3.52 \(\int \frac{1}{x \sqrt{a^2-x^2}} \, dx\)

Optimal. Leaf size=23 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a^2-x^2}}{a}\right )}{a} \]

[Out]

-(ArcTanh[Sqrt[a^2 - x^2]/a]/a)

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Rubi [A]  time = 0.0141744, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {266, 63, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a^2-x^2}}{a}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a^2 - x^2]),x]

[Out]

-(ArcTanh[Sqrt[a^2 - x^2]/a]/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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0038571, size = 23, normalized size = 1. \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a^2-x^2}}{a}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[a^2 - x^2]),x]

[Out]

-(ArcTanh[Sqrt[a^2 - x^2]/a]/a)

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Maple [A]  time = 0.005, size = 37, normalized size = 1.6 \begin{align*} -{\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2\,\sqrt{{a}^{2}}\sqrt{{a}^{2}-{x}^{2}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a^2-x^2)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88047, size = 45, normalized size = 1.96 \begin{align*} \frac{\log \left (-\frac{a - \sqrt{a^{2} - x^{2}}}{x}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-(a - sqrt(a^2 - x^2))/x)/a

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Sympy [A]  time = 1.07572, size = 24, normalized size = 1.04 \begin{align*} \begin{cases} - \frac{\operatorname{acosh}{\left (\frac{a}{x} \right )}}{a} & \text{for}\: \frac{\left |{a^{2}}\right |}{\left |{x^{2}}\right |} > 1 \\\frac{i \operatorname{asin}{\left (\frac{a}{x} \right )}}{a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-acosh(a/x)/a, Abs(a**2)/Abs(x**2) > 1), (I*asin(a/x)/a, True))

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Giac [B]  time = 1.07154, size = 58, normalized size = 2.52 \begin{align*} -\frac{\log \left ({\left | a + \sqrt{a^{2} - x^{2}} \right |}\right )}{2 \, a} + \frac{\log \left ({\left | -a + \sqrt{a^{2} - x^{2}} \right |}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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