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

Optimal. Leaf size=14 \[ -F\left (\left .\cos ^{-1}\left (\sqrt {\frac {2}{3}} x\right )\right |3\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1095, 420} \[ -F\left (\left .\cos ^{-1}\left (\sqrt {\frac {2}{3}} x\right )\right |3\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 + 5*x^2 - 2*x^4],x]

[Out]

-EllipticF[ArcCos[Sqrt[2/3]*x], 3]

Rule 420

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> -Simp[EllipticF[ArcCos[Rt[-(d/c), 2]
*x], (b*c)/(b*c - a*d)]/(Sqrt[c]*Rt[-(d/c), 2]*Sqrt[a - (b*c)/d]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &
& GtQ[c, 0] && GtQ[a - (b*c)/d, 0]

Rule 1095

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

Rubi steps

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

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Mathematica [B]  time = 0.02, size = 53, normalized size = 3.79 \[ \frac {\sqrt {3-2 x^2} \sqrt {1-x^2} F\left (\sin ^{-1}\left (\sqrt {\frac {2}{3}} x\right )|\frac {3}{2}\right )}{\sqrt {-4 x^4+10 x^2-6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-2 \, x^{4} + 5 \, x^{2} - 3}}{2 \, x^{4} - 5 \, x^{2} + 3}, x\right ) \]

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(-sqrt(-2*x^4 + 5*x^2 - 3)/(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 [A]  time = 0.01, size = 50, normalized size = 3.57 \[ \frac {\sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \EllipticF \left (\frac {\sqrt {6}\, x}{3}, \frac {\sqrt {6}}{2}\right )}{6 \sqrt {-2 x^{4}+5 x^{2}-3}} \]

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*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*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 = -0.07 \[ \int \frac {1}{\sqrt {-2\,x^4+5\,x^2-3}} \,d x \]

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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