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

Optimal. Leaf size=36 \[ -\frac{\sqrt{\frac{b x^2}{a}+1} \text{EllipticF}\left (\sin ^{-1}(x),-\frac{b}{a}\right )}{\sqrt{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0248665, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {21, 421, 419} \[ -\frac{\sqrt{\frac{b x^2}{a}+1} F\left (\sin ^{-1}(x)|-\frac{b}{a}\right )}{\sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - x^2]/((-1 + x^2)*Sqrt[a + b*x^2]),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

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

Mathematica [A]  time = 0.0420609, size = 37, normalized size = 1.03 \[ -\frac{\sqrt{\frac{a+b x^2}{a}} \text{EllipticF}\left (\sin ^{-1}(x),-\frac{b}{a}\right )}{\sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[(a + b*x^2)/a]*EllipticF[ArcSin[x], -(b/a)])/Sqrt[a + b*x^2])

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Maple [A]  time = 0.026, size = 35, normalized size = 1. \begin{align*} -{\sqrt{{\frac{b{x}^{2}+a}{a}}}{\it EllipticF} \left ( x,\sqrt{-{\frac{b}{a}}} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{-x^{2} + 1}}{b x^{4} +{\left (a - b\right )} x^{2} - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(-x^2 + 1)/(b*x^4 + (a - b)*x^2 - a), x)

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Sympy [A]  time = 5.7869, size = 19, normalized size = 0.53 \begin{align*} \begin{cases} - \frac{F\left (\operatorname{asin}{\left (x \right )}\middle | - \frac{b}{a}\right )}{\sqrt{a}} & \text{for}\: x > -1 \wedge x < 1 \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-elliptic_f(asin(x), -b/a)/sqrt(a), (x > -1) & (x < 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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