3.1.97 \(\int \frac {\sqrt {1-x^2}}{(-1+x^2) \sqrt {a+b x^2}} \, dx\) [97]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 432, 430} \begin {gather*} -\frac {\sqrt {\frac {b x^2}{a}+1} F\left (\text {ArcSin}(x)\left |-\frac {b}{a}\right .\right )}{\sqrt {a+b x^2}} \end {gather*}

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 430

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

Rule 432

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

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*}

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Mathematica [A]
time = 0.49, size = 37, normalized size = 1.03 \begin {gather*} -\frac {\sqrt {\frac {a+b x^2}{a}} F\left (\sin ^{-1}(x)|-\frac {b}{a}\right )}{\sqrt {a+b x^2}} \end {gather*}

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 = 0.16, size = 35, normalized size = 0.97

method	result	size
default	\(-\frac {\sqrt {\frac {b \,x^{2}+a}{a}}\, \EllipticF \left (x , \sqrt {-\frac {b}{a}}\right )}{\sqrt {b \,x^{2}+a}}\)	\(35\)
elliptic	\(-\frac {\sqrt {-\left (x^{2}-1\right ) \left (b \,x^{2}+a \right )}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \EllipticF \left (x , \sqrt {-1-\frac {-a +b}{a}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{4}-a \,x^{2}+b \,x^{2}+a}}\)	\(77\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [A]
time = 0.38, size = 13, normalized size = 0.36 \begin {gather*} -\frac {{\rm ellipticF}\left (x, -\frac {b}{a}\right )}{\sqrt {a}} \end {gather*}

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]

-ellipticF(x, -b/a)/sqrt(a)

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Sympy [A]
time = 2.07, size = 19, normalized size = 0.53 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {1}{\sqrt {1-x^2}\,\sqrt {b\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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