3.933 \(\int (1-x^2) \sqrt {\frac {1}{2-x^2}} \, dx\)

Optimal. Leaf size=18 \[ \frac {x}{2 \sqrt {\frac {1}{2-x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 18, 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 = {6720, 383} \[ \frac {x}{2 \sqrt {\frac {1}{2-x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x^2)*Sqrt[(2 - x^2)^(-1)],x]

[Out]

x/(2*Sqrt[(2 - x^2)^(-1)])

Rule 383

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*x*(a + b*x^n)^(p + 1))/a, x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a*d - b*c*(n*(p + 1) + 1), 0]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ \frac {x}{2 \sqrt {\frac {1}{2-x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(2*Sqrt[(2 - x^2)^(-1)])

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fricas [A]  time = 0.79, size = 20, normalized size = 1.11 \[ -\frac {1}{2} \, {\left (x^{3} - 2 \, x\right )} \sqrt {-\frac {1}{x^{2} - 2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(x^3 - 2*x)*sqrt(-1/(x^2 - 2))

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giac [A]  time = 0.48, size = 18, normalized size = 1.00 \[ -\frac {1}{2} \, \sqrt {-x^{2} + 2} x \mathrm {sgn}\left (x^{2} - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-x^2 + 2)*x*sgn(x^2 - 2)

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maple [A]  time = 0.01, size = 20, normalized size = 1.11 \[ -\frac {\left (x^{2}-2\right ) \sqrt {-\frac {1}{x^{2}-2}}\, x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int {\left (x^{2} - 1\right )} \sqrt {-\frac {1}{x^{2} - 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.50, size = 19, normalized size = 1.06 \[ -\frac {x\,\left (x^2-2\right )\,\sqrt {-\frac {1}{x^2-2}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x*(x^2 - 2)*(-1/(x^2 - 2))^(1/2))/2

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sympy [B]  time = 0.50, size = 26, normalized size = 1.44 \[ - \frac {x^{3} \sqrt {\frac {1}{2 - x^{2}}}}{2} + x \sqrt {\frac {1}{2 - x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3*sqrt(1/(2 - x**2))/2 + x*sqrt(1/(2 - x**2))

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