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

Optimal. Leaf size=99 \[ \frac {32 \left (3 x^2+2\right )^{3/4} x}{1053}-\frac {128 x}{1053 \sqrt [4]{3 x^2+2}}+\frac {2}{39} \left (3 x^2+2\right )^{3/4} x^5-\frac {40 \left (3 x^2+2\right )^{3/4} x^3}{1053}+\frac {128 \sqrt [4]{2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{1053 \sqrt {3}} \]

[Out]

-128/1053*x/(3*x^2+2)^(1/4)+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
/3159*2^(1/4)*(cos(1/2*arctan(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arctan(1/2*x*6^(1/2)))*EllipticE(sin(1/2*arctan
(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {321, 227, 196} \[ \frac {2}{39} \left (3 x^2+2\right )^{3/4} x^5-\frac {40 \left (3 x^2+2\right )^{3/4} x^3}{1053}+\frac {32 \left (3 x^2+2\right )^{3/4} x}{1053}-\frac {128 x}{1053 \sqrt [4]{3 x^2+2}}+\frac {128 \sqrt [4]{2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{1053 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(2 + 3*x^2)^(1/4),x]

[Out]

(-128*x)/(1053*(2 + 3*x^2)^(1/4)) + (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*2^(1/4)*EllipticE[ArcTan[Sqrt[3/2]*x]/2, 2])/(1053*Sqrt[3])

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]

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 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]

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 {128 x}{1053 \sqrt [4]{2+3 x^2}}+\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 \int \frac {1}{\left (2+3 x^2\right )^{5/4}} \, dx}{1053}\\ &=-\frac {128 x}{1053 \sqrt [4]{2+3 x^2}}+\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 \sqrt [4]{2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{1053 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 54, normalized size = 0.55 \[ \frac {2 x \left (\left (3 x^2+2\right )^{3/4} \left (27 x^4-20 x^2+16\right )-16\ 2^{3/4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {3 x^2}{2}\right )\right )}{1053} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(2 + 3*x^2)^(1/4),x]

[Out]

(2*x*((2 + 3*x^2)^(3/4)*(16 - 20*x^2 + 27*x^4) - 16*2^(3/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (-3*x^2)/2]))/105
3

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fricas [F]  time = 0.90, 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 = 0.30, size = 43, normalized size = 0.43 \[ -\frac {32 \,2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{1053}+\frac {2 \left (27 x^{4}-20 x^{2}+16\right ) \left (3 x^{2}+2\right )^{\frac {3}{4}} x}{1053} \]

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)*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 = -0.01 \[ \int \frac {x^6}{{\left (3\,x^2+2\right )}^{1/4}} \,d x \]

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 = 0.83, size = 27, normalized size = 0.27 \[ \frac {2^{\frac {3}{4}} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{14} \]

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

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