3.911 \(\int \frac{1}{x^6 (-2+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=140 \[ \frac{27 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac{1}{2}\right )}{64 \sqrt [4]{2} x}+\frac{27 \sqrt [4]{3 x^2-2}}{32 x}+\frac{9 \sqrt [4]{3 x^2-2}}{40 x^3}+\frac{\sqrt [4]{3 x^2-2}}{10 x^5} \]

[Out]

(-2 + 3*x^2)^(1/4)/(10*x^5) + (9*(-2 + 3*x^2)^(1/4))/(40*x^3) + (27*(-2 + 3*x^2)^(1/4))/(32*x) + (27*Sqrt[3]*S
qrt[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])/(64*2^(1/4)*x)

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Rubi [A]  time = 0.056775, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {325, 234, 220} \[ \frac{27 \sqrt [4]{3 x^2-2}}{32 x}+\frac{9 \sqrt [4]{3 x^2-2}}{40 x^3}+\frac{\sqrt [4]{3 x^2-2}}{10 x^5}+\frac{27 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{3 x^2-2}+\sqrt{2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{64 \sqrt [4]{2} x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(-2 + 3*x^2)^(3/4)),x]

[Out]

(-2 + 3*x^2)^(1/4)/(10*x^5) + (9*(-2 + 3*x^2)^(1/4))/(40*x^3) + (27*(-2 + 3*x^2)^(1/4))/(32*x) + (27*Sqrt[3]*S
qrt[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])/(64*2^(1/4)*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 234

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/(b*x), Subst[Int[1/Sqrt[1 - x^4/a],
x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[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]

Rubi steps

\begin{align*} \int \frac{1}{x^6 \left (-2+3 x^2\right )^{3/4}} \, dx &=\frac{\sqrt [4]{-2+3 x^2}}{10 x^5}+\frac{27}{20} \int \frac{1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx\\ &=\frac{\sqrt [4]{-2+3 x^2}}{10 x^5}+\frac{9 \sqrt [4]{-2+3 x^2}}{40 x^3}+\frac{27}{16} \int \frac{1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx\\ &=\frac{\sqrt [4]{-2+3 x^2}}{10 x^5}+\frac{9 \sqrt [4]{-2+3 x^2}}{40 x^3}+\frac{27 \sqrt [4]{-2+3 x^2}}{32 x}+\frac{81}{64} \int \frac{1}{\left (-2+3 x^2\right )^{3/4}} \, dx\\ &=\frac{\sqrt [4]{-2+3 x^2}}{10 x^5}+\frac{9 \sqrt [4]{-2+3 x^2}}{40 x^3}+\frac{27 \sqrt [4]{-2+3 x^2}}{32 x}+\frac{\left (27 \sqrt{\frac{3}{2}} \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 )}{32 x}\\ &=\frac{\sqrt [4]{-2+3 x^2}}{10 x^5}+\frac{9 \sqrt [4]{-2+3 x^2}}{40 x^3}+\frac{27 \sqrt [4]{-2+3 x^2}}{32 x}+\frac{27 \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 )}{64 \sqrt [4]{2} x}\\ \end{align*}

Mathematica [C]  time = 0.0073827, size = 48, normalized size = 0.34 \[ -\frac{\left (1-\frac{3 x^2}{2}\right )^{3/4} \, _2F_1\left (-\frac{5}{2},\frac{3}{4};-\frac{3}{2};\frac{3 x^2}{2}\right )}{5 x^5 \left (3 x^2-2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - (3*x^2)/2)^(3/4)*Hypergeometric2F1[-5/2, 3/4, -3/2, (3*x^2)/2])/(5*x^5*(-2 + 3*x^2)^(3/4))

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Maple [C]  time = 0.039, size = 72, normalized size = 0.5 \begin{align*}{\frac{405\,{x}^{6}-162\,{x}^{4}-24\,{x}^{2}-32}{160\,{x}^{5}} \left ( 3\,{x}^{2}-2 \right ) ^{-{\frac{3}{4}}}}+{\frac{81\,\sqrt [4]{2}x}{128} \left ( -{\it signum} \left ( -1+{\frac{3\,{x}^{2}}{2}} \right ) \right ) ^{{\frac{3}{4}}}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{3}{4}};\,{\frac{3}{2}};\,{\frac{3\,{x}^{2}}{2}})} \left ({\it signum} \left ( -1+{\frac{3\,{x}^{2}}{2}} \right ) \right ) ^{-{\frac{3}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(3*x^2-2)^(3/4),x)

[Out]

1/160*(405*x^6-162*x^4-24*x^2-32)/x^5/(3*x^2-2)^(3/4)+81/128*2^(1/4)/signum(-1+3/2*x^2)^(3/4)*(-signum(-1+3/2*
x^2))^(3/4)*x*hypergeom([1/2,3/4],[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{3}{4}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(3*x^2-2)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((3*x^2 - 2)^(3/4)*x^6), 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{1}{4}}}{3 \, x^{8} - 2 \, x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(3*x^2-2)^(3/4),x, algorithm="fricas")

[Out]

integral((3*x^2 - 2)^(1/4)/(3*x^8 - 2*x^6), x)

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Sympy [C]  time = 1.20733, size = 32, normalized size = 0.23 \begin{align*} \frac{\sqrt [4]{2} e^{\frac{i \pi }{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, \frac{3}{4} \\ - \frac{3}{2} \end{matrix}\middle |{\frac{3 x^{2}}{2}} \right )}}{10 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(3*x**2-2)**(3/4),x)

[Out]

2**(1/4)*exp(I*pi/4)*hyper((-5/2, 3/4), (-3/2,), 3*x**2/2)/(10*x**5)

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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{3}{4}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(3*x^2-2)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((3*x^2 - 2)^(3/4)*x^6), x)