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

Optimal. Leaf size=155 \[ \frac{\sqrt{-\left (2-\sqrt{10}\right ) x^2-3} \sqrt{\frac{\left (2+\sqrt{10}\right ) x^2+3}{\left (2-\sqrt{10}\right ) x^2+3}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{2^{3/4} \sqrt [4]{5} x}{\sqrt{-\left (2-\sqrt{10}\right ) x^2-3}}\right ),\frac{1}{10} \left (5-\sqrt{10}\right )\right )}{2^{3/4} \sqrt{3} \sqrt [4]{5} \sqrt{\frac{1}{\left (2-\sqrt{10}\right ) x^2+3}} \sqrt{2 x^4-4 x^2-3}} \]

[Out]

(Sqrt[-3 - (2 - Sqrt[10])*x^2]*Sqrt[(3 + (2 + Sqrt[10])*x^2)/(3 + (2 - Sqrt[10])*x^2)]*EllipticF[ArcSin[(2^(3/
4)*5^(1/4)*x)/Sqrt[-3 - (2 - Sqrt[10])*x^2]], (5 - Sqrt[10])/10])/(2^(3/4)*Sqrt[3]*5^(1/4)*Sqrt[(3 + (2 - Sqrt
[10])*x^2)^(-1)]*Sqrt[-3 - 4*x^2 + 2*x^4])

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Rubi [A]  time = 0.0236497, antiderivative size = 155, 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} \[ \frac{\sqrt{-\left (2-\sqrt{10}\right ) x^2-3} \sqrt{\frac{\left (2+\sqrt{10}\right ) x^2+3}{\left (2-\sqrt{10}\right ) x^2+3}} F\left (\sin ^{-1}\left (\frac{2^{3/4} \sqrt [4]{5} x}{\sqrt{-\left (2-\sqrt{10}\right ) x^2-3}}\right )|\frac{1}{10} \left (5-\sqrt{10}\right )\right )}{2^{3/4} \sqrt{3} \sqrt [4]{5} \sqrt{\frac{1}{\left (2-\sqrt{10}\right ) x^2+3}} \sqrt{2 x^4-4 x^2-3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 - 4*x^2 + 2*x^4],x]

[Out]

(Sqrt[-3 - (2 - Sqrt[10])*x^2]*Sqrt[(3 + (2 + Sqrt[10])*x^2)/(3 + (2 - Sqrt[10])*x^2)]*EllipticF[ArcSin[(2^(3/
4)*5^(1/4)*x)/Sqrt[-3 - (2 - Sqrt[10])*x^2]], (5 - Sqrt[10])/10])/(2^(3/4)*Sqrt[3]*5^(1/4)*Sqrt[(3 + (2 - Sqrt
[10])*x^2)^(-1)]*Sqrt[-3 - 4*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{-3-4 x^2+2 x^4}} \, dx &=\frac{\sqrt{-3-\left (2-\sqrt{10}\right ) x^2} \sqrt{\frac{3+\left (2+\sqrt{10}\right ) x^2}{3+\left (2-\sqrt{10}\right ) x^2}} F\left (\sin ^{-1}\left (\frac{2^{3/4} \sqrt [4]{5} x}{\sqrt{-3-\left (2-\sqrt{10}\right ) x^2}}\right )|\frac{1}{10} \left (5-\sqrt{10}\right )\right )}{2^{3/4} \sqrt{3} \sqrt [4]{5} \sqrt{\frac{1}{3+\left (2-\sqrt{10}\right ) x^2}} \sqrt{-3-4 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0652713, size = 83, normalized size = 0.54 \[ -\frac{i \sqrt{-2 x^4+4 x^2+3} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{2}{\sqrt{10}-2}} x\right ),\frac{2 \sqrt{10}}{3}-\frac{7}{3}\right )}{\sqrt{2+\sqrt{10}} \sqrt{2 x^4-4 x^2-3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*Sqrt[3 + 4*x^2 - 2*x^4]*EllipticF[I*ArcSinh[Sqrt[2/(-2 + Sqrt[10])]*x], -7/3 + (2*Sqrt[10])/3])/(Sqrt[2
+ Sqrt[10]]*Sqrt[-3 - 4*x^2 + 2*x^4])

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Maple [C]  time = 0.183, size = 84, normalized size = 0.5 \begin{align*} 3\,{\frac{\sqrt{1- \left ( -2/3-1/3\,\sqrt{10} \right ){x}^{2}}\sqrt{1- \left ( -2/3+1/3\,\sqrt{10} \right ){x}^{2}}{\it EllipticF} \left ( 1/3\,\sqrt{-6-3\,\sqrt{10}}x,i/3\sqrt{15}-i/3\sqrt{6} \right ) }{\sqrt{-6-3\,\sqrt{10}}\sqrt{2\,{x}^{4}-4\,{x}^{2}-3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(-6-3*10^(1/2))^(1/2)*(1-(-2/3-1/3*10^(1/2))*x^2)^(1/2)*(1-(-2/3+1/3*10^(1/2))*x^2)^(1/2)/(2*x^4-4*x^2-3)^(1
/2)*EllipticF(1/3*(-6-3*10^(1/2))^(1/2)*x,1/3*I*15^(1/2)-1/3*I*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} - 4 \, x^{2} - 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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