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

Optimal. Leaf size=52 \[ -\frac{\sqrt{-3 x^2-2} \text{EllipticF}\left (\tan ^{-1}(x),-\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^2+1} \sqrt{\frac{3 x^2+2}{x^2+1}}} \]

[Out]

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

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Rubi [A]  time = 0.0167469, antiderivative size = 52, normalized size of antiderivative = 1., 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, 418} \[ -\frac{\sqrt{-3 x^2-2} F\left (\tan ^{-1}(x)|-\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^2+1} \sqrt{\frac{3 x^2+2}{x^2+1}}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-2 - 5*x^2 - 3*x^4],x]

[Out]

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

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]

Rule 418

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

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

Mathematica [C]  time = 0.024037, size = 63, normalized size = 1.21 \[ -\frac{i \sqrt{x^2+1} \sqrt{3 x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right ),\frac{2}{3}\right )}{\sqrt{3} \sqrt{-3 x^4-5 x^2-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 50, normalized size = 1. \begin{align*}{-{\frac{i}{6}}\sqrt{6}\sqrt{6\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{6},{\frac{\sqrt{6}}{3}} \right ){\frac{1}{\sqrt{-3\,{x}^{4}-5\,{x}^{2}-2}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-3 \, x^{4} - 5 \, x^{2} - 2}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-3 \, x^{4} - 5 \, x^{2} - 2}}{3 \, x^{4} + 5 \, x^{2} + 2}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- 3 x^{4} - 5 x^{2} - 2}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-3 \, x^{4} - 5 \, x^{2} - 2}}\,{d x} \end{align*}

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)