3.1.84 \(\int \frac {1}{\sqrt {3+3 x^2+2 x^4}} \, dx\) [84]

Optimal. Leaf size=92 \[ \frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3+3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3+3 x^2+2 x^4}} \]

[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 = 92, 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 = {1117} \begin {gather*} \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 \text {ArcTan}\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}} \end {gather*}

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 1117

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)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], 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] Result contains complex when optimal does not.
time = 10.08, size = 142, normalized size = 1.54 \begin {gather*} -\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 {3+3 x^2+2 x^4}} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.05, size = 87, normalized size = 0.95

method	result	size
default	\(\frac {6 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{\sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}}\)	\(87\)
elliptic	\(\frac {6 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{\sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4+3*x^2+3)^(1/2),x,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [A]
time = 0.08, size = 51, normalized size = 0.55 \begin {gather*} -\frac {1}{24} \, \sqrt {6} \sqrt {\sqrt {3} \sqrt {-5} - 3} {\left (\sqrt {3} + \sqrt {-5}\right )} {\rm ellipticF}\left (\frac {1}{6} \, \sqrt {6} \sqrt {\sqrt {3} \sqrt {-5} - 3} x, \frac {1}{4} \, \sqrt {3} \sqrt {-5} - \frac {1}{4}\right ) \end {gather*}

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]

-1/24*sqrt(6)*sqrt(sqrt(3)*sqrt(-5) - 3)*(sqrt(3) + sqrt(-5))*ellipticF(1/6*sqrt(6)*sqrt(sqrt(3)*sqrt(-5) - 3)
*x, 1/4*sqrt(3)*sqrt(-5) - 1/4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {2 x^{4} + 3 x^{2} + 3}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2 + 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(3*x^2 + 2*x^4 + 3)^(1/2), x)

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