3.1049 \(\int \frac{x^2}{(-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\)

Optimal. Leaf size=224 \[ \frac{\sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right ),\frac{1}{2}\right )}{3 \sqrt{3} x}+\frac{2 \sqrt [4]{3 x^2-1} x}{3 \left (\sqrt{3 x^2-1}+1\right )}-\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{3 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{3 \sqrt{6}}-\frac{2 \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac{1}{2}\right )}{3 \sqrt{3} x} \]

[Out]

(2*x*(-1 + 3*x^2)^(1/4))/(3*(1 + Sqrt[-1 + 3*x^2])) - ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(3*Sqrt[6]) - A
rcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(3*Sqrt[6]) - (2*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*
x^2])*EllipticE[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(3*Sqrt[3]*x) + (Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + S
qrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(3*Sqrt[3]*x)

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Rubi [A]  time = 0.104839, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {440, 230, 305, 220, 1196, 398} \[ \frac{2 \sqrt [4]{3 x^2-1} x}{3 \left (\sqrt{3 x^2-1}+1\right )}-\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{3 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{3 \sqrt{6}}+\frac{\sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac{1}{2}\right )}{3 \sqrt{3} x}-\frac{2 \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac{1}{2}\right )}{3 \sqrt{3} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-1 + 3*x^2)^(1/4))/(3*(1 + Sqrt[-1 + 3*x^2])) - ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(3*Sqrt[6]) - A
rcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(3*Sqrt[6]) - (2*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*
x^2])*EllipticE[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(3*Sqrt[3]*x) + (Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + S
qrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(3*Sqrt[3]*x)

Rule 440

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(1/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

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]

Rule 398

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b^2/a), 4]}, Simp[(b*Ar
cTan[(q*x)/(Sqrt[2]*(a + b*x^2)^(1/4))])/(2*Sqrt[2]*a*d*q), x] + Simp[(b*ArcTanh[(q*x)/(Sqrt[2]*(a + b*x^2)^(1
/4))])/(2*Sqrt[2]*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && NegQ[b^2/a]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx &=\int \left (\frac{1}{3 \sqrt [4]{-1+3 x^2}}+\frac{2}{3 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}}\right ) \, dx\\ &=\frac{1}{3} \int \frac{1}{\sqrt [4]{-1+3 x^2}} \, dx+\frac{2}{3} \int \frac{1}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{3 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{3 \sqrt{6}}+\frac{\left (2 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{3 \sqrt{3} x}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{3 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{3 \sqrt{6}}+\frac{\left (2 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{3 \sqrt{3} x}-\frac{\left (2 \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{\sqrt{1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{3 \sqrt{3} x}\\ &=\frac{2 x \sqrt [4]{-1+3 x^2}}{3 \left (1+\sqrt{-1+3 x^2}\right )}-\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{3 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{3 \sqrt{6}}-\frac{2 \sqrt{\frac{x^2}{\left (1+\sqrt{-1+3 x^2}\right )^2}} \left (1+\sqrt{-1+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac{1}{2}\right )}{3 \sqrt{3} x}+\frac{\sqrt{\frac{x^2}{\left (1+\sqrt{-1+3 x^2}\right )^2}} \left (1+\sqrt{-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac{1}{2}\right )}{3 \sqrt{3} x}\\ \end{align*}

Mathematica [C]  time = 0.0203964, size = 52, normalized size = 0.23 \[ -\frac{x^3 \sqrt [4]{1-3 x^2} F_1\left (\frac{3}{2};\frac{1}{4},1;\frac{5}{2};3 x^2,\frac{3 x^2}{2}\right )}{6 \sqrt [4]{3 x^2-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(x^3*(1 - 3*x^2)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, 3*x^2, (3*x^2)/2])/(6*(-1 + 3*x^2)^(1/4))

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{3\,{x}^{2}-2}{\frac{1}{\sqrt [4]{3\,{x}^{2}-1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3 \, x^{2} - 1\right )}^{\frac{3}{4}} x^{2}}{9 \, x^{4} - 9 \, x^{2} + 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^2 - 1)^(3/4)*x^2/(9*x^4 - 9*x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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