ( b{x}^{2}-a \right ) }}-{\frac{\ln \left ( b{x}^{2}-a \right ) }{2\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20726, size = 55, normalized size = 1.38 \begin{align*} -\frac{1}{2 \,{\left (a b x^{2} - a^{2}\right )}} - \frac{\log \left (b x^{2} - a\right )}{2 \, a^{2}} + \frac{\log \left (x^{2}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.24117, size = 108, normalized size = 2.7 \begin{align*} -\frac{{\left (b x^{2} - a\right )} \log \left (b x^{2} - a\right ) - 2 \,{\left (b x^{2} - a\right )} \log \left (x\right ) + a}{2 \,{\left (a^{2} b x^{2} - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((b*x^2 - a)*log(b*x^2 - a) - 2*(b*x^2 - a)*log(x) + a)/(a^2*b*x^2 - a^3)

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Sympy [A]  time = 0.447608, size = 34, normalized size = 0.85 \begin{align*} - \frac{1}{- 2 a^{2} + 2 a b x^{2}} + \frac{\log{\left (x \right )}}{a^{2}} - \frac{\log{\left (- \frac{a}{b} + x^{2} \right )}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.76258, size = 69, normalized size = 1.72 \begin{align*} \frac{\log \left (x^{2}\right )}{2 \, a^{2}} - \frac{\log \left ({\left | b x^{2} - a \right |}\right )}{2 \, a^{2}} + \frac{b x^{2} - 2 \, a}{2 \,{\left (b x^{2} - a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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