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

Optimal. Leaf size=260 \[ -\frac {63 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{160\ 2^{3/4} x}-\frac {189 \sqrt [4]{3 x^2-2} x}{160 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}+\frac {63 \left (3 x^2-2\right )^{3/4}}{160 x}+\frac {63 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{80\ 2^{3/4} x}+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}+\frac {7 \left (3 x^2-2\right )^{3/4}}{40 x^3} \]

[Out]

1/10*(3*x^2-2)^(3/4)/x^5+7/40*(3*x^2-2)^(3/4)/x^3+63/160*(3*x^2-2)^(3/4)/x-189/160*x*(3*x^2-2)^(1/4)/(2^(1/2)+
(3*x^2-2)^(1/2))+63/160*2^(1/4)*(cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arctan(1/2*(3*x^2-2
)^(1/4)*2^(3/4)))*EllipticE(sin(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*
(x^2/(2^(1/2)+(3*x^2-2)^(1/2))^2)^(1/2)/x*3^(1/2)-63/320*2^(1/4)*(cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2
)^(1/2)/cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))*EllipticF(sin(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2^
(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*(x^2/(2^(1/2)+(3*x^2-2)^(1/2))^2)^(1/2)/x*3^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {189 \sqrt [4]{3 x^2-2} x}{160 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}+\frac {63 \left (3 x^2-2\right )^{3/4}}{160 x}+\frac {7 \left (3 x^2-2\right )^{3/4}}{40 x^3}+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}-\frac {63 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{160\ 2^{3/4} x}+\frac {63 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{80\ 2^{3/4} x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(-2 + 3*x^2)^(1/4)),x]

[Out]

(-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) - (189*x*(-2 +
3*x^2)^(1/4))/(160*(Sqrt[2] + Sqrt[-2 + 3*x^2])) + (63*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])/(80*2^(3/4)*x) - (63*Sqrt[3]*Sqr
t[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])/(160*2^(3/4)*x)

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 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 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 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^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 {\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 {\left (63 \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 )}{160 x}\\ &=\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 {\left (63 \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 )}{160 x}+\frac {\left (63 \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 )}{160 x}\\ &=\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 x \sqrt [4]{-2+3 x^2}}{160 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}+\frac {63 \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 )}{80\ 2^{3/4} x}-\frac {63 \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 )}{160\ 2^{3/4} x}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.18 \[ -\frac {\sqrt [4]{1-\frac {3 x^2}{2}} \, _2F_1\left (-\frac {5}{2},\frac {1}{4};-\frac {3}{2};\frac {3 x^2}{2}\right )}{5 x^5 \sqrt [4]{3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{3 \, x^{8} - 2 \, x^{6}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{6}}\,{d x} \]

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)

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maple [C]  time = 0.29, size = 72, normalized size = 0.28 \[ -\frac {189 \,2^{\frac {3}{4}} \left (-\mathrm {signum}\left (\frac {3 x^{2}}{2}-1\right )\right )^{\frac {1}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{640 \mathrm {signum}\left (\frac {3 x^{2}}{2}-1\right )^{\frac {1}{4}}}+\frac {189 x^{6}-42 x^{4}-8 x^{2}-32}{160 \left (3 x^{2}-2\right )^{\frac {1}{4}} x^{5}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{6}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^6\,{\left (3\,x^2-2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(3*x^2 - 2)^(1/4)),x)

[Out]

int(1/(x^6*(3*x^2 - 2)^(1/4)), x)

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sympy [C]  time = 0.98, size = 34, normalized size = 0.13 \[ \frac {2^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{10 x^{5}} \]

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

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