3.80 \(\int \frac {1}{\sqrt {3+7 x^2+2 x^4}} \, dx\)

Optimal. Leaf size=60 \[ \frac {\sqrt {\frac {x^2+3}{2 x^2+1}} \left (2 x^2+1\right ) F\left (\tan ^{-1}\left (\sqrt {2} x\right )|\frac {5}{6}\right )}{\sqrt {6} \sqrt {2 x^4+7 x^2+3}} \]

[Out]

1/6*(2*x^2+1)^(3/2)*(1/(2*x^2+1))^(1/2)*EllipticF(x*2^(1/2)/(2*x^2+1)^(1/2),1/6*30^(1/2))*((x^2+3)/(2*x^2+1))^
(1/2)*6^(1/2)/(2*x^4+7*x^2+3)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 60, 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} \[ \frac {\sqrt {\frac {x^2+3}{2 x^2+1}} \left (2 x^2+1\right ) F\left (\tan ^{-1}\left (\sqrt {2} x\right )|\frac {5}{6}\right )}{\sqrt {6} \sqrt {2 x^4+7 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[3 + 7*x^2 + 2*x^4],x]

[Out]

(Sqrt[(3 + x^2)/(1 + 2*x^2)]*(1 + 2*x^2)*EllipticF[ArcTan[Sqrt[2]*x], 5/6])/(Sqrt[6]*Sqrt[3 + 7*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+7 x^2+2 x^4}} \, dx &=\frac {\sqrt {\frac {3+x^2}{1+2 x^2}} \left (1+2 x^2\right ) F\left (\tan ^{-1}\left (\sqrt {2} x\right )|\frac {5}{6}\right )}{\sqrt {6} \sqrt {3+7 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 1.02 \[ -\frac {i \sqrt {x^2+3} \sqrt {2 x^2+1} F\left (i \sinh ^{-1}\left (\sqrt {2} x\right )|\frac {1}{6}\right )}{\sqrt {6} \sqrt {2 x^4+7 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[3 + 7*x^2 + 2*x^4],x]

[Out]

((-I)*Sqrt[3 + x^2]*Sqrt[1 + 2*x^2]*EllipticF[I*ArcSinh[Sqrt[2]*x], 1/6])/(Sqrt[6]*Sqrt[3 + 7*x^2 + 2*x^4])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {2 \, x^{4} + 7 \, x^{2} + 3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^4+7*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

integral(1/sqrt(2*x^4 + 7*x^2 + 3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} + 7 \, x^{2} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^4+7*x^2+3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(2*x^4 + 7*x^2 + 3), x)

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maple [C]  time = 0.02, size = 50, normalized size = 0.83 \[ -\frac {i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {2 x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {3}\, x}{3}, \sqrt {6}\right )}{3 \sqrt {2 x^{4}+7 x^{2}+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4+7*x^2+3)^(1/2),x)

[Out]

-1/3*I*3^(1/2)*(3*x^2+9)^(1/2)*(2*x^2+1)^(1/2)/(2*x^4+7*x^2+3)^(1/2)*EllipticF(1/3*I*3^(1/2)*x,6^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} + 7 \, x^{2} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^4+7*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 + 7*x^2 + 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {2\,x^4+7\,x^2+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(7*x^2 + 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(7*x^2 + 2*x^4 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 x^{4} + 7 x^{2} + 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x**4+7*x**2+3)**(1/2),x)

[Out]

Integral(1/sqrt(2*x**4 + 7*x**2 + 3), x)

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