3.1399 \(\int \frac {x^6}{\sqrt {2+x^6}} \, dx\)

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

[Out]

1/4*x*(x^6+2)^(1/2)-1/24*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 = 0.05, 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 = {321, 225} \[ \frac {1}{4} x \sqrt {x^6+2}-\frac {x \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [3]{2} \sqrt [4]{3} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[2 + x^6],x]

[Out]

(x*Sqrt[2 + x^6])/4 - (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]*E
llipticF[ArcCos[(2^(1/3) + (1 - Sqrt[3])*x^2)/(2^(1/3) + (1 + Sqrt[3])*x^2)], (2 + Sqrt[3])/4])/(4*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 225

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]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(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]), x]] /; FreeQ[{a, b}, x]

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 {2+x^6}} \, dx &=\frac {1}{4} x \sqrt {2+x^6}-\frac {1}{2} \int \frac {1}{\sqrt {2+x^6}} \, dx\\ &=\frac {1}{4} x \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 )}{4 \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]  time = 0.01, size = 39, normalized size = 0.22 \[ \frac {1}{4} x \left (\sqrt {x^6+2}-\sqrt {2} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\frac {x^6}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[2 + x^6],x]

[Out]

(x*(Sqrt[2 + x^6] - Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, -1/2*x^6]))/4

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6}}{\sqrt {x^{6} + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^6+2)^(1/2),x, algorithm="fricas")

[Out]

integral(x^6/sqrt(x^6 + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {x^{6} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^6+2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(x^6 + 2), x)

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maple [C]  time = 0.14, size = 29, normalized size = 0.16 \[ -\frac {\sqrt {2}\, x \hypergeom \left (\left [\frac {1}{6}, \frac {1}{2}\right ], \left [\frac {7}{6}\right ], -\frac {x^{6}}{2}\right )}{4}+\frac {\sqrt {x^{6}+2}\, x}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^6+2)^(1/2),x)

[Out]

1/4*x*(x^6+2)^(1/2)-1/4*2^(1/2)*x*hypergeom([1/6,1/2],[7/6],-1/2*x^6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {x^{6} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^6+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(x^6 + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6}{\sqrt {x^6+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^6 + 2)^(1/2),x)

[Out]

int(x^6/(x^6 + 2)^(1/2), x)

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sympy [C]  time = 1.45, size = 36, normalized size = 0.20 \[ \frac {\sqrt {2} x^{7} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{6} \\ \frac {13}{6} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{12 \Gamma \left (\frac {13}{6}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(x**6+2)**(1/2),x)

[Out]

sqrt(2)*x**7*gamma(7/6)*hyper((1/2, 7/6), (13/6,), x**6*exp_polar(I*pi)/2)/(12*gamma(13/6))

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