\(\int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx\) [226]

Optimal result

Integrand size = 23, antiderivative size = 10 \[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\frac {\operatorname {EllipticF}(\arcsin (x),2)}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {430} \[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\frac {\operatorname {EllipticF}(\arcsin (x),2)}{\sqrt {2}} \]

[In]

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

[Out]

EllipticF[ArcSin[x], 2]/Sqrt[2]

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rubi steps \begin{align*} \text {integral}& = \frac {F\left (\left .\sin ^{-1}(x)\right |2\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\frac {\operatorname {EllipticF}(\arcsin (x),2)}{\sqrt {2}} \]

[In]

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

[Out]

EllipticF[ArcSin[x], 2]/Sqrt[2]

Maple [A] (verified)

Time = 2.62 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {F\left (x , \sqrt {2}\right ) \sqrt {2}}{2}\)	\(11\)
elliptic	\(\frac {\sqrt {\left (2 x^{2}-1\right ) \left (x^{2}-1\right )}\, F\left (x , \sqrt {2}\right )}{\sqrt {4 x^{4}-6 x^{2}+2}}\)	\(36\)

[In]

int(1/(-4*x^2+2)^(1/2)/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\frac {1}{2} \, \sqrt {2} F(\arcsin \left (x\right )\,|\,2) \]

[In]

integrate(1/(-4*x^2+2)^(1/2)/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*elliptic_f(arcsin(x), 2)

Sympy [A] (verification not implemented)

Time = 1.88 (sec) , antiderivative size = 39, normalized size of antiderivative = 3.90 \[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\frac {\sqrt {2} \left (\begin {cases} \frac {\sqrt {2} F\left (\operatorname {asin}{\left (\sqrt {2} x \right )}\middle | \frac {1}{2}\right )}{2} & \text {for}\: x > - \frac {\sqrt {2}}{2} \wedge x < \frac {\sqrt {2}}{2} \end {cases}\right )}{2} \]

[In]

integrate(1/(-4*x**2+2)**(1/2)/(-x**2+1)**(1/2),x)

[Out]

sqrt(2)*Piecewise((sqrt(2)*elliptic_f(asin(sqrt(2)*x), 1/2)/2, (x > -sqrt(2)/2) & (x < sqrt(2)/2)))/2

Maxima [F]

\[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\int { \frac {1}{\sqrt {-x^{2} + 1} \sqrt {-4 \, x^{2} + 2}} \,d x } \]

[In]

integrate(1/(-4*x^2+2)^(1/2)/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\int { \frac {1}{\sqrt {-x^{2} + 1} \sqrt {-4 \, x^{2} + 2}} \,d x } \]

[In]

integrate(1/(-4*x^2+2)^(1/2)/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {2-4 x^2} \sqrt {1-x^2}} \, dx=\int \frac {1}{\sqrt {1-x^2}\,\sqrt {2-4\,x^2}} \,d x \]

[In]

int(1/((1 - x^2)^(1/2)*(2 - 4*x^2)^(1/2)),x)

[Out]

int(1/((1 - x^2)^(1/2)*(2 - 4*x^2)^(1/2)), x)