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

Optimal. Leaf size=90 \[ \frac {\left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2 x^4-3 x^2+3}} \]

[Out]

1/12*(cos(2*arctan(1/3*2^(1/4)*3^(3/4)*x))^2)^(1/2)/cos(2*arctan(1/3*2^(1/4)*3^(3/4)*x))*EllipticF(sin(2*arcta
n(1/3*2^(1/4)*3^(3/4)*x)),1/4*(8+2*6^(1/2))^(1/2))*(3+x^2*6^(1/2))*((2*x^4-3*x^2+3)/(3+x^2*6^(1/2))^2)^(1/2)*6
^(3/4)/(2*x^4-3*x^2+3)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 = {1103} \[ \frac {\left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2 x^4-3 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[3 - 3*x^2 + 2*x^4],x]

[Out]

((3 + Sqrt[6]*x^2)*Sqrt[(3 - 3*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(2/3)^(1/4)*x], (4 + Sqrt[
6])/8])/(2*6^(1/4)*Sqrt[3 - 3*x^2 + 2*x^4])

Rule 1103

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] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx &=\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-3 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}{8} \left (4+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-3 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 142, normalized size = 1.58 \[ -\frac {i \sqrt {1-\frac {4 x^2}{3-i \sqrt {15}}} \sqrt {1-\frac {4 x^2}{3+i \sqrt {15}}} F\left (i \sinh ^{-1}\left (2 \sqrt {-\frac {1}{3-i \sqrt {15}}} x\right )|\frac {3-i \sqrt {15}}{3+i \sqrt {15}}\right )}{2 \sqrt {-\frac {1}{3-i \sqrt {15}}} \sqrt {2 x^4-3 x^2+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*Sqrt[1 - (4*x^2)/(3 - I*Sqrt[15])]*Sqrt[1 - (4*x^2)/(3 + I*Sqrt[15])]*EllipticF[I*ArcSinh[2*Sqrt[-(3
- I*Sqrt[15])^(-1)]*x], (3 - I*Sqrt[15])/(3 + I*Sqrt[15])])/(Sqrt[-(3 - I*Sqrt[15])^(-1)]*Sqrt[3 - 3*x^2 + 2*
x^4])

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {2 \, x^{4} - 3 \, x^{2} + 3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} - 3 \, x^{2} + 3}}\,{d x} \]

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)

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maple [C]  time = 0.10, size = 87, normalized size = 0.97 \[ \frac {6 \sqrt {-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}+1}\, \sqrt {-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {18+6 i \sqrt {15}}\, x}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )}{\sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}} \]

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*I*15^(1/2))^(1/2)*(1-(1/2+1/6*I*15^(1/2))*x^2)^(1/2)*(1-(1/2-1/6*I*15^(1/2))*x^2)^(1/2)/(2*x^4-3*x^2+3
)^(1/2)*EllipticF(1/6*x*(18+6*I*15^(1/2))^(1/2),1/2*(-1-I*15^(1/2))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} - 3 \, x^{2} + 3}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {2\,x^4-3\,x^2+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4 - 3*x^2 + 3)^(1/2),x)

[Out]

int(1/(2*x^4 - 3*x^2 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 x^{4} - 3 x^{2} + 3}}\, dx \]

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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