\(\int \frac {1}{\sqrt {-1+x^2} \sqrt {2+4 x^2}} \, dx\) [239]

Optimal result

Integrand size = 21, antiderivative size = 30 \[ \int \frac {1}{\sqrt {-1+x^2} \sqrt {2+4 x^2}} \, dx=\frac {\sqrt {1-x^2} \operatorname {EllipticF}(\arcsin (x),-2)}{\sqrt {2} \sqrt {-1+x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {432, 430} \[ \int \frac {1}{\sqrt {-1+x^2} \sqrt {2+4 x^2}} \, dx=\frac {\sqrt {1-x^2} \operatorname {EllipticF}(\arcsin (x),-2)}{\sqrt {2} \sqrt {x^2-1}} \]

[In]

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

[Out]

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

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {-1+x^2} \sqrt {2+4 x^2}} \, dx=\frac {\sqrt {1-x^2} \operatorname {EllipticF}(\arcsin (x),-2)}{\sqrt {2} \sqrt {-1+x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {i F\left (i x \sqrt {2}, \frac {i \sqrt {2}}{2}\right ) \sqrt {-x^{2}+1}}{2 \sqrt {x^{2}-1}}\)	\(34\)
elliptic	\(-\frac {i \sqrt {\left (x^{2}-1\right ) \left (2 x^{2}+1\right )}\, \sqrt {2}\, \sqrt {-x^{2}+1}\, F\left (i x \sqrt {2}, \frac {i \sqrt {2}}{2}\right )}{2 \sqrt {x^{2}-1}\, \sqrt {4 x^{4}-2 x^{2}-2}}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt {-1+x^2} \sqrt {2+4 x^2}} \, dx=-\frac {1}{2} \, \sqrt {-2} F(\arcsin \left (x\right )\,|\,-2) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

sqrt(2)*Integral(1/(sqrt(x**2 - 1)*sqrt(2*x**2 + 1)), x)/2

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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