3.3.21 \(\int \frac {1}{\sqrt {1-x^2} \sqrt {2+2 x^2}} \, dx\) [221]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {254, 227} \begin {gather*} \frac {F(\text {ArcSin}(x)|-1)}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticF[ArcSin[x], -1]/Sqrt[2]

Rule 227

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]

Rule 254

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]))

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-x^2} \sqrt {2+2 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*}

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Mathematica [A]
time = 10.02, size = 10, normalized size = 1.00 \begin {gather*} \frac {F\left (\left .\sin ^{-1}(x)\right |-1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticF[ArcSin[x], -1]/Sqrt[2]

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Maple [A]
time = 0.09, size = 10, normalized size = 1.00

method	result	size
default	\(\frac {\EllipticF \left (x , i\right ) \sqrt {2}}{2}\)	\(10\)
elliptic	\(\frac {\sqrt {-x^{4}+1}\, \EllipticF \left (x , i\right )}{\sqrt {-2 x^{4}+2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.12, size = 8, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, \sqrt {2} {\rm ellipticF}\left (x, -1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*ellipticF(x, -1)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (8) = 16\).
time = 10.95, size = 73, normalized size = 7.30 \begin {gather*} - \frac {\sqrt {2} {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 {e^{- 2 i \pi }}{x^{4}}} \right )}}{16 \pi ^{\frac {3}{2}}} + \frac {\sqrt {2} {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 {1}{x^{4}}} \right )}}{16 \pi ^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.10 \begin {gather*} \int \frac {1}{\sqrt {1-x^2}\,\sqrt {2\,x^2+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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