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

Optimal. Leaf size=92 \[ \frac{\left (\sqrt{6} x^2+3\right ) \sqrt{\frac{2 x^4-5 x^2+3}{\left (\sqrt{6} x^2+3\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{\frac{2}{3}} x\right ),\frac{1}{24} \left (12+5 \sqrt{6}\right )\right )}{2 \sqrt [4]{6} \sqrt{2 x^4-5 x^2+3}} \]

[Out]

((3 + Sqrt[6]*x^2)*Sqrt[(3 - 5*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(2/3)^(1/4)*x], (12 + 5*Sq
rt[6])/24])/(2*6^(1/4)*Sqrt[3 - 5*x^2 + 2*x^4])

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Rubi [A]  time = 0.0091571, antiderivative size = 92, 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 = {1096} \[ \frac{\left (\sqrt{6} x^2+3\right ) \sqrt{\frac{2 x^4-5 x^2+3}{\left (\sqrt{6} x^2+3\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac{2}{3}} x\right )|\frac{1}{24} \left (12+5 \sqrt{6}\right )\right )}{2 \sqrt [4]{6} \sqrt{2 x^4-5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[3 - 5*x^2 + 2*x^4],x]

[Out]

((3 + Sqrt[6]*x^2)*Sqrt[(3 - 5*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(2/3)^(1/4)*x], (12 + 5*Sq
rt[6])/24])/(2*6^(1/4)*Sqrt[3 - 5*x^2 + 2*x^4])

Rule 1096

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{3-5 x^2+2 x^4}} \, dx &=\frac{\left (3+\sqrt{6} x^2\right ) \sqrt{\frac{3-5 x^2+2 x^4}{\left (3+\sqrt{6} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac{2}{3}} x\right )|\frac{1}{24} \left (12+5 \sqrt{6}\right )\right )}{2 \sqrt [4]{6} \sqrt{3-5 x^2+2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0257186, size = 53, normalized size = 0.58 \[ \frac{\sqrt{3-2 x^2} \sqrt{1-x^2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{2}{3}} x\right ),\frac{3}{2}\right )}{\sqrt{4 x^4-10 x^2+6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 - 2*x^2]*Sqrt[1 - x^2]*EllipticF[ArcSin[Sqrt[2/3]*x], 3/2])/Sqrt[6 - 10*x^2 + 4*x^4]

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Maple [A]  time = 0.048, size = 42, normalized size = 0.5 \begin{align*}{\frac{1}{3}\sqrt{-{x}^{2}+1}\sqrt{-6\,{x}^{2}+9}{\it EllipticF} \left ( x,{\frac{\sqrt{6}}{3}} \right ){\frac{1}{\sqrt{2\,{x}^{4}-5\,{x}^{2}+3}}}} \end{align*}

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

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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} + 3}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{2 \, x^{4} - 5 \, x^{2} + 3}}, x\right ) \end{align*}

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(1/sqrt(2*x^4 - 5*x^2 + 3), 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} + 3}}\, dx \end{align*}

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

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)