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

Optimal. Leaf size=67 \[ \frac {\sqrt {x^2+3} \sqrt {2 x^2-1} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {7}{3}} x}{\sqrt {2 x^2-1}}\right )|\frac {6}{7}\right )}{\sqrt {7} \sqrt {2 x^4+5 x^2-3}} \]

[Out]

1/7*EllipticF(1/3*x*21^(1/2)/(2*x^2-1)^(1/2),1/7*42^(1/2))*(x^2+3)^(1/2)*(2*x^2-1)^(1/2)*7^(1/2)/(2*x^4+5*x^2-
3)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 67, 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 = {1097} \[ \frac {\sqrt {x^2+3} \sqrt {2 x^2-1} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {7}{3}} x}{\sqrt {2 x^2-1}}\right )|\frac {6}{7}\right )}{\sqrt {7} \sqrt {2 x^4+5 x^2-3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 + 5*x^2 + 2*x^4],x]

[Out]

(Sqrt[3 + x^2]*Sqrt[-1 + 2*x^2]*EllipticF[ArcSin[(Sqrt[7/3]*x)/Sqrt[-1 + 2*x^2]], 6/7])/(Sqrt[7]*Sqrt[-3 + 5*x
^2 + 2*x^4])

Rule 1097

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-3+5 x^2+2 x^4}} \, dx &=\frac {\sqrt {3+x^2} \sqrt {-1+2 x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {7}{3}} x}{\sqrt {-1+2 x^2}}\right )|\frac {6}{7}\right )}{\sqrt {7} \sqrt {-3+5 x^2+2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 54, normalized size = 0.81 \[ \frac {\sqrt {1-2 x^2} \sqrt {x^2+3} F\left (\sin ^{-1}\left (\sqrt {2} x\right )|-\frac {1}{6}\right )}{\sqrt {6} \sqrt {2 x^4+5 x^2-3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x^2]*Sqrt[3 + x^2]*EllipticF[ArcSin[Sqrt[2]*x], -1/6])/(Sqrt[6]*Sqrt[-3 + 5*x^2 + 2*x^4])

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fricas [F]  time = 0.79, 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 = 0.01, size = 53, normalized size = 0.79 \[ -\frac {i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {3}\, x}{3}, i \sqrt {6}\right )}{3 \sqrt {2 x^{4}+5 x^{2}-3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*3^(1/2)*(3*x^2+9)^(1/2)*(-2*x^2+1)^(1/2)/(2*x^4+5*x^2-3)^(1/2)*EllipticF(1/3*I*3^(1/2)*x,I*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 = -0.01 \[ \int \frac {1}{\sqrt {2\,x^4+5\,x^2-3}} \,d x \]

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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