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

Optimal. Leaf size=18 \[ \frac {F\left (\sin ^{-1}\left (\sqrt {3} x\right )|-\frac {1}{6}\right )}{\sqrt {6}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 18, 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} \[ \frac {F\left (\sin ^{-1}\left (\sqrt {3} x\right )|-\frac {1}{6}\right )}{\sqrt {6}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[2 - 5*x^2 - 3*x^4],x]

[Out]

EllipticF[ArcSin[Sqrt[3]*x], -1/6]/Sqrt[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 {2-5 x^2-3 x^4}} \, dx &=\left (2 \sqrt {3}\right ) \int \frac {1}{\sqrt {2-6 x^2} \sqrt {12+6 x^2}} \, dx\\ &=\frac {F\left (\sin ^{-1}\left (\sqrt {3} x\right )|-\frac {1}{6}\right )}{\sqrt {6}}\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 54, normalized size = 3.00 \[ \frac {\sqrt {1-3 x^2} \sqrt {x^2+2} F\left (\sin ^{-1}\left (\sqrt {3} x\right )|-\frac {1}{6}\right )}{\sqrt {6} \sqrt {-3 x^4-5 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-3 \, x^{4} - 5 \, x^{2} + 2}}{3 \, x^{4} + 5 \, x^{2} - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 50, normalized size = 2.78 \[ \frac {\sqrt {3}\, \sqrt {-3 x^{2}+1}\, \sqrt {2 x^{2}+4}\, \EllipticF \left (\sqrt {3}\, x , \frac {i \sqrt {6}}{6}\right )}{6 \sqrt {-3 x^{4}-5 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int \frac {1}{\sqrt {-3\,x^4-5\,x^2+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2 - 3*x^4 - 5*x^2)^(1/2),x)

[Out]

int(1/(2 - 3*x^4 - 5*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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