3.14.97 \(\int \frac {x}{\sqrt {2+x^6}} \, dx\) [1397]

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

[Out]

1/6*2^(5/6)*(2^(1/3)+x^2)*EllipticF((x^2+2^(1/3)*(1-3^(1/2)))/(x^2+2^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^
(1/2)+1/2*2^(1/2))*((2^(2/3)-2^(1/3)*x^2+x^4)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/(x^6+2)^(1/2)/((2^(1/
3)+x^2)/(x^2+2^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 = {281, 224} \begin {gather*} \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 (\text {ArcSin}\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}} \end {gather*}

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 224

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

Rule 281

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]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {2+x^6}} \, dx &=\frac {1}{2} \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 29, normalized size = 0.17 \begin {gather*} \frac {x^2 \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {x^6}{2}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[2 + x^6],x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.15, size = 20, normalized size = 0.12

method	result	size
meijerg	\(\frac {\sqrt {2}\, x^{2} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], -\frac {x^{6}}{2}\right )}{4}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^6+2)^(1/2),x,method=_RETURNVERBOSE)

[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 = 0.00, 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/(x^6+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/sqrt(x^6 + 2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.07, size = 6, normalized size = 0.04 \begin {gather*} {\rm weierstrassPInverse}\left (0, -8, x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

weierstrassPInverse(0, -8, x^2)

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Sympy [A]
time = 0.30, size = 36, normalized size = 0.22 \begin {gather*} \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 {gather*}

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 = 0.00, 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/(x^6+2)^(1/2),x, algorithm="giac")

[Out]

integrate(x/sqrt(x^6 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^6 + 2)^(1/2),x)

[Out]

int(x/(x^6 + 2)^(1/2), x)

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