3.1405 \(\int \frac{x^{10}}{\sqrt{2+x^6}} \, dx\)

Optimal. Leaf size=394 \[ \frac{5 \left (1-\sqrt{3}\right ) \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} x \text{EllipticF}\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{8\ 2^{2/3} \sqrt [4]{3} \sqrt{\frac{x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt{x^6+2}}+\frac{1}{8} \sqrt{x^6+2} x^5-\frac{5 \left (1+\sqrt{3}\right ) \sqrt{x^6+2} x}{8 \left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )}+\frac{5 \sqrt [4]{3} \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} x E\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{4\ 2^{2/3} \sqrt{\frac{x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt{x^6+2}} \]

[Out]

(x^5*Sqrt[2 + x^6])/8 - (5*(1 + Sqrt[3])*x*Sqrt[2 + x^6])/(8*(2^(1/3) + (1 + Sqrt[3])*x^2)) + (5*3^(1/4)*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]*EllipticE[ArcCos[(2^(1/3) + (
1 - Sqrt[3])*x^2)/(2^(1/3) + (1 + Sqrt[3])*x^2)], (2 + Sqrt[3])/4])/(4*2^(2/3)*Sqrt[(x^2*(2^(1/3) + x^2))/(2^(
1/3) + (1 + Sqrt[3])*x^2)^2]*Sqrt[2 + x^6]) + (5*(1 - Sqrt[3])*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]*EllipticF[ArcCos[(2^(1/3) + (1 - Sqrt[3])*x^2)/(2^(1/3) + (1 + Sqrt[3])
*x^2)], (2 + Sqrt[3])/4])/(8*2^(2/3)*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 = 0.130567, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {321, 308, 225, 1881} \[ \frac{1}{8} \sqrt{x^6+2} x^5-\frac{5 \left (1+\sqrt{3}\right ) \sqrt{x^6+2} x}{8 \left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )}+\frac{5 \left (1-\sqrt{3}\right ) \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} x F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{8\ 2^{2/3} \sqrt [4]{3} \sqrt{\frac{x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt{x^6+2}}+\frac{5 \sqrt [4]{3} \left (x^2+\sqrt [3]{2}\right ) \sqrt{\frac{x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} x E\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{4\ 2^{2/3} \sqrt{\frac{x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt{3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt{x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[x^10/Sqrt[2 + x^6],x]

[Out]

(x^5*Sqrt[2 + x^6])/8 - (5*(1 + Sqrt[3])*x*Sqrt[2 + x^6])/(8*(2^(1/3) + (1 + Sqrt[3])*x^2)) + (5*3^(1/4)*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]*EllipticE[ArcCos[(2^(1/3) + (
1 - Sqrt[3])*x^2)/(2^(1/3) + (1 + Sqrt[3])*x^2)], (2 + Sqrt[3])/4])/(4*2^(2/3)*Sqrt[(x^2*(2^(1/3) + x^2))/(2^(
1/3) + (1 + Sqrt[3])*x^2)^2]*Sqrt[2 + x^6]) + (5*(1 - Sqrt[3])*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]*EllipticF[ArcCos[(2^(1/3) + (1 - Sqrt[3])*x^2)/(2^(1/3) + (1 + Sqrt[3])
*x^2)], (2 + Sqrt[3])/4])/(8*2^(2/3)*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 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]

Rule 308

Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(
(Sqrt[3] - 1)*s^2)/(2*r^2), Int[1/Sqrt[a + b*x^6], x], x] - Dist[1/(2*r^2), Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4
)/Sqrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]

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 1881

Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/
a, 3]]}, Simp[((1 + Sqrt[3])*d*s^3*x*Sqrt[a + b*x^6])/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2)), x] - Simp[(3^(1/4)*
d*s*x*(s + r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticE[ArcCos[(s + (1 - Sqrt[
3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*
x^2)^2]*Sqrt[a + b*x^6]), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 - Sqrt[3])*d, 0]

Rubi steps

\begin{align*} \int \frac{x^{10}}{\sqrt{2+x^6}} \, dx &=\frac{1}{8} x^5 \sqrt{2+x^6}-\frac{5}{4} \int \frac{x^4}{\sqrt{2+x^6}} \, dx\\ &=\frac{1}{8} x^5 \sqrt{2+x^6}+\frac{5}{8} \int \frac{2^{2/3} \left (-1+\sqrt{3}\right )-2 x^4}{\sqrt{2+x^6}} \, dx+\frac{\left (5 \left (1-\sqrt{3}\right )\right ) \int \frac{1}{\sqrt{2+x^6}} \, dx}{4 \sqrt [3]{2}}\\ &=\frac{1}{8} x^5 \sqrt{2+x^6}-\frac{5 \left (1+\sqrt{3}\right ) x \sqrt{2+x^6}}{8 \left (\sqrt [3]{2}+\left (1+\sqrt{3}\right ) x^2\right )}+\frac{5 \sqrt [4]{3} 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}} E\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\ 2^{2/3} \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}}+\frac{5 \left (1-\sqrt{3}\right ) 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 )}{8\ 2^{2/3} \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 = 0.0067514, size = 41, normalized size = 0.1 \[ \frac{1}{8} x^5 \left (\sqrt{x^6+2}-\sqrt{2} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};-\frac{x^6}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/Sqrt[2 + x^6],x]

[Out]

(x^5*(Sqrt[2 + x^6] - Sqrt[2]*Hypergeometric2F1[1/2, 5/6, 11/6, -x^6/2]))/8

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Maple [C]  time = 0.032, size = 33, normalized size = 0.1 \begin{align*}{\frac{{x}^{5}}{8}\sqrt{{x}^{6}+2}}-{\frac{{x}^{5}\sqrt{2}}{8}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{5}{6}};\,{\frac{11}{6}};\,-{\frac{{x}^{6}}{2}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(x^6+2)^(1/2),x)

[Out]

1/8*x^5*(x^6+2)^(1/2)-1/8*2^(1/2)*x^5*hypergeom([1/2,5/6],[11/6],-1/2*x^6)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{\sqrt{x^{6} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^6+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^10/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^{10}}{\sqrt{x^{6} + 2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^6+2)^(1/2),x, algorithm="fricas")

[Out]

integral(x^10/sqrt(x^6 + 2), x)

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Sympy [C]  time = 0.970959, size = 36, normalized size = 0.09 \begin{align*} \frac{\sqrt{2} x^{11} \Gamma \left (\frac{11}{6}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{11}{6} \\ \frac{17}{6} \end{matrix}\middle |{\frac{x^{6} e^{i \pi }}{2}} \right )}}{12 \Gamma \left (\frac{17}{6}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(x**6+2)**(1/2),x)

[Out]

sqrt(2)*x**11*gamma(11/6)*hyper((1/2, 11/6), (17/6,), x**6*exp_polar(I*pi)/2)/(12*gamma(17/6))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{\sqrt{x^{6} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^6+2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^10/sqrt(x^6 + 2), x)