3.1426 \(\int \frac {1}{(2+x^6)^{3/2}} \, dx\)

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

[Out]

1/6*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 = 0.03, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {199, 225} \[ \frac {x}{6 \sqrt {x^6+2}}+\frac {\left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} x F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{6 \sqrt [3]{2} \sqrt [4]{3} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + x^6)^(-3/2),x]

[Out]

x/(6*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]*E
llipticF[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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (2+x^6\right )^{3/2}} \, dx &=\frac {x}{6 \sqrt {2+x^6}}+\frac {1}{3} \int \frac {1}{\sqrt {2+x^6}} \, dx\\ &=\frac {x}{6 \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]  time = 0.02, size = 38, normalized size = 0.21 \[ \frac {1}{6} x \left (\sqrt {2} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\frac {x^6}{2}\right )+\frac {1}{\sqrt {x^6+2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x^6)^(-3/2),x]

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{6} + 2}}{x^{12} + 4 \, x^{6} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6+2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{6} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6+2)^(3/2),x, algorithm="giac")

[Out]

integrate((x^6 + 2)^(-3/2), x)

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maple [C]  time = 0.14, size = 18, normalized size = 0.10 \[ \frac {\sqrt {2}\, x \hypergeom \left (\left [\frac {1}{6}, \frac {3}{2}\right ], \left [\frac {7}{6}\right ], -\frac {x^{6}}{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6+2)^(3/2),x)

[Out]

1/4*2^(1/2)*x*hypergeom([1/6,3/2],[7/6],-1/2*x^6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{6} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6+2)^(3/2),x, algorithm="maxima")

[Out]

integrate((x^6 + 2)^(-3/2), x)

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mupad [B]  time = 0.08, size = 16, normalized size = 0.09 \[ \frac {\sqrt {2}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{6},\frac {3}{2};\ \frac {7}{6};\ -\frac {x^6}{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6 + 2)^(3/2),x)

[Out]

(2^(1/2)*x*hypergeom([1/6, 3/2], 7/6, -x^6/2))/4

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sympy [C]  time = 0.74, size = 34, normalized size = 0.19 \[ \frac {\sqrt {2} x \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{6}, \frac {3}{2} \\ \frac {7}{6} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{24 \Gamma \left (\frac {7}{6}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**6+2)**(3/2),x)

[Out]

sqrt(2)*x*gamma(1/6)*hyper((1/6, 3/2), (7/6,), x**6*exp_polar(I*pi)/2)/(24*gamma(7/6))

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