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

Optimal. Leaf size=146 \[ \frac{\sqrt{\frac{6-\left (3-\sqrt{33}\right ) x^2}{6-\left (3+\sqrt{33}\right ) x^2}} \sqrt{\left (3+\sqrt{33}\right ) x^2-6} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{33} x}{\sqrt{\left (3+\sqrt{33}\right ) x^2-6}}\right ),\frac{1}{22} \left (11+\sqrt{33}\right )\right )}{2\ 3^{3/4} \sqrt [4]{11} \sqrt{\frac{1}{6-\left (3+\sqrt{33}\right ) x^2}} \sqrt{2 x^4+3 x^2-3}} \]

[Out]

(Sqrt[(6 - (3 - Sqrt[33])*x^2)/(6 - (3 + Sqrt[33])*x^2)]*Sqrt[-6 + (3 + Sqrt[33])*x^2]*EllipticF[ArcSin[(Sqrt[
2]*33^(1/4)*x)/Sqrt[-6 + (3 + Sqrt[33])*x^2]], (11 + Sqrt[33])/22])/(2*3^(3/4)*11^(1/4)*Sqrt[(6 - (3 + Sqrt[33
])*x^2)^(-1)]*Sqrt[-3 + 3*x^2 + 2*x^4])

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Rubi [A]  time = 0.0266953, antiderivative size = 146, 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 = {1098} \[ \frac{\sqrt{\frac{6-\left (3-\sqrt{33}\right ) x^2}{6-\left (3+\sqrt{33}\right ) x^2}} \sqrt{\left (3+\sqrt{33}\right ) x^2-6} F\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{33} x}{\sqrt{\left (3+\sqrt{33}\right ) x^2-6}}\right )|\frac{1}{22} \left (11+\sqrt{33}\right )\right )}{2\ 3^{3/4} \sqrt [4]{11} \sqrt{\frac{1}{6-\left (3+\sqrt{33}\right ) x^2}} \sqrt{2 x^4+3 x^2-3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 + 3*x^2 + 2*x^4],x]

[Out]

(Sqrt[(6 - (3 - Sqrt[33])*x^2)/(6 - (3 + Sqrt[33])*x^2)]*Sqrt[-6 + (3 + Sqrt[33])*x^2]*EllipticF[ArcSin[(Sqrt[
2]*33^(1/4)*x)/Sqrt[-6 + (3 + Sqrt[33])*x^2]], (11 + Sqrt[33])/22])/(2*3^(3/4)*11^(1/4)*Sqrt[(6 - (3 + Sqrt[33
])*x^2)^(-1)]*Sqrt[-3 + 3*x^2 + 2*x^4])

Rule 1098

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-3+3 x^2+2 x^4}} \, dx &=\frac{\sqrt{\frac{6-\left (3-\sqrt{33}\right ) x^2}{6-\left (3+\sqrt{33}\right ) x^2}} \sqrt{-6+\left (3+\sqrt{33}\right ) x^2} F\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{33} x}{\sqrt{-6+\left (3+\sqrt{33}\right ) x^2}}\right )|\frac{1}{22} \left (11+\sqrt{33}\right )\right )}{2\ 3^{3/4} \sqrt [4]{11} \sqrt{\frac{1}{6-\left (3+\sqrt{33}\right ) x^2}} \sqrt{-3+3 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0584732, size = 80, normalized size = 0.55 \[ -\frac{i \sqrt{-4 x^4-6 x^2+6} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{2 x}{\sqrt{3+\sqrt{33}}}\right ),-\frac{7}{4}-\frac{\sqrt{33}}{4}\right )}{\sqrt{\sqrt{33}-3} \sqrt{2 x^4+3 x^2-3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*Sqrt[6 - 6*x^2 - 4*x^4]*EllipticF[I*ArcSinh[(2*x)/Sqrt[3 + Sqrt[33]]], -7/4 - Sqrt[33]/4])/(Sqrt[-3 + Sq
rt[33]]*Sqrt[-3 + 3*x^2 + 2*x^4])

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Maple [C]  time = 0.188, size = 84, normalized size = 0.6 \begin{align*} 6\,{\frac{\sqrt{1- \left ( -1/6\,\sqrt{33}+1/2 \right ){x}^{2}}\sqrt{1- \left ( 1/6\,\sqrt{33}+1/2 \right ){x}^{2}}{\it EllipticF} \left ( 1/6\,\sqrt{18-6\,\sqrt{33}}x,i/4\sqrt{6}+i/4\sqrt{22} \right ) }{\sqrt{18-6\,\sqrt{33}}\sqrt{2\,{x}^{4}+3\,{x}^{2}-3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/(18-6*33^(1/2))^(1/2)*(1-(-1/6*33^(1/2)+1/2)*x^2)^(1/2)*(1-(1/6*33^(1/2)+1/2)*x^2)^(1/2)/(2*x^4+3*x^2-3)^(1/
2)*EllipticF(1/6*(18-6*33^(1/2))^(1/2)*x,1/4*I*6^(1/2)+1/4*I*22^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(2*x**4 + 3*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} + 3 \, x^{2} - 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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