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

Optimal. Leaf size=242 \[ -\frac{3 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac{1}{2}\right )}{8\ 2^{3/4} x}-\frac{9 \sqrt [4]{3 x^2-2} x}{8 \left (\sqrt{3 x^2-2}+\sqrt{2}\right )}+\frac{3 \left (3 x^2-2\right )^{3/4}}{8 x}+\frac{\left (3 x^2-2\right )^{3/4}}{6 x^3}+\frac{3 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{4\ 2^{3/4} x} \]

[Out]

(-2 + 3*x^2)^(3/4)/(6*x^3) + (3*(-2 + 3*x^2)^(3/4))/(8*x) - (9*x*(-2 + 3*x^2)^(1/4))/(8*(Sqrt[2] + Sqrt[-2 + 3
*x^2])) + (3*Sqrt[3]*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])/(4*2^(3/4)*x) - (3*Sqrt[3]*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2
] + Sqrt[-2 + 3*x^2])*EllipticF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(8*2^(3/4)*x)

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Rubi [A]  time = 0.109254, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {325, 230, 305, 220, 1196} \[ -\frac{9 \sqrt [4]{3 x^2-2} x}{8 \left (\sqrt{3 x^2-2}+\sqrt{2}\right )}+\frac{3 \left (3 x^2-2\right )^{3/4}}{8 x}+\frac{\left (3 x^2-2\right )^{3/4}}{6 x^3}-\frac{3 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{8\ 2^{3/4} x}+\frac{3 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{4\ 2^{3/4} x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(-2 + 3*x^2)^(1/4)),x]

[Out]

(-2 + 3*x^2)^(3/4)/(6*x^3) + (3*(-2 + 3*x^2)^(3/4))/(8*x) - (9*x*(-2 + 3*x^2)^(1/4))/(8*(Sqrt[2] + Sqrt[-2 + 3
*x^2])) + (3*Sqrt[3]*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])/(4*2^(3/4)*x) - (3*Sqrt[3]*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2
] + Sqrt[-2 + 3*x^2])*EllipticF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(8*2^(3/4)*x)

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx &=\frac{\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac{3}{4} \int \frac{1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx\\ &=\frac{\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac{3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac{9}{16} \int \frac{1}{\sqrt [4]{-2+3 x^2}} \, dx\\ &=\frac{\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac{3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac{\left (3 \sqrt{\frac{3}{2}} \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 )}{8 x}\\ &=\frac{\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac{3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac{\left (3 \sqrt{3} \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 )}{8 x}+\frac{\left (3 \sqrt{3} \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 )}{8 x}\\ &=\frac{\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac{3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac{9 x \sqrt [4]{-2+3 x^2}}{8 \left (\sqrt{2}+\sqrt{-2+3 x^2}\right )}+\frac{3 \sqrt{3} \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 )}{4\ 2^{3/4} x}-\frac{3 \sqrt{3} \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 )}{8\ 2^{3/4} x}\\ \end{align*}

Mathematica [C]  time = 0.0072733, size = 48, normalized size = 0.2 \[ -\frac{\sqrt [4]{1-\frac{3 x^2}{2}} \, _2F_1\left (-\frac{3}{2},\frac{1}{4};-\frac{1}{2};\frac{3 x^2}{2}\right )}{3 x^3 \sqrt [4]{3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.037, size = 67, normalized size = 0.3 \begin{align*}{\frac{27\,{x}^{4}-6\,{x}^{2}-8}{24\,{x}^{3}}{\frac{1}{\sqrt [4]{3\,{x}^{2}-2}}}}-{\frac{9\,{2}^{3/4}x}{32}\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(1/x^4/(3*x^2-2)^(1/4),x)

[Out]

1/24*(27*x^4-6*x^2-8)/x^3/(3*x^2-2)^(1/4)-9/32*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{1}{{\left (3 \, x^{2} - 2\right )}^{\frac{1}{4}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3 \, x^{2} - 2\right )}^{\frac{3}{4}}}{3 \, x^{6} - 2 \, x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^2 - 2)^(3/4)/(3*x^6 - 2*x^4), x)

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Sympy [C]  time = 0.82427, size = 34, normalized size = 0.14 \begin{align*} \frac{2^{\frac{3}{4}} e^{\frac{3 i \pi }{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{1}{4} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{3 x^{2}}{2}} \right )}}{6 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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