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

Optimal. Leaf size=23 \[ \frac{\log (x)}{a}-\frac{\log \left (a-b x^2\right )}{2 a} \]

[Out]

Log[x]/a - Log[a - b*x^2]/(2*a)

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Rubi [A]  time = 0.0122526, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {266, 36, 29, 31} \[ \frac{\log (x)}{a}-\frac{\log \left (a-b x^2\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a - b*x^2)),x]

[Out]

Log[x]/a - Log[a - b*x^2]/(2*a)

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

Mathematica [A]  time = 0.006051, size = 23, normalized size = 1. \[ \frac{\log (x)}{a}-\frac{\log \left (a-b x^2\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a - b*x^2)),x]

[Out]

Log[x]/a - Log[a - b*x^2]/(2*a)

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Maple [A]  time = 0.005, size = 23, normalized size = 1. \begin{align*}{\frac{\ln \left ( x \right ) }{a}}-{\frac{\ln \left ( b{x}^{2}-a \right ) }{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.50863, size = 34, normalized size = 1.48 \begin{align*} -\frac{\log \left (b x^{2} - a\right )}{2 \, a} + \frac{\log \left (x^{2}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.26249, size = 49, normalized size = 2.13 \begin{align*} -\frac{\log \left (b x^{2} - a\right ) - 2 \, \log \left (x\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(log(b*x^2 - a) - 2*log(x))/a

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Sympy [A]  time = 0.192758, size = 15, normalized size = 0.65 \begin{align*} \frac{\log{\left (x \right )}}{a} - \frac{\log{\left (- \frac{a}{b} + x^{2} \right )}}{2 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/a - log(-a/b + x**2)/(2*a)

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Giac [A]  time = 1.8525, size = 35, normalized size = 1.52 \begin{align*} \frac{\log \left (x^{2}\right )}{2 \, a} - \frac{\log \left ({\left | b x^{2} - a \right |}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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