3.1.63 \(\int \frac {1}{\sqrt {-3-3 x^2+2 x^4}} \, dx\) [63]

Optimal. Leaf size=153 \[ \frac {\sqrt {-6-\left (3-\sqrt {33}\right ) x^2} \sqrt {\frac {6+\left (3+\sqrt {33}\right ) x^2}{6+\left (3-\sqrt {33}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{33} x}{\sqrt {-6-\left (3-\sqrt {33}\right ) x^2}}\right )|\frac {1}{22} \left (11-\sqrt {33}\right )\right )}{2\ 3^{3/4} \sqrt [4]{11} \sqrt {\frac {1}{6+\left (3-\sqrt {33}\right ) x^2}} \sqrt {-3-3 x^2+2 x^4}} \]

[Out]

1/66*EllipticF(33^(1/4)*x*2^(1/2)/(-6-x^2*(3-33^(1/2)))^(1/2),1/22*(242-22*33^(1/2))^(1/2))*(-6-x^2*(3-33^(1/2
)))^(1/2)*((6+x^2*(3+33^(1/2)))/(6+x^2*(3-33^(1/2))))^(1/2)*3^(1/4)*11^(3/4)/(2*x^4-3*x^2-3)^(1/2)/(1/(6+x^2*(
3-33^(1/2))))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 153, 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 = {1112} \begin {gather*} \frac {\sqrt {-\left (\left (3-\sqrt {33}\right ) x^2\right )-6} \sqrt {\frac {\left (3+\sqrt {33}\right ) x^2+6}{\left (3-\sqrt {33}\right ) x^2+6}} F\left (\text {ArcSin}\left (\frac {\sqrt {2} \sqrt [4]{33} x}{\sqrt {-\left (\left (3-\sqrt {33}\right ) x^2\right )-6}}\right )|\frac {1}{22} \left (11-\sqrt {33}\right )\right )}{2\ 3^{3/4} \sqrt [4]{11} \sqrt {\frac {1}{\left (3-\sqrt {33}\right ) x^2+6}} \sqrt {2 x^4-3 x^2-3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 - 3*x^2 + 2*x^4],x]

[Out]

(Sqrt[-6 - (3 - Sqrt[33])*x^2]*Sqrt[(6 + (3 + Sqrt[33])*x^2)/(6 + (3 - Sqrt[33])*x^2)]*EllipticF[ArcSin[(Sqrt[
2]*33^(1/4)*x)/Sqrt[-6 - (3 - Sqrt[33])*x^2]], (11 - Sqrt[33])/22])/(2*3^(3/4)*11^(1/4)*Sqrt[(6 + (3 - Sqrt[33
])*x^2)^(-1)]*Sqrt[-3 - 3*x^2 + 2*x^4])

Rule 1112

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)/(2*a + (b + q)*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)
*x^2)]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x]] /; FreeQ[{a, b, c}, x] && Gt
Q[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-3-3 x^2+2 x^4}} \, dx &=\frac {\sqrt {-6-\left (3-\sqrt {33}\right ) x^2} \sqrt {\frac {6+\left (3+\sqrt {33}\right ) x^2}{6+\left (3-\sqrt {33}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{33} x}{\sqrt {-6-\left (3-\sqrt {33}\right ) x^2}}\right )|\frac {1}{22} \left (11-\sqrt {33}\right )\right )}{2\ 3^{3/4} \sqrt [4]{11} \sqrt {\frac {1}{6+\left (3-\sqrt {33}\right ) x^2}} \sqrt {-3-3 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.04, size = 78, normalized size = 0.51 \begin {gather*} -\frac {i \sqrt {6+6 x^2-4 x^4} F\left (i \sinh ^{-1}\left (\frac {2 x}{\sqrt {-3+\sqrt {33}}}\right )|\frac {1}{4} \left (-7+\sqrt {33}\right )\right )}{\sqrt {3+\sqrt {33}} \sqrt {-3-3 x^2+2 x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*Sqrt[6 + 6*x^2 - 4*x^4]*EllipticF[I*ArcSinh[(2*x)/Sqrt[-3 + Sqrt[33]]], (-7 + Sqrt[33])/4])/(Sqrt[3 + Sq
rt[33]]*Sqrt[-3 - 3*x^2 + 2*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.04, size = 84, normalized size = 0.55

method	result	size
default	\(\frac {6 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {33}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}+\frac {\sqrt {33}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {\sqrt {-18-6 \sqrt {33}}\, x}{6}, \frac {i \sqrt {22}}{4}-\frac {i \sqrt {6}}{4}\right )}{\sqrt {-18-6 \sqrt {33}}\, \sqrt {2 x^{4}-3 x^{2}-3}}\)	\(84\)
elliptic	\(\frac {6 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {33}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}+\frac {\sqrt {33}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {\sqrt {-18-6 \sqrt {33}}\, x}{6}, \frac {i \sqrt {22}}{4}-\frac {i \sqrt {6}}{4}\right )}{\sqrt {-18-6 \sqrt {33}}\, \sqrt {2 x^{4}-3 x^{2}-3}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4-3*x^2-3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

6/(-18-6*33^(1/2))^(1/2)*(1-(-1/2-1/6*33^(1/2))*x^2)^(1/2)*(1-(-1/2+1/6*33^(1/2))*x^2)^(1/2)/(2*x^4-3*x^2-3)^(
1/2)*EllipticF(1/6*(-18-6*33^(1/2))^(1/2)*x,1/4*I*22^(1/2)-1/4*I*6^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {2 x^{4} - 3 x^{2} - 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x**4-3*x**2-3)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {2\,x^4-3\,x^2-3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4 - 3*x^2 - 3)^(1/2),x)

[Out]

int(1/(2*x^4 - 3*x^2 - 3)^(1/2), x)

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