3.1.56 \(\int \frac {1}{\sqrt {-3+4 x^2+2 x^4}} \, dx\) [56]

Optimal. Leaf size=148 \[ \frac {\sqrt {\frac {3-\left (2-\sqrt {10}\right ) x^2}{3-\left (2+\sqrt {10}\right ) x^2}} \sqrt {-3+\left (2+\sqrt {10}\right ) x^2} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-3+\left (2+\sqrt {10}\right ) x^2}}\right )|\frac {1}{10} \left (5+\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (2+\sqrt {10}\right ) x^2}} \sqrt {-3+4 x^2+2 x^4}} \]

[Out]

1/30*EllipticF(2^(3/4)*5^(1/4)*x/(-3+x^2*(2+10^(1/2)))^(1/2),1/10*(50+10*10^(1/2))^(1/2))*((3-x^2*(2-10^(1/2))
)/(3-x^2*(2+10^(1/2))))^(1/2)*(-3+x^2*(2+10^(1/2)))^(1/2)*2^(1/4)*5^(3/4)*3^(1/2)/(2*x^4+4*x^2-3)^(1/2)/(1/(3-
x^2*(2+10^(1/2))))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 148, 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 {\frac {3-\left (2-\sqrt {10}\right ) x^2}{3-\left (2+\sqrt {10}\right ) x^2}} \sqrt {\left (2+\sqrt {10}\right ) x^2-3} F\left (\text {ArcSin}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {\left (2+\sqrt {10}\right ) x^2-3}}\right )|\frac {1}{10} \left (5+\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (2+\sqrt {10}\right ) x^2}} \sqrt {2 x^4+4 x^2-3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 + 4*x^2 + 2*x^4],x]

[Out]

(Sqrt[(3 - (2 - Sqrt[10])*x^2)/(3 - (2 + Sqrt[10])*x^2)]*Sqrt[-3 + (2 + Sqrt[10])*x^2]*EllipticF[ArcSin[(2^(3/
4)*5^(1/4)*x)/Sqrt[-3 + (2 + Sqrt[10])*x^2]], (5 + Sqrt[10])/10])/(2^(3/4)*Sqrt[3]*5^(1/4)*Sqrt[(3 - (2 + Sqrt
[10])*x^2)^(-1)]*Sqrt[-3 + 4*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+4 x^2+2 x^4}} \, dx &=\frac {\sqrt {\frac {3-\left (2-\sqrt {10}\right ) x^2}{3-\left (2+\sqrt {10}\right ) x^2}} \sqrt {-3+\left (2+\sqrt {10}\right ) x^2} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-3+\left (2+\sqrt {10}\right ) x^2}}\right )|\frac {1}{10} \left (5+\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (2+\sqrt {10}\right ) x^2}} \sqrt {-3+4 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.05, size = 83, normalized size = 0.56 \begin {gather*} -\frac {i \sqrt {3-4 x^2-2 x^4} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{2+\sqrt {10}}} x\right )|-\frac {7}{3}-\frac {2 \sqrt {10}}{3}\right )}{\sqrt {-2+\sqrt {10}} \sqrt {-3+4 x^2+2 x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*Sqrt[3 - 4*x^2 - 2*x^4]*EllipticF[I*ArcSinh[Sqrt[2/(2 + Sqrt[10])]*x], -7/3 - (2*Sqrt[10])/3])/(Sqrt[-2
+ Sqrt[10]]*Sqrt[-3 + 4*x^2 + 2*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.05, size = 84, normalized size = 0.57

method	result	size
default	\(\frac {3 \sqrt {1-\left (\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \EllipticF \left (\frac {\sqrt {6-3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )}{\sqrt {6-3 \sqrt {10}}\, \sqrt {2 x^{4}+4 x^{2}-3}}\)	\(84\)
elliptic	\(\frac {3 \sqrt {1-\left (\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \EllipticF \left (\frac {\sqrt {6-3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )}{\sqrt {6-3 \sqrt {10}}\, \sqrt {2 x^{4}+4 x^{2}-3}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(6-3*10^(1/2))^(1/2)*(1-(2/3-1/3*10^(1/2))*x^2)^(1/2)*(1-(2/3+1/3*10^(1/2))*x^2)^(1/2)/(2*x^4+4*x^2-3)^(1/2)
*EllipticF(1/3*(6-3*10^(1/2))^(1/2)*x,1/3*I*6^(1/2)+1/3*I*15^(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+4*x^2-3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 + 4*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+4*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} + 4 x^{2} - 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/sqrt(2*x^4 + 4*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+4\,x^2-3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x^2 + 2*x^4 - 3)^(1/2),x)

[Out]

int(1/(4*x^2 + 2*x^4 - 3)^(1/2), x)

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