3.1432 \(\int \frac {x^{10}}{(2+x^6)^{3/2}} \, dx\)

Optimal. Leaf size=392 \[ -\frac {x^5}{3 \sqrt {x^6+2}}+\frac {5 \left (1+\sqrt {3}\right ) \sqrt {x^6+2} x}{6 \left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )}-\frac {5 \left (1-\sqrt {3}\right ) \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} x F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{6\ 2^{2/3} \sqrt [4]{3} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}}-\frac {5 \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} x E\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2^{2/3} 3^{3/4} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}} \]

[Out]

-1/3*x^5/(x^6+2)^(1/2)+5/6*x*(1+3^(1/2))*(x^6+2)^(1/2)/(2^(1/3)+x^2*(1+3^(1/2)))-5/6*3^(1/4)*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)))*EllipticE((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^(1/3)/(x^6+2)^(1/2)/(x^2*(2^(1/3)+x^2)/(2^(1/3)+x^
2*(1+3^(1/2)))^2)^(1/2)-5/36*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))*(1-3^(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^(1/3)*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 = 0.11, antiderivative size = 392, normalized size of antiderivative = 1.00, 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 = {288, 308, 225, 1881} \[ -\frac {x^5}{3 \sqrt {x^6+2}}+\frac {5 \left (1+\sqrt {3}\right ) \sqrt {x^6+2} x}{6 \left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )}-\frac {5 \left (1-\sqrt {3}\right ) \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} x F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{6\ 2^{2/3} \sqrt [4]{3} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}}-\frac {5 \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} x E\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) x^2+\sqrt [3]{2}}{\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2^{2/3} 3^{3/4} \sqrt {\frac {x^2 \left (x^2+\sqrt [3]{2}\right )}{\left (\left (1+\sqrt {3}\right ) x^2+\sqrt [3]{2}\right )^2}} \sqrt {x^6+2}} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(2 + x^6)^(3/2),x]

[Out]

-x^5/(3*Sqrt[2 + x^6]) + (5*(1 + Sqrt[3])*x*Sqrt[2 + x^6])/(6*(2^(1/3) + (1 + Sqrt[3])*x^2)) - (5*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 - Sqr
t[3])*x^2)/(2^(1/3) + (1 + Sqrt[3])*x^2)], (2 + Sqrt[3])/4])/(2^(2/3)*3^(3/4)*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])/(6*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 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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 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}}{\left (2+x^6\right )^{3/2}} \, dx &=-\frac {x^5}{3 \sqrt {2+x^6}}+\frac {5}{3} \int \frac {x^4}{\sqrt {2+x^6}} \, dx\\ &=-\frac {x^5}{3 \sqrt {2+x^6}}-\frac {5}{6} \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}{3 \sqrt [3]{2}}\\ &=-\frac {x^5}{3 \sqrt {2+x^6}}+\frac {5 \left (1+\sqrt {3}\right ) x \sqrt {2+x^6}}{6 \left (\sqrt [3]{2}+\left (1+\sqrt {3}\right ) x^2\right )}-\frac {5 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 )}{2^{2/3} 3^{3/4} \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 )}{6\ 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*}

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Mathematica [C]  time = 0.02, size = 43, normalized size = 0.11 \[ \frac {1}{4} x^5 \left (\frac {2}{\sqrt {x^6+2}}-\sqrt {2} \, _2F_1\left (\frac {5}{6},\frac {3}{2};\frac {11}{6};-\frac {x^6}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(2 + x^6)^(3/2),x]

[Out]

(x^5*(2/Sqrt[2 + x^6] - Sqrt[2]*Hypergeometric2F1[5/6, 3/2, 11/6, -1/2*x^6]))/4

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{6} + 2} x^{10}}{x^{12} + 4 \, x^{6} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^6+2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^6 + 2)*x^10/(x^12 + 4*x^6 + 4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{{\left (x^{6} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^6+2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^10/(x^6 + 2)^(3/2), x)

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maple [C]  time = 0.13, size = 33, normalized size = 0.08 \[ \frac {\sqrt {2}\, x^{5} \hypergeom \left (\left [\frac {1}{2}, \frac {5}{6}\right ], \left [\frac {11}{6}\right ], -\frac {x^{6}}{2}\right )}{6}-\frac {x^{5}}{3 \sqrt {x^{6}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(x^6+2)^(3/2),x)

[Out]

-1/3*x^5/(x^6+2)^(1/2)+1/6*2^(1/2)*x^5*hypergeom([1/2,5/6],[11/6],-1/2*x^6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{{\left (x^{6} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(x^6+2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^10/(x^6 + 2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^{10}}{{\left (x^6+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(x^6 + 2)^(3/2),x)

[Out]

int(x^10/(x^6 + 2)^(3/2), x)

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sympy [C]  time = 1.98, size = 36, normalized size = 0.09 \[ \frac {\sqrt {2} x^{11} \Gamma \left (\frac {11}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {11}{6} \\ \frac {17}{6} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{24 \Gamma \left (\frac {17}{6}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(x**6+2)**(3/2),x)

[Out]

sqrt(2)*x**11*gamma(11/6)*hyper((3/2, 11/6), (17/6,), x**6*exp_polar(I*pi)/2)/(24*gamma(17/6))

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