3.320 \(\int \sqrt {2+x^2-x^4} \, dx\)

Optimal. Leaf size=44 \[ \frac {1}{3} \sqrt {-x^4+x^2+2} x+F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {1}{3} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1091, 1180, 524, 424, 419} \[ \frac {1}{3} \sqrt {-x^4+x^2+2} x+F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {1}{3} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + x^2 - x^4],x]

[Out]

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

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)])

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 90, normalized size = 2.05 \[ \frac {-x^5+x^3-3 i \sqrt {-2 x^4+2 x^2+4} F\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+i \sqrt {-2 x^4+2 x^2+4} E\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+2 x}{3 \sqrt {-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + x^2 - x^4],x]

[Out]

(2*x + x^3 - x^5 + I*Sqrt[4 + 2*x^2 - 2*x^4]*EllipticE[I*ArcSinh[x], -1/2] - (3*I)*Sqrt[4 + 2*x^2 - 2*x^4]*Ell
ipticF[I*ArcSinh[x], -1/2])/(3*Sqrt[2 + x^2 - x^4])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-x^{4} + x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-x^4 + x^2 + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-x^{4} + x^{2} + 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-x^4 + x^2 + 2), x)

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maple [B]  time = 0.00, size = 125, normalized size = 2.84 \[ \frac {\sqrt {-x^{4}+x^{2}+2}\, x}{3}+\frac {2 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{3 \sqrt {-x^{4}+x^{2}+2}}-\frac {\sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )+\EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )\right )}{6 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-x^4+x^2+2)^(1/2)*x+2/3*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticF(1/2*2^(1/2)*x
,I*2^(1/2))-1/6*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*(EllipticF(1/2*2^(1/2)*x,I*2^(1/2))-
EllipticE(1/2*2^(1/2)*x,I*2^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-x^{4} + x^{2} + 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-x^4 + x^2 + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {-x^4+x^2+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - x^4 + 2)^(1/2),x)

[Out]

int((x^2 - x^4 + 2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- x^{4} + x^{2} + 2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-x**4 + x**2 + 2), x)

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