3.29 \(\int \frac {1}{\sqrt {3+5 x^2-2 x^4}} \, dx\)

Optimal. Leaf size=10 \[ F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {3}}\right )\right |-6\right ) \]

[Out]

EllipticF(1/3*x*3^(1/2),I*6^(1/2))

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1095, 419} \[ F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {3}}\right )\right |-6\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[3 + 5*x^2 - 2*x^4],x]

[Out]

EllipticF[ArcSin[x/Sqrt[3]], -6]

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 1095

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[2*Sqrt[-c], I
nt[1/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0] &&
LtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {3+5 x^2-2 x^4}} \, dx &=\left (2 \sqrt {2}\right ) \int \frac {1}{\sqrt {12-4 x^2} \sqrt {2+4 x^2}} \, dx\\ &=F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {3}}\right )\right |-6\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 65, normalized size = 6.50 \[ -\frac {i \sqrt {1-\frac {x^2}{3}} \sqrt {2 x^2+1} F\left (i \sinh ^{-1}\left (\sqrt {2} x\right )|-\frac {1}{6}\right )}{\sqrt {2} \sqrt {-2 x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[3 + 5*x^2 - 2*x^4],x]

[Out]

((-I)*Sqrt[1 - x^2/3]*Sqrt[1 + 2*x^2]*EllipticF[I*ArcSinh[Sqrt[2]*x], -1/6])/(Sqrt[2]*Sqrt[3 + 5*x^2 - 2*x^4])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-2 \, x^{4} + 5 \, x^{2} + 3}}{2 \, x^{4} - 5 \, x^{2} - 3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x^4+5*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-2*x^4 + 5*x^2 + 3)/(2*x^4 - 5*x^2 - 3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-2 \, x^{4} + 5 \, x^{2} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x^4+5*x^2+3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(-2*x^4 + 5*x^2 + 3), x)

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maple [B]  time = 0.03, size = 51, normalized size = 5.10 \[ \frac {\sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {2 x^{2}+1}\, \EllipticF \left (\frac {\sqrt {3}\, x}{3}, i \sqrt {6}\right )}{3 \sqrt {-2 x^{4}+5 x^{2}+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-2*x^4+5*x^2+3)^(1/2),x)

[Out]

1/3*3^(1/2)*(-3*x^2+9)^(1/2)*(2*x^2+1)^(1/2)/(-2*x^4+5*x^2+3)^(1/2)*EllipticF(1/3*3^(1/2)*x,I*6^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-2 \, x^{4} + 5 \, x^{2} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x^4+5*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(-2*x^4 + 5*x^2 + 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.10 \[ \int \frac {1}{\sqrt {-2\,x^4+5\,x^2+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^2 - 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(5*x^2 - 2*x^4 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- 2 x^{4} + 5 x^{2} + 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2*x**4+5*x**2+3)**(1/2),x)

[Out]

Integral(1/sqrt(-2*x**4 + 5*x**2 + 3), x)

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