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

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

[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])

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Rubi [A]  time = 0.0270916, antiderivative size = 179, normalized size of antiderivative = 1., 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*}

Mathematica [C]  time = 0.0075764, 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, -x^6/2]))/6

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Maple [C]  time = 0.017, size = 18, normalized size = 0.1 \begin{align*}{\frac{x\sqrt{2}}{4}{\mbox{$_2$F$_1$}({\frac{1}{6}},{\frac{3}{2}};\,{\frac{7}{6}};\,-{\frac{{x}^{6}}{2}})}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{6} + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{6} + 2}}{x^{12} + 4 \, x^{6} + 4}, x\right ) \end{align*}

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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Sympy [C]  time = 0.591547, size = 34, normalized size = 0.19 \begin{align*} \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 )} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{6} + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)