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

Optimal. Leaf size=141 \[ \frac{\sqrt{\frac{2-\left (1-\sqrt{7}\right ) x^2}{2-\left (1+\sqrt{7}\right ) x^2}} \sqrt{\left (1+\sqrt{7}\right ) x^2-2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{7} x}{\sqrt{\left (1+\sqrt{7}\right ) x^2-2}}\right ),\frac{1}{14} \left (7+\sqrt{7}\right )\right )}{2 \sqrt [4]{7} \sqrt{\frac{1}{2-\left (1+\sqrt{7}\right ) x^2}} \sqrt{3 x^4+2 x^2-2}} \]

[Out]

(Sqrt[(2 - (1 - Sqrt[7])*x^2)/(2 - (1 + Sqrt[7])*x^2)]*Sqrt[-2 + (1 + Sqrt[7])*x^2]*EllipticF[ArcSin[(Sqrt[2]*
7^(1/4)*x)/Sqrt[-2 + (1 + Sqrt[7])*x^2]], (7 + Sqrt[7])/14])/(2*7^(1/4)*Sqrt[(2 - (1 + Sqrt[7])*x^2)^(-1)]*Sqr
t[-2 + 2*x^2 + 3*x^4])

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Rubi [A]  time = 0.0302293, antiderivative size = 141, normalized size of antiderivative = 1., 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 = {1098} \[ \frac{\sqrt{\frac{2-\left (1-\sqrt{7}\right ) x^2}{2-\left (1+\sqrt{7}\right ) x^2}} \sqrt{\left (1+\sqrt{7}\right ) x^2-2} F\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{7} x}{\sqrt{\left (1+\sqrt{7}\right ) x^2-2}}\right )|\frac{1}{14} \left (7+\sqrt{7}\right )\right )}{2 \sqrt [4]{7} \sqrt{\frac{1}{2-\left (1+\sqrt{7}\right ) x^2}} \sqrt{3 x^4+2 x^2-2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-2 + 2*x^2 + 3*x^4],x]

[Out]

(Sqrt[(2 - (1 - Sqrt[7])*x^2)/(2 - (1 + Sqrt[7])*x^2)]*Sqrt[-2 + (1 + Sqrt[7])*x^2]*EllipticF[ArcSin[(Sqrt[2]*
7^(1/4)*x)/Sqrt[-2 + (1 + Sqrt[7])*x^2]], (7 + Sqrt[7])/14])/(2*7^(1/4)*Sqrt[(2 - (1 + Sqrt[7])*x^2)^(-1)]*Sqr
t[-2 + 2*x^2 + 3*x^4])

Rule 1098

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]*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q
)]], (b + q)/(2*q)])/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)*x^2)]), 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{-2+2 x^2+3 x^4}} \, dx &=\frac{\sqrt{\frac{2-\left (1-\sqrt{7}\right ) x^2}{2-\left (1+\sqrt{7}\right ) x^2}} \sqrt{-2+\left (1+\sqrt{7}\right ) x^2} F\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{7} x}{\sqrt{-2+\left (1+\sqrt{7}\right ) x^2}}\right )|\frac{1}{14} \left (7+\sqrt{7}\right )\right )}{2 \sqrt [4]{7} \sqrt{\frac{1}{2-\left (1+\sqrt{7}\right ) x^2}} \sqrt{-2+2 x^2+3 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0515857, size = 83, normalized size = 0.59 \[ -\frac{i \sqrt{-3 x^4-2 x^2+2} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{3}{1+\sqrt{7}}} x\right ),-\frac{4}{3}-\frac{\sqrt{7}}{3}\right )}{\sqrt{\sqrt{7}-1} \sqrt{3 x^4+2 x^2-2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.178, size = 84, normalized size = 0.6 \begin{align*} 2\,{\frac{\sqrt{1- \left ( -1/2\,\sqrt{7}+1/2 \right ){x}^{2}}\sqrt{1- \left ( 1/2\,\sqrt{7}+1/2 \right ){x}^{2}}{\it EllipticF} \left ( 1/2\,\sqrt{2-2\,\sqrt{7}}x,i/6\sqrt{6}+i/6\sqrt{42} \right ) }{\sqrt{2-2\,\sqrt{7}}\sqrt{3\,{x}^{4}+2\,{x}^{2}-2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(2-2*7^(1/2))^(1/2)*(1-(-1/2*7^(1/2)+1/2)*x^2)^(1/2)*(1-(1/2*7^(1/2)+1/2)*x^2)^(1/2)/(3*x^4+2*x^2-2)^(1/2)*E
llipticF(1/2*(2-2*7^(1/2))^(1/2)*x,1/6*I*6^(1/2)+1/6*I*42^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{3 \, x^{4} + 2 \, x^{2} - 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{3 \, x^{4} + 2 \, x^{2} - 2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{3 \, x^{4} + 2 \, x^{2} - 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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