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

Optimal. Leaf size=222 \[ \frac{4 \sqrt [4]{2} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac{1}{2}\right )}{15 \sqrt{3} x}+\frac{2}{15} \left (3 x^2-2\right )^{3/4} x+\frac{8 \sqrt [4]{3 x^2-2} x}{15 \left (\sqrt{3 x^2-2}+\sqrt{2}\right )}-\frac{8 \sqrt [4]{2} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{15 \sqrt{3} x} \]

[Out]

(2*x*(-2 + 3*x^2)^(3/4))/15 + (8*x*(-2 + 3*x^2)^(1/4))/(15*(Sqrt[2] + Sqrt[-2 + 3*x^2])) - (8*2^(1/4)*Sqrt[x^2
/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*EllipticE[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)],
1/2])/(15*Sqrt[3]*x) + (4*2^(1/4)*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*Ellipt
icF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(15*Sqrt[3]*x)

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Rubi [A]  time = 0.090511, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {321, 230, 305, 220, 1196} \[ \frac{2}{15} \left (3 x^2-2\right )^{3/4} x+\frac{8 \sqrt [4]{3 x^2-2} x}{15 \left (\sqrt{3 x^2-2}+\sqrt{2}\right )}+\frac{4 \sqrt [4]{2} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{15 \sqrt{3} x}-\frac{8 \sqrt [4]{2} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{15 \sqrt{3} x} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(-2 + 3*x^2)^(1/4),x]

[Out]

(2*x*(-2 + 3*x^2)^(3/4))/15 + (8*x*(-2 + 3*x^2)^(1/4))/(15*(Sqrt[2] + Sqrt[-2 + 3*x^2])) - (8*2^(1/4)*Sqrt[x^2
/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*EllipticE[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)],
1/2])/(15*Sqrt[3]*x) + (4*2^(1/4)*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*Ellipt
icF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(15*Sqrt[3]*x)

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 230

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

Rule 305

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

Rule 220

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt [4]{-2+3 x^2}} \, dx &=\frac{2}{15} x \left (-2+3 x^2\right )^{3/4}+\frac{4}{15} \int \frac{1}{\sqrt [4]{-2+3 x^2}} \, dx\\ &=\frac{2}{15} x \left (-2+3 x^2\right )^{3/4}+\frac{\left (4 \sqrt{\frac{2}{3}} \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+\frac{x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{15 x}\\ &=\frac{2}{15} x \left (-2+3 x^2\right )^{3/4}+\frac{\left (8 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{15 \sqrt{3} x}-\frac{\left (8 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{x^2}{\sqrt{2}}}{\sqrt{1+\frac{x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{15 \sqrt{3} x}\\ &=\frac{2}{15} x \left (-2+3 x^2\right )^{3/4}+\frac{8 x \sqrt [4]{-2+3 x^2}}{15 \left (\sqrt{2}+\sqrt{-2+3 x^2}\right )}-\frac{8 \sqrt [4]{2} \sqrt{\frac{x^2}{\left (\sqrt{2}+\sqrt{-2+3 x^2}\right )^2}} \left (\sqrt{2}+\sqrt{-2+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{15 \sqrt{3} x}+\frac{4 \sqrt [4]{2} \sqrt{\frac{x^2}{\left (\sqrt{2}+\sqrt{-2+3 x^2}\right )^2}} \left (\sqrt{2}+\sqrt{-2+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{15 \sqrt{3} x}\\ \end{align*}

Mathematica [C]  time = 0.0113824, size = 57, normalized size = 0.26 \[ \frac{2 x \left (2^{3/4} \sqrt [4]{2-3 x^2} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{3 x^2}{2}\right )+3 x^2-2\right )}{15 \sqrt [4]{3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(-2 + 3*x^2)^(1/4),x]

[Out]

(2*x*(-2 + 3*x^2 + 2^(3/4)*(2 - 3*x^2)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (3*x^2)/2]))/(15*(-2 + 3*x^2)^(1
/4))

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Maple [C]  time = 0.035, size = 53, normalized size = 0.2 \begin{align*}{\frac{2\,x}{15} \left ( 3\,{x}^{2}-2 \right ) ^{{\frac{3}{4}}}}+{\frac{2\,{2}^{3/4}x}{15}\sqrt [4]{-{\it signum} \left ( -1+{\frac{3\,{x}^{2}}{2}} \right ) }{\mbox{$_2$F$_1$}({\frac{1}{4}},{\frac{1}{2}};\,{\frac{3}{2}};\,{\frac{3\,{x}^{2}}{2}})}{\frac{1}{\sqrt [4]{{\it signum} \left ( -1+{\frac{3\,{x}^{2}}{2}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*x*(3*x^2-2)^(3/4)+2/15*2^(3/4)/signum(-1+3/2*x^2)^(1/4)*(-signum(-1+3/2*x^2))^(1/4)*x*hypergeom([1/4,1/2]
,[3/2],3/2*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(3*x^2 - 2)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(3*x^2 - 2)^(1/4), x)

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Sympy [C]  time = 0.620308, size = 29, normalized size = 0.13 \begin{align*} \frac{2^{\frac{3}{4}} x^{3} e^{- \frac{i \pi }{4}}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{3 x^{2}}{2}} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(3/4)*x**3*exp(-I*pi/4)*hyper((1/4, 3/2), (5/2,), 3*x**2/2)/6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(3*x^2 - 2)^(1/4), x)