\(\int \frac {1}{\sqrt {2-x^2} \sqrt {1+x^2}} \, dx\) [233]

Optimal result

Integrand size = 21, antiderivative size = 10 \[ \int \frac {1}{\sqrt {2-x^2} \sqrt {1+x^2}} \, dx=\operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {2}}\right ),-2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {430} \[ \int \frac {1}{\sqrt {2-x^2} \sqrt {1+x^2}} \, dx=\operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {2}}\right ),-2\right ) \]

[In]

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

[Out]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\sqrt {2-x^2} \sqrt {1+x^2}} \, dx=-\frac {i \operatorname {EllipticF}\left (i \text {arcsinh}(x),-\frac {1}{2}\right )}{\sqrt {2}} \]

[In]

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

[Out]

((-I)*EllipticF[I*ArcSinh[x], -1/2])/Sqrt[2]

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.40

method	result	size
default	\(F\left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\)	\(14\)
elliptic	\(\frac {\sqrt {-\left (x^{2}-2\right ) \left (x^{2}+1\right )}\, \sqrt {2}\, \sqrt {-2 x^{2}+4}\, F\left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{2 \sqrt {-x^{2}+2}\, \sqrt {-x^{4}+x^{2}+2}}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {2-x^2} \sqrt {1+x^2}} \, dx=F(\arcsin \left (\frac {1}{2} \, \sqrt {2} x\right )\,|\,-2) \]

[In]

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

[Out]

elliptic_f(arcsin(1/2*sqrt(2)*x), -2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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