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

Optimal. Leaf size=10 \[ \frac{\text{EllipticF}\left (\sin ^{-1}(x),-1\right )}{\sqrt{2}} \]

[Out]

EllipticF[ArcSin[x], -1]/Sqrt[2]

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Rubi [A]  time = 0.0047369, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {248, 221} \[ \frac{F\left (\left .\sin ^{-1}(x)\right |-1\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticF[ArcSin[x], -1]/Sqrt[2]

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

Mathematica [A]  time = 0.0118819, size = 10, normalized size = 1. \[ \frac{\text{EllipticF}\left (\sin ^{-1}(x),-1\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticF[ArcSin[x], -1]/Sqrt[2]

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Maple [A]  time = 0.006, size = 10, normalized size = 1. \begin{align*}{\frac{{\it EllipticF} \left ( x,i \right ) \sqrt{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*EllipticF(x,I)*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x^2 + 1)*sqrt(-2*x^2 + 2)), 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{-2 \, x^{2} + 2}}{2 \,{\left (x^{4} - 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 7.75348, size = 76, normalized size = 7.6 \begin{align*} \frac{\sqrt{2} i{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{1}{2}, 1, 1 & \frac{3}{4}, \frac{3}{4}, \frac{5}{4} \\\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4} & 0 \end{matrix} \middle |{\frac{1}{x^{4}}} \right )}}{16 \pi ^{\frac{3}{2}}} - \frac{\sqrt{2} i{G_{6, 6}^{3, 5}\left (\begin{matrix} - \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4} & 1 \\0, \frac{1}{2}, 0 & - \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \end{matrix} \middle |{\frac{e^{- 2 i \pi }}{x^{4}}} \right )}}{16 \pi ^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*I*meijerg(((1/2, 1, 1), (3/4, 3/4, 5/4)), ((1/4, 1/2, 3/4, 1, 5/4), (0,)), x**(-4))/(16*pi**(3/2)) - s
qrt(2)*I*meijerg(((-1/4, 0, 1/4, 1/2, 3/4), (1,)), ((0, 1/2, 0), (-1/4, 1/4, 1/4)), exp_polar(-2*I*pi)/x**4)/(
16*pi**(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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