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

Optimal. Leaf size=258 \[ \frac {64 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{1053 \sqrt {3} x}+\frac {32 \left (3 x^2-2\right )^{3/4} x}{1053}+\frac {128 \sqrt [4]{3 x^2-2} x}{1053 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}-\frac {128 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{1053 \sqrt {3} x}+\frac {2}{39} \left (3 x^2-2\right )^{3/4} x^5+\frac {40 \left (3 x^2-2\right )^{3/4} x^3}{1053} \]

[Out]

32/1053*x*(3*x^2-2)^(3/4)+40/1053*x^3*(3*x^2-2)^(3/4)+2/39*x^5*(3*x^2-2)^(3/4)+128/1053*x*(3*x^2-2)^(1/4)/(2^(
1/2)+(3*x^2-2)^(1/2))-128/3159*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)+64/3159*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 = 0.14, antiderivative size = 258, 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 = {321, 230, 305, 220, 1196} \[ \frac {2}{39} \left (3 x^2-2\right )^{3/4} x^5+\frac {40 \left (3 x^2-2\right )^{3/4} x^3}{1053}+\frac {32 \left (3 x^2-2\right )^{3/4} x}{1053}+\frac {128 \sqrt [4]{3 x^2-2} x}{1053 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}+\frac {64 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{1053 \sqrt {3} x}-\frac {128 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{1053 \sqrt {3} x} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(-2 + 3*x^2)^(1/4),x]

[Out]

(32*x*(-2 + 3*x^2)^(3/4))/1053 + (40*x^3*(-2 + 3*x^2)^(3/4))/1053 + (2*x^5*(-2 + 3*x^2)^(3/4))/39 + (128*x*(-2
+ 3*x^2)^(1/4))/(1053*(Sqrt[2] + Sqrt[-2 + 3*x^2])) - (128*2^(1/4)*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])/(1053*Sqrt[3]*x) + (64*2^(1/
4)*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])/(1053*Sqrt[3]*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 {x^6}{\sqrt [4]{-2+3 x^2}} \, dx &=\frac {2}{39} x^5 \left (-2+3 x^2\right )^{3/4}+\frac {20}{39} \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx\\ &=\frac {40 x^3 \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {2}{39} x^5 \left (-2+3 x^2\right )^{3/4}+\frac {80}{351} \int \frac {x^2}{\sqrt [4]{-2+3 x^2}} \, dx\\ &=\frac {32 x \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {40 x^3 \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {2}{39} x^5 \left (-2+3 x^2\right )^{3/4}+\frac {64 \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx}{1053}\\ &=\frac {32 x \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {40 x^3 \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {2}{39} x^5 \left (-2+3 x^2\right )^{3/4}+\frac {\left (64 \sqrt {\frac {2}{3}} \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 )}{1053 x}\\ &=\frac {32 x \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {40 x^3 \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {2}{39} x^5 \left (-2+3 x^2\right )^{3/4}+\frac {\left (128 \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 )}{1053 \sqrt {3} x}-\frac {\left (128 \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 )}{1053 \sqrt {3} x}\\ &=\frac {32 x \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {40 x^3 \left (-2+3 x^2\right )^{3/4}}{1053}+\frac {2}{39} x^5 \left (-2+3 x^2\right )^{3/4}+\frac {128 x \sqrt [4]{-2+3 x^2}}{1053 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}-\frac {128 \sqrt [4]{2} \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 )}{1053 \sqrt {3} x}+\frac {64 \sqrt [4]{2} \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 )}{1053 \sqrt {3} x}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 68, normalized size = 0.26 \[ \frac {2 x \left (16\ 2^{3/4} \sqrt [4]{2-3 x^2} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^2}{2}\right )+81 x^6+6 x^4+8 x^2-32\right )}{1053 \sqrt [4]{3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(-2 + 3*x^2)^(1/4),x]

[Out]

(2*x*(-32 + 8*x^2 + 6*x^4 + 81*x^6 + 16*2^(3/4)*(2 - 3*x^2)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (3*x^2)/2])
)/(1053*(-2 + 3*x^2)^(1/4))

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6}}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^6/(3*x^2 - 2)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(3*x^2 - 2)^(1/4), x)

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maple [C]  time = 0.29, size = 65, normalized size = 0.25 \[ \frac {32 \,2^{\frac {3}{4}} \left (-\mathrm {signum}\left (\frac {3 x^{2}}{2}-1\right )\right )^{\frac {1}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{1053 \mathrm {signum}\left (\frac {3 x^{2}}{2}-1\right )^{\frac {1}{4}}}+\frac {2 \left (27 x^{4}+20 x^{2}+16\right ) \left (3 x^{2}-2\right )^{\frac {3}{4}} x}{1053} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1053*x*(27*x^4+20*x^2+16)*(3*x^2-2)^(3/4)+32/1053*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(3*x^2 - 2)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(3*x^2 - 2)^(1/4),x)

[Out]

int(x^6/(3*x^2 - 2)^(1/4), x)

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sympy [C]  time = 0.85, size = 29, normalized size = 0.11 \[ \frac {2^{\frac {3}{4}} x^{7} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(3/4)*x**7*exp(-I*pi/4)*hyper((1/4, 7/2), (9/2,), 3*x**2/2)/14

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