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

Optimal. Leaf size=58 \[ \frac{\sqrt{\frac{x^2+2}{2 x^2+1}} \left (2 x^2+1\right ) \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{2} x\right ),\frac{3}{4}\right )}{2 \sqrt{2 x^4+5 x^2+2}} \]

[Out]

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

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Rubi [A]  time = 0.0071399, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {1099} \[ \frac{\sqrt{\frac{x^2+2}{2 x^2+1}} \left (2 x^2+1\right ) F\left (\tan ^{-1}\left (\sqrt{2} x\right )|\frac{3}{4}\right )}{2 \sqrt{2 x^4+5 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[2 + 5*x^2 + 2*x^4],x]

[Out]

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

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

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

Mathematica [C]  time = 0.0224686, size = 58, normalized size = 1. \[ -\frac{i \sqrt{x^2+2} \sqrt{2 x^2+1} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x\right ),\frac{1}{4}\right )}{2 \sqrt{2 x^4+5 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.053, size = 48, normalized size = 0.8 \begin{align*}{-{\frac{i}{2}}\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},2 \right ) \sqrt{2\,{x}^{2}+4}\sqrt{2\,{x}^{2}+1}{\frac{1}{\sqrt{2\,{x}^{4}+5\,{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(2*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{1}{\sqrt{2 \, x^{4} + 5 \, x^{2} + 2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(2*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{2 x^{4} + 5 x^{2} + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(2*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{2 \, x^{4} + 5 \, x^{2} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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