3.260 \(\int \sqrt{(1-x^2) (3+x^2)} \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{3} \sqrt{-x^4-2 x^2+3} x+\frac{4 F\left (\sin ^{-1}(x)|-\frac{1}{3}\right )}{\sqrt{3}}-\frac{2 E\left (\sin ^{-1}(x)|-\frac{1}{3}\right )}{\sqrt{3}} \]

[Out]

(x*Sqrt[3 - 2*x^2 - x^4])/3 - (2*EllipticE[ArcSin[x], -1/3])/Sqrt[3] + (4*EllipticF[ArcSin[x], -1/3])/Sqrt[3]

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Rubi [A]  time = 0.0428554, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {1988, 1091, 1180, 524, 424, 419} \[ \frac{1}{3} \sqrt{-x^4-2 x^2+3} x+\frac{4 F\left (\sin ^{-1}(x)|-\frac{1}{3}\right )}{\sqrt{3}}-\frac{2 E\left (\sin ^{-1}(x)|-\frac{1}{3}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 - x^2)*(3 + x^2)],x]

[Out]

(x*Sqrt[3 - 2*x^2 - x^4])/3 - (2*EllipticE[ArcSin[x], -1/3])/Sqrt[3] + (4*EllipticF[ArcSin[x], -1/3])/Sqrt[3]

Rule 1988

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && TrinomialQ[u, x] &&  !TrinomialMatch
Q[u, x]

Rule 1091

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a + b*x^2 + c*x^4)^p)/(4*p + 1), x] + Dis
t[(2*p)/(4*p + 1), Int[(2*a + b*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4
*a*c, 0] && GtQ[p, 0] && IntegerQ[2*p]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[2*Sqrt[-c], Int[(d + e*x^2)/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c,
d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 424

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

Rule 419

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

Rubi steps

\begin{align*} \int \sqrt{\left (1-x^2\right ) \left (3+x^2\right )} \, dx &=\int \sqrt{3-2 x^2-x^4} \, dx\\ &=\frac{1}{3} x \sqrt{3-2 x^2-x^4}+\frac{1}{3} \int \frac{6-2 x^2}{\sqrt{3-2 x^2-x^4}} \, dx\\ &=\frac{1}{3} x \sqrt{3-2 x^2-x^4}+\frac{2}{3} \int \frac{6-2 x^2}{\sqrt{2-2 x^2} \sqrt{6+2 x^2}} \, dx\\ &=\frac{1}{3} x \sqrt{3-2 x^2-x^4}-\frac{2}{3} \int \frac{\sqrt{6+2 x^2}}{\sqrt{2-2 x^2}} \, dx+8 \int \frac{1}{\sqrt{2-2 x^2} \sqrt{6+2 x^2}} \, dx\\ &=\frac{1}{3} x \sqrt{3-2 x^2-x^4}-\frac{2 E\left (\sin ^{-1}(x)|-\frac{1}{3}\right )}{\sqrt{3}}+\frac{4 F\left (\sin ^{-1}(x)|-\frac{1}{3}\right )}{\sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.0571395, size = 59, normalized size = 1.23 \[ \frac{1}{3} \left (\sqrt{-x^4-2 x^2+3} x-4 i F\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{3}}\right )\right |-3\right )-2 i E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{3}}\right )\right |-3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 - x^2)*(3 + x^2)],x]

[Out]

(x*Sqrt[3 - 2*x^2 - x^4] - (2*I)*EllipticE[I*ArcSinh[x/Sqrt[3]], -3] - (4*I)*EllipticF[I*ArcSinh[x/Sqrt[3]], -
3])/3

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Maple [B]  time = 0.017, size = 114, normalized size = 2.4 \begin{align*}{\frac{x}{3}\sqrt{-{x}^{4}-2\,{x}^{2}+3}}+{\frac{2\,{\it EllipticF} \left ( x,i/3\sqrt{3} \right ) }{3}\sqrt{-{x}^{2}+1}\sqrt{3\,{x}^{2}+9}{\frac{1}{\sqrt{-{x}^{4}-2\,{x}^{2}+3}}}}+{\frac{2\,{\it EllipticF} \left ( x,i/3\sqrt{3} \right ) -2\,{\it EllipticE} \left ( x,i/3\sqrt{3} \right ) }{3}\sqrt{-{x}^{2}+1}\sqrt{3\,{x}^{2}+9}{\frac{1}{\sqrt{-{x}^{4}-2\,{x}^{2}+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+1)*(x^2+3))^(1/2),x)

[Out]

1/3*x*(-x^4-2*x^2+3)^(1/2)+2/3*(-x^2+1)^(1/2)*(3*x^2+9)^(1/2)/(-x^4-2*x^2+3)^(1/2)*EllipticF(x,1/3*I*3^(1/2))+
2/3*(-x^2+1)^(1/2)*(3*x^2+9)^(1/2)/(-x^4-2*x^2+3)^(1/2)*(EllipticF(x,1/3*I*3^(1/2))-EllipticE(x,1/3*I*3^(1/2))
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-{\left (x^{2} + 3\right )}{\left (x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+1)*(x^2+3))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-(x^2 + 3)*(x^2 - 1)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-x^{4} - 2 \, x^{2} + 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+1)*(x^2+3))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-x^4 - 2*x^2 + 3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (1 - x^{2}\right ) \left (x^{2} + 3\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+1)*(x**2+3))**(1/2),x)

[Out]

Integral(sqrt((1 - x**2)*(x**2 + 3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-{\left (x^{2} + 3\right )}{\left (x^{2} - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+1)*(x^2+3))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-(x^2 + 3)*(x^2 - 1)), x)