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

Optimal. Leaf size=35 \[ -\frac{b \log \left (a-b x^2\right )}{2 a^2}+\frac{b \log (x)}{a^2}-\frac{1}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.0235559, antiderivative size = 35, 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} \[ -\frac{b \log \left (a-b x^2\right )}{2 a^2}+\frac{b \log (x)}{a^2}-\frac{1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^2)),x]

[Out]

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

Mathematica [A]  time = 0.0082722, size = 35, normalized size = 1. \[ -\frac{b \log \left (a-b x^2\right )}{2 a^2}+\frac{b \log (x)}{a^2}-\frac{1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a - b*x^2)),x]

[Out]

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

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Maple [A]  time = 0.005, size = 33, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,a{x}^{2}}}+{\frac{b\ln \left ( x \right ) }{{a}^{2}}}-{\frac{b\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^3/(-b*x^2+a),x)

[Out]

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

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Maxima [A]  time = 1.58463, size = 47, normalized size = 1.34 \begin{align*} -\frac{b \log \left (b x^{2} - a\right )}{2 \, a^{2}} + \frac{b \log \left (x^{2}\right )}{2 \, a^{2}} - \frac{1}{2 \, a x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.26507, size = 81, normalized size = 2.31 \begin{align*} -\frac{b x^{2} \log \left (b x^{2} - a\right ) - 2 \, b x^{2} \log \left (x\right ) + a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.417309, size = 31, normalized size = 0.89 \begin{align*} - \frac{1}{2 a x^{2}} + \frac{b \log{\left (x \right )}}{a^{2}} - \frac{b \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**3/(-b*x**2+a),x)

[Out]

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

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Giac [A]  time = 2.96638, size = 58, normalized size = 1.66 \begin{align*} \frac{b \log \left (x^{2}\right )}{2 \, a^{2}} - \frac{b \log \left ({\left | b x^{2} - a \right |}\right )}{2 \, a^{2}} - \frac{b x^{2} + a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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