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

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

[Out]

1/3*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/3*3^(1/2))*((2*x^2+3)/(x^2+1))^(1/2)*3^(1/2)/(
2*x^4+5*x^2+3)^(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 = {1099} \[ \frac {\left (x^2+1\right ) \sqrt {\frac {2 x^2+3}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{3}\right )}{\sqrt {3} \sqrt {2 x^4+5 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[3 + 5*x^2 + 2*x^4],x]

[Out]

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

Rule 1099

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] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 58, normalized size = 1.12 \[ -\frac {i \sqrt {x^2+1} \sqrt {2 x^2+3} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{3}} x\right )|\frac {3}{2}\right )}{\sqrt {4 x^4+10 x^2+6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + x^2]*Sqrt[3 + 2*x^2]*EllipticF[I*ArcSinh[Sqrt[2/3]*x], 3/2])/Sqrt[6 + 10*x^2 + 4*x^4]

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {2 \, x^{4} + 5 \, x^{2} + 3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 50, normalized size = 0.96 \[ -\frac {i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {6}\, x}{3}, \frac {\sqrt {6}}{2}\right )}{6 \sqrt {2 x^{4}+5 x^{2}+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*I*6^(1/2)*(6*x^2+9)^(1/2)*(x^2+1)^(1/2)/(2*x^4+5*x^2+3)^(1/2)*EllipticF(1/3*I*6^(1/2)*x,1/2*6^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^2 + 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(5*x^2 + 2*x^4 + 3)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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