3.869 \(\int \frac{1}{x^6 \sqrt [4]{2+3 x^2}} \, dx\)

Optimal. Leaf size=101 \[ \frac{189 x}{160 \sqrt [4]{3 x^2+2}}-\frac{63 \left (3 x^2+2\right )^{3/4}}{160 x}+\frac{7 \left (3 x^2+2\right )^{3/4}}{40 x^3}-\frac{\left (3 x^2+2\right )^{3/4}}{10 x^5}-\frac{63 \sqrt{3} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{80\ 2^{3/4}} \]

[Out]

(189*x)/(160*(2 + 3*x^2)^(1/4)) - (2 + 3*x^2)^(3/4)/(10*x^5) + (7*(2 + 3*x^2)^(3/4))/(40*x^3) - (63*(2 + 3*x^2
)^(3/4))/(160*x) - (63*Sqrt[3]*EllipticE[ArcTan[Sqrt[3/2]*x]/2, 2])/(80*2^(3/4))

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Rubi [A]  time = 0.0295966, antiderivative size = 101, 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, 227, 196} \[ \frac{189 x}{160 \sqrt [4]{3 x^2+2}}-\frac{63 \left (3 x^2+2\right )^{3/4}}{160 x}+\frac{7 \left (3 x^2+2\right )^{3/4}}{40 x^3}-\frac{\left (3 x^2+2\right )^{3/4}}{10 x^5}-\frac{63 \sqrt{3} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{80\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(2 + 3*x^2)^(1/4)),x]

[Out]

(189*x)/(160*(2 + 3*x^2)^(1/4)) - (2 + 3*x^2)^(3/4)/(10*x^5) + (7*(2 + 3*x^2)^(3/4))/(40*x^3) - (63*(2 + 3*x^2
)^(3/4))/(160*x) - (63*Sqrt[3]*EllipticE[ArcTan[Sqrt[3/2]*x]/2, 2])/(80*2^(3/4))

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 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{1}{x^6 \sqrt [4]{2+3 x^2}} \, dx &=-\frac{\left (2+3 x^2\right )^{3/4}}{10 x^5}-\frac{21}{20} \int \frac{1}{x^4 \sqrt [4]{2+3 x^2}} \, dx\\ &=-\frac{\left (2+3 x^2\right )^{3/4}}{10 x^5}+\frac{7 \left (2+3 x^2\right )^{3/4}}{40 x^3}+\frac{63}{80} \int \frac{1}{x^2 \sqrt [4]{2+3 x^2}} \, dx\\ &=-\frac{\left (2+3 x^2\right )^{3/4}}{10 x^5}+\frac{7 \left (2+3 x^2\right )^{3/4}}{40 x^3}-\frac{63 \left (2+3 x^2\right )^{3/4}}{160 x}+\frac{189}{320} \int \frac{1}{\sqrt [4]{2+3 x^2}} \, dx\\ &=\frac{189 x}{160 \sqrt [4]{2+3 x^2}}-\frac{\left (2+3 x^2\right )^{3/4}}{10 x^5}+\frac{7 \left (2+3 x^2\right )^{3/4}}{40 x^3}-\frac{63 \left (2+3 x^2\right )^{3/4}}{160 x}-\frac{189}{160} \int \frac{1}{\left (2+3 x^2\right )^{5/4}} \, dx\\ &=\frac{189 x}{160 \sqrt [4]{2+3 x^2}}-\frac{\left (2+3 x^2\right )^{3/4}}{10 x^5}+\frac{7 \left (2+3 x^2\right )^{3/4}}{40 x^3}-\frac{63 \left (2+3 x^2\right )^{3/4}}{160 x}-\frac{63 \sqrt{3} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{80\ 2^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0050861, size = 29, normalized size = 0.29 \[ -\frac{\, _2F_1\left (-\frac{5}{2},\frac{1}{4};-\frac{3}{2};-\frac{3 x^2}{2}\right )}{5 \sqrt [4]{2} x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.022, size = 50, normalized size = 0.5 \begin{align*} -{\frac{189\,{x}^{6}+42\,{x}^{4}-8\,{x}^{2}+32}{160\,{x}^{5}}{\frac{1}{\sqrt [4]{3\,{x}^{2}+2}}}}+{\frac{189\,{2}^{3/4}x}{640}{\mbox{$_2$F$_1$}({\frac{1}{4}},{\frac{1}{2}};\,{\frac{3}{2}};\,-{\frac{3\,{x}^{2}}{2}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/160*(189*x^6+42*x^4-8*x^2+32)/x^5/(3*x^2+2)^(1/4)+189/640*2^(3/4)*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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Sympy [C]  time = 1.02844, size = 32, normalized size = 0.32 \begin{align*} - \frac{2^{\frac{3}{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, \frac{1}{4} \\ - \frac{3}{2} \end{matrix}\middle |{\frac{3 x^{2} e^{i \pi }}{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)**(1/4),x)

[Out]

-2**(3/4)*hyper((-5/2, 1/4), (-3/2,), 3*x**2*exp_polar(I*pi)/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{1}{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)^(1/4),x, algorithm="giac")

[Out]

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