3.984 \(\int \frac{x^2}{\sqrt{3-b x^4}} \, dx\)

Optimal. Leaf size=54 \[ \frac{\sqrt [4]{3} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{3}}\right )\right |-1\right )}{b^{3/4}}-\frac{\sqrt [4]{3} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{3}}\right ),-1\right )}{b^{3/4}} \]

[Out]

(3^(1/4)*EllipticE[ArcSin[(b^(1/4)*x)/3^(1/4)], -1])/b^(3/4) - (3^(1/4)*EllipticF[ArcSin[(b^(1/4)*x)/3^(1/4)],
-1])/b^(3/4)

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Rubi [A]  time = 0.0467568, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {307, 221, 1199, 424} \[ \frac{\sqrt [4]{3} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{3}}\right )\right |-1\right )}{b^{3/4}}-\frac{\sqrt [4]{3} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{3}}\right )\right |-1\right )}{b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[3 - b*x^4],x]

[Out]

(3^(1/4)*EllipticE[ArcSin[(b^(1/4)*x)/3^(1/4)], -1])/b^(3/4) - (3^(1/4)*EllipticF[ArcSin[(b^(1/4)*x)/3^(1/4)],
-1])/b^(3/4)

Rule 307

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 424

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

Rubi steps

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

Mathematica [C]  time = 0.0060864, size = 30, normalized size = 0.56 \[ \frac{x^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{b x^4}{3}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[3 - b*x^4],x]

[Out]

(x^3*Hypergeometric2F1[1/2, 3/4, 7/4, (b*x^4)/3])/(3*Sqrt[3])

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Maple [B]  time = 0.01, size = 94, normalized size = 1.7 \begin{align*} -{\frac{1}{3}\sqrt{9-3\,\sqrt{3}\sqrt{b}{x}^{2}}\sqrt{9+3\,\sqrt{3}\sqrt{b}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{3}}{3}\sqrt{\sqrt{3}\sqrt{b}}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{3}}{3}\sqrt{\sqrt{3}\sqrt{b}}},i \right ) \right ){\frac{1}{\sqrt{\sqrt{3}\sqrt{b}}}}{\frac{1}{\sqrt{-b{x}^{4}+3}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/(3^(1/2)*b^(1/2))^(1/2)*(9-3*3^(1/2)*b^(1/2)*x^2)^(1/2)*(9+3*3^(1/2)*b^(1/2)*x^2)^(1/2)/(-b*x^4+3)^(1/2)/
b^(1/2)*(EllipticF(1/3*x*3^(1/2)*(3^(1/2)*b^(1/2))^(1/2),I)-EllipticE(1/3*x*3^(1/2)*(3^(1/2)*b^(1/2))^(1/2),I)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(-b*x^4 + 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.839415, size = 39, normalized size = 0.72 \begin{align*} \frac{\sqrt{3} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{3}} \right )}}{12 \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(3)*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(2*I*pi)/3)/(12*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(-b*x^4 + 3), x)