3.15.25 \(\int \frac {x^6}{(2+x^6)^{3/2}} \, dx\) [1425]

Optimal. Leaf size=179 \[ -\frac {x}{3 \sqrt {2+x^6}}+\frac {x \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}+\left (1-\sqrt {3}\right ) x^2}{\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{6 \sqrt [3]{2} \sqrt [4]{3} \sqrt {\frac {x^2 \left (\sqrt [3]{2}+x^2\right )}{\left (\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {2+x^6}} \]

[Out]

-1/3*x/(x^6+2)^(1/2)+1/36*x*(2^(1/3)+x^2)*((2^(1/3)+x^2*(1-3^(1/2)))^2/(2^(1/3)+x^2*(1+3^(1/2)))^2)^(1/2)/(2^(
1/3)+x^2*(1-3^(1/2)))*(2^(1/3)+x^2*(1+3^(1/2)))*EllipticF((1-(2^(1/3)+x^2*(1-3^(1/2)))^2/(2^(1/3)+x^2*(1+3^(1/
2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*((2^(2/3)-2^(1/3)*x^2+x^4)/(2^(1/3)+x^2*(1+3^(1/2)))^2)^(1/2)*2^(2/3)*3
^(3/4)/(x^6+2)^(1/2)/(x^2*(2^(1/3)+x^2)/(2^(1/3)+x^2*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {294, 231} \begin {gather*} \frac {x \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} F\left (\text {ArcCos}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{6 \sqrt [3]{2} \sqrt [4]{3} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}}-\frac {x}{3 \sqrt {x^6+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/(2 + x^6)^(3/2),x]

[Out]

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

Rule 231

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s +
r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*(
(s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (2+x^6\right )^{3/2}} \, dx &=-\frac {x}{3 \sqrt {2+x^6}}+\frac {1}{3} \int \frac {1}{\sqrt {2+x^6}} \, dx\\ &=-\frac {x}{3 \sqrt {2+x^6}}+\frac {x \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2\right )^2}} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}+\left (1-\sqrt {3}\right ) x^2}{\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{6 \sqrt [3]{2} \sqrt [4]{3} \sqrt {\frac {x^2 \left (\sqrt [3]{2}+x^2\right )}{\left (\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {2+x^6}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 5.01, size = 42, normalized size = 0.23 \begin {gather*} -\frac {x}{3 \sqrt {2+x^6}}+\frac {x \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\frac {x^6}{2}\right )}{3 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/(2 + x^6)^(3/2),x]

[Out]

-1/3*x/Sqrt[2 + x^6] + (x*Hypergeometric2F1[1/6, 1/2, 7/6, -1/2*x^6])/(3*Sqrt[2])

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Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.19, size = 20, normalized size = 0.11

method	result	size
meijerg	\(\frac {\sqrt {2}\, x^{7} \hypergeom \left (\left [\frac {7}{6}, \frac {3}{2}\right ], \left [\frac {13}{6}\right ], -\frac {x^{6}}{2}\right )}{28}\)	\(20\)
risch	\(-\frac {x}{3 \sqrt {x^{6}+2}}+\frac {\sqrt {2}\, x \hypergeom \left (\left [\frac {1}{6}, \frac {1}{2}\right ], \left [\frac {7}{6}\right ], -\frac {x^{6}}{2}\right )}{6}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^6+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/28*2^(1/2)*x^7*hypergeom([7/6,3/2],[13/6],-1/2*x^6)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(x^6 + 2)^(3/2), x)

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Fricas [F]
time = 0.07, size = 25, normalized size = 0.14 \begin {gather*} {\rm integral}\left (\frac {\sqrt {x^{6} + 2} x^{6}}{x^{12} + 4 \, x^{6} + 4}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^6 + 2)*x^6/(x^12 + 4*x^6 + 4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.39, size = 36, normalized size = 0.20 \begin {gather*} \frac {\sqrt {2} x^{7} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{6}, \frac {3}{2} \\ \frac {13}{6} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{24 \Gamma \left (\frac {13}{6}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*x**7*gamma(7/6)*hyper((7/6, 3/2), (13/6,), x**6*exp_polar(I*pi)/2)/(24*gamma(13/6))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(x^6 + 2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^6 + 2)^(3/2),x)

[Out]

int(x^6/(x^6 + 2)^(3/2), x)

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