3.3.41 \(\int \frac {1}{\sqrt {-1+x^2} \sqrt {2+2 x^2}} \, dx\) [241]

Optimal. Leaf size=25 \[ \frac {1}{2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 = {259, 228} \begin {gather*} \frac {1}{2} F\left (\text {ArcSin}\left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right )|\frac {1}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 228

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*b, 2]}, Simp[Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2
)/q]/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]))*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 259

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.05, size = 46, normalized size = 1.84 \begin {gather*} \frac {x \sqrt {1-x^4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};x^4\right )}{\sqrt {-1+x^2} \sqrt {2+2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 - x^4]*Hypergeometric2F1[1/4, 1/2, 5/4, x^4])/(Sqrt[-1 + x^2]*Sqrt[2 + 2*x^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.10, size = 30, normalized size = 1.20

method	result	size
default	\(-\frac {i \EllipticF \left (i x , i\right ) \sqrt {-x^{2}+1}\, \sqrt {2}}{2 \sqrt {x^{2}-1}}\)	\(30\)
elliptic	\(-\frac {i \sqrt {x^{4}-1}\, \sqrt {-x^{2}+1}\, \EllipticF \left (i x , i\right )}{\sqrt {x^{2}-1}\, \sqrt {2 x^{4}-2}}\)	\(43\)

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*I*EllipticF(I*x,I)*(-x^2+1)^(1/2)*2^(1/2)/(x^2-1)^(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 [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="fricas")

[Out]

0

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Sympy [C] Result contains complex when optimal does not.
time = 10.97, size = 75, normalized size = 3.00 \begin {gather*} \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 {e^{2 i \pi }}{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 {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)*I*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)*I*meijerg(((-1/4, 0, 1/4, 1/2, 3/4), (1,)), ((0, 1/2, 0), (-1/4, 1/4, 1/4)), x**(-4))/(1
6*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 = -0.04 \begin {gather*} \int \frac {1}{\sqrt {x^2-1}\,\sqrt {2\,x^2+2}} \,d x \end {gather*}

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)

[Out]

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

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