3.1397 \(\int \frac{x}{\sqrt{2+x^6}} \, dx\)

Optimal. Leaf size=166 \[ \frac{\sqrt{2+\sqrt{3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x^2+\sqrt [3]{2} \left (1-\sqrt{3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )}\right ),-7-4 \sqrt{3}\right )}{\sqrt [6]{2} \sqrt [4]{3} \sqrt{\frac{x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \sqrt{x^6+2}} \]

[Out]

(Sqrt[2 + Sqrt[3]]*(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]*Ellipti
cF[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(2^(1/6)*3^(1/4)*Sqrt
[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])

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Rubi [A]  time = 0.0696663, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {275, 218} \[ \frac{\sqrt{2+\sqrt{3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} F\left (\sin ^{-1}\left (\frac{x^2+\sqrt [3]{2} \left (1-\sqrt{3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )}\right )|-7-4 \sqrt{3}\right )}{\sqrt [6]{2} \sqrt [4]{3} \sqrt{\frac{x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt{3}\right )\right )^2}} \sqrt{x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[2 + x^6],x]

[Out]

(Sqrt[2 + Sqrt[3]]*(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]*Ellipti
cF[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(2^(1/6)*3^(1/4)*Sqrt
[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6])

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

\begin{align*} \int \frac{x}{\sqrt{2+x^6}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+x^3}} \, dx,x,x^2\right )\\ &=\frac{\sqrt{2+\sqrt{3}} \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 (\sin ^{-1}\left (\frac{\sqrt [3]{2} \left (1-\sqrt{3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt{3}\right )+x^2}\right )|-7-4 \sqrt{3}\right )}{\sqrt [6]{2} \sqrt [4]{3} \sqrt{\frac{\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt{3}\right )+x^2\right )^2}} \sqrt{2+x^6}}\\ \end{align*}

Mathematica [C]  time = 0.0046612, size = 29, normalized size = 0.17 \[ \frac{x^2 \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{x^6}{2}\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[2 + x^6],x]

[Out]

(x^2*Hypergeometric2F1[1/3, 1/2, 4/3, -x^6/2])/(2*Sqrt[2])

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Maple [C]  time = 0.017, size = 20, normalized size = 0.1 \begin{align*}{\frac{{x}^{2}\sqrt{2}}{4}{\mbox{$_2$F$_1$}({\frac{1}{3}},{\frac{1}{2}};\,{\frac{4}{3}};\,-{\frac{{x}^{6}}{2}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*2^(1/2)*x^2*hypergeom([1/3,1/2],[4/3],-1/2*x^6)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{x^{6} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(x^6 + 2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\sqrt{x^{6} + 2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/sqrt(x^6 + 2), x)

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Sympy [A]  time = 0.532103, size = 36, normalized size = 0.22 \begin{align*} \frac{\sqrt{2} x^{2} \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{4}{3} \end{matrix}\middle |{\frac{x^{6} e^{i \pi }}{2}} \right )}}{12 \Gamma \left (\frac{4}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*x**2*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**6*exp_polar(I*pi)/2)/(12*gamma(4/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{x^{6} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(x^6 + 2), x)