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

Optimal. Leaf size=21 \[ \frac{2 \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{a} \sqrt{x}\right ),-1\right )}{\sqrt{a}} \]

[Out]

(2*EllipticF[ArcSin[Sqrt[a]*Sqrt[x]], -1])/Sqrt[a]

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Rubi [A]  time = 0.008636, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {329, 221} \[ \frac{2 F\left (\left .\sin ^{-1}\left (\sqrt{a} \sqrt{x}\right )\right |-1\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*Sqrt[1 - a^2*x^2]),x]

[Out]

(2*EllipticF[ArcSin[Sqrt[a]*Sqrt[x]], -1])/Sqrt[a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{x} \sqrt{1-a^2 x^2}} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-a^2 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 F\left (\left .\sin ^{-1}\left (\sqrt{a} \sqrt{x}\right )\right |-1\right )}{\sqrt{a}}\\ \end{align*}

Mathematica [C]  time = 0.0065838, size = 24, normalized size = 1.14 \[ 2 \sqrt{x} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};a^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*Sqrt[1 - a^2*x^2]),x]

[Out]

2*Sqrt[x]*Hypergeometric2F1[1/4, 1/2, 5/4, a^2*x^2]

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Maple [B]  time = 0.045, size = 66, normalized size = 3.1 \begin{align*} -{\frac{1}{a \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{ax+1}\sqrt{-2\,ax+2}\sqrt{-ax}{\it EllipticF} \left ( \sqrt{ax+1},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-a^{2} x^{2} + 1} \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} \sqrt{x}}{a^{2} x^{3} - x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*sqrt(x)/(a^2*x^3 - x), x)

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Sympy [B]  time = 0.67367, size = 36, normalized size = 1.71 \begin{align*} \frac{\sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{2} \\ \frac{5}{4} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x)*gamma(1/4)*hyper((1/4, 1/2), (5/4,), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(5/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-a^{2} x^{2} + 1} \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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