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

Optimal. Leaf size=13 \[ 2 \text{EllipticF}\left (\sin ^{-1}(x),-1\right )-E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

-EllipticE[ArcSin[x], -1] + 2*EllipticF[ArcSin[x], -1]

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Rubi [A]  time = 0.0137832, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {423, 424, 248, 221} \[ 2 F\left (\left .\sin ^{-1}(x)\right |-1\right )-E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - x^2]/Sqrt[1 + x^2],x]

[Out]

-EllipticE[ArcSin[x], -1] + 2*EllipticF[ArcSin[x], -1]

Rule 423

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

Rule 424

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

Rule 248

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_.)*((a2_.) + (b2_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[(a1*a2 + b1*b2*x^(2*
n))^p, x] /; FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a
2, 0]))

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

Mathematica [C]  time = 0.0031746, size = 12, normalized size = 0.92 \[ -i E\left (\left .i \sinh ^{-1}(x)\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - x^2]/Sqrt[1 + x^2],x]

[Out]

(-I)*EllipticE[I*ArcSinh[x], -1]

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Maple [A]  time = 0.008, size = 14, normalized size = 1.1 \begin{align*} -{\it EllipticE} \left ( x,i \right ) +2\,{\it EllipticF} \left ( x,i \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-EllipticE(x,I)+2*EllipticF(x,I)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(x - 1)*(x + 1))/sqrt(x**2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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