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

Optimal. Leaf size=83 \[ -\frac {320\ 2^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right ),2\right )}{2079 \sqrt {3}}+\frac {160 \sqrt [4]{3 x^2+2} x}{2079}+\frac {2}{33} \sqrt [4]{3 x^2+2} x^5-\frac {40}{693} \sqrt [4]{3 x^2+2} x^3 \]

[Out]

160/2079*x*(3*x^2+2)^(1/4)-40/693*x^3*(3*x^2+2)^(1/4)+2/33*x^5*(3*x^2+2)^(1/4)-320/6237*2^(3/4)*(cos(1/2*arcta
n(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arctan(1/2*x*6^(1/2)))*EllipticF(sin(1/2*arctan(1/2*x*6^(1/2))),2^(1/2))*3^
(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {321, 231} \[ \frac {2}{33} \sqrt [4]{3 x^2+2} x^5-\frac {40}{693} \sqrt [4]{3 x^2+2} x^3+\frac {160 \sqrt [4]{3 x^2+2} x}{2079}-\frac {320\ 2^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2079 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(2 + 3*x^2)^(3/4),x]

[Out]

(160*x*(2 + 3*x^2)^(1/4))/2079 - (40*x^3*(2 + 3*x^2)^(1/4))/693 + (2*x^5*(2 + 3*x^2)^(1/4))/33 - (320*2^(3/4)*
EllipticF[ArcTan[Sqrt[3/2]*x]/2, 2])/(2079*Sqrt[3])

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), 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}{\left (2+3 x^2\right )^{3/4}} \, dx &=\frac {2}{33} x^5 \sqrt [4]{2+3 x^2}-\frac {20}{33} \int \frac {x^4}{\left (2+3 x^2\right )^{3/4}} \, dx\\ &=-\frac {40}{693} x^3 \sqrt [4]{2+3 x^2}+\frac {2}{33} x^5 \sqrt [4]{2+3 x^2}+\frac {80}{231} \int \frac {x^2}{\left (2+3 x^2\right )^{3/4}} \, dx\\ &=\frac {160 x \sqrt [4]{2+3 x^2}}{2079}-\frac {40}{693} x^3 \sqrt [4]{2+3 x^2}+\frac {2}{33} x^5 \sqrt [4]{2+3 x^2}-\frac {320 \int \frac {1}{\left (2+3 x^2\right )^{3/4}} \, dx}{2079}\\ &=\frac {160 x \sqrt [4]{2+3 x^2}}{2079}-\frac {40}{693} x^3 \sqrt [4]{2+3 x^2}+\frac {2}{33} x^5 \sqrt [4]{2+3 x^2}-\frac {320\ 2^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2079 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.65 \[ \frac {2 x \left (\sqrt [4]{3 x^2+2} \left (63 x^4-60 x^2+80\right )-80 \sqrt [4]{2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\frac {3 x^2}{2}\right )\right )}{2079} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(2 + 3*x^2)^(3/4),x]

[Out]

(2*x*((2 + 3*x^2)^(1/4)*(80 - 60*x^2 + 63*x^4) - 80*2^(1/4)*Hypergeometric2F1[1/2, 3/4, 3/2, (-3*x^2)/2]))/207
9

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6}}{{\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^6/(3*x^2 + 2)^(3/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(3*x^2 + 2)^(3/4), x)

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maple [C]  time = 0.29, size = 43, normalized size = 0.52 \[ -\frac {160 \,2^{\frac {1}{4}} x \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{2079}+\frac {2 \left (63 x^{4}-60 x^{2}+80\right ) \left (3 x^{2}+2\right )^{\frac {1}{4}} x}{2079} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/2079*x*(63*x^4-60*x^2+80)*(3*x^2+2)^(1/4)-160/2079*2^(1/4)*x*hypergeom([1/2,3/4],[3/2],-3/2*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(3*x^2 + 2)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6}{{\left (3\,x^2+2\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(3*x^2 + 2)^(3/4),x)

[Out]

int(x^6/(3*x^2 + 2)^(3/4), x)

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sympy [C]  time = 0.82, size = 27, normalized size = 0.33 \[ \frac {\sqrt [4]{2} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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