\(\int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx\) [254]

Optimal result

Integrand size = 23, antiderivative size = 42 \[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=-\frac {\sqrt {1-\frac {1}{x^4}} x^2 \operatorname {EllipticF}\left (\csc ^{-1}(x),-1\right )}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \]

[Out]

-x^2*EllipticF(1/x,I)*(1-1/x^4)^(1/2)/(-2*x^2+2)^(1/2)/(-x^2-1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.55, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {259, 228} \[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=\frac {\sqrt {x^2-1} \sqrt {x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right ),\frac {1}{2}\right )}{2 \sqrt {-x^2-1} \sqrt {1-x^2}} \]

[In]

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

[Out]

(Sqrt[-1 + x^2]*Sqrt[1 + x^2]*EllipticF[ArcSin[(Sqrt[2]*x)/Sqrt[-1 + x^2]], 1/2])/(2*Sqrt[-1 - x^2]*Sqrt[1 - x
^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*} \text {integral}& = \frac {\sqrt {-2+2 x^4} \int \frac {1}{\sqrt {-2+2 x^4}} \, dx}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \\ & = \frac {\sqrt {-1+x^2} \sqrt {1+x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {-1+x^2}}\right )|\frac {1}{2}\right )}{2 \sqrt {-1-x^2} \sqrt {1-x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=\frac {x \sqrt {1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},x^4\right )}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=-\frac {1}{2} \, \sqrt {-2} F(\arcsin \left (x\right )\,|\,-1) \]

[In]

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

[Out]

-1/2*sqrt(-2)*elliptic_f(arcsin(x), -1)

Sympy [A] (verification not implemented)

Time = 11.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.74 \[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=\frac {\sqrt {2} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{2}, 1, 1 & \frac {3}{4}, \frac {3}{4}, \frac {5}{4} \\\frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4} & 0 \end {matrix} \middle | {\frac {1}{x^{4}}} \right )}}{16 \pi ^{\frac {3}{2}}} - \frac {\sqrt {2} {G_{6, 6}^{3, 5}\left (\begin {matrix} - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4} & 1 \\0, \frac {1}{2}, 0 & - \frac {1}{4}, \frac {1}{4}, \frac {1}{4} \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{x^{4}}} \right )}}{16 \pi ^{\frac {3}{2}}} \]

[In]

integrate(1/(-2*x**2+2)**(1/2)/(-x**2-1)**(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,)), x**(-4))/(16*pi**(3/2)) - sqr
t(2)*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*p
i**(3/2))

Maxima [F]

\[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=\int { \frac {1}{\sqrt {-x^{2} - 1} \sqrt {-2 \, x^{2} + 2}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=\int { \frac {1}{\sqrt {-x^{2} - 1} \sqrt {-2 \, x^{2} + 2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {2-2 x^2} \sqrt {-1-x^2}} \, dx=\int \frac {1}{\sqrt {-x^2-1}\,\sqrt {2-2\,x^2}} \,d x \]

[In]

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

[Out]

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