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

Optimal. Leaf size=52 \[ \frac {\left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}} \]

[Out]

1/2*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*I*2^(1/2))*((3*x^2+2)/(x^2+1))^(1/2)*2^(1/2)
/(3*x^4+5*x^2+2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 52, 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 = {1100} \[ \frac {\left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[2 + 5*x^2 + 3*x^4],x]

[Out]

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

Rule 1100

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

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 58, normalized size = 1.12 \[ -\frac {i \sqrt {x^2+1} \sqrt {3 x^2+2} F\left (i \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )}{\sqrt {9 x^4+15 x^2+6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/Sqrt[6 + 15*x^2 + 9*x^4]

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^4+5*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

integral(1/sqrt(3*x^4 + 5*x^2 + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^4+5*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(3*x^4 + 5*x^2 + 2), x)

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maple [A]  time = 0.02, size = 44, normalized size = 0.85 \[ -\frac {i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \EllipticF \left (i x , \frac {\sqrt {6}}{2}\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^4+5*x^2+2)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^4+5*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(3*x^4 + 5*x^2 + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {3\,x^4+5\,x^2+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^2 + 3*x^4 + 2)^(1/2),x)

[Out]

int(1/(5*x^2 + 3*x^4 + 2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {3 x^{4} + 5 x^{2} + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x**4+5*x**2+2)**(1/2),x)

[Out]

Integral(1/sqrt(3*x**4 + 5*x**2 + 2), x)

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