\(\int \frac {x^9}{\sqrt {2+x^6}} \, dx\) [1402]

Optimal result

Integrand size = 13, antiderivative size = 378 \[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\frac {1}{7} x^4 \sqrt {2+x^6}-\frac {8 \sqrt {2+x^6}}{7 \left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )}+\frac {4 \sqrt [6]{2} \sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{7 \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}}-\frac {8\ 2^{2/3} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right ),-7-4 \sqrt {3}\right )}{7 \sqrt [4]{3} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}} \]

[Out]

1/7*x^4*(x^6+2)^(1/2)-8/7*(x^6+2)^(1/2)/(x^2+2^(1/3)*(1+3^(1/2)))-8/21*2^(2/3)*(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)*((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)+4/7*2^(1/6)*3^(1/4)*(2^(1/
3)+x^2)*EllipticE((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)/(x^6+2)^(1/2)/((2^(1/3)+x^2)/(x^2+2^(1/3)*(1+3^
(1/2)))^2)^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {281, 327, 309, 224, 1891} \[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=-\frac {8\ 2^{2/3} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right ),-7-4 \sqrt {3}\right )}{7 \sqrt [4]{3} \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}+\frac {4 \sqrt [6]{2} \sqrt [4]{3} \sqrt {2-\sqrt {3}} \left (x^2+\sqrt [3]{2}\right ) \sqrt {\frac {x^4-\sqrt [3]{2} x^2+2^{2/3}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} E\left (\arcsin \left (\frac {x^2+\sqrt [3]{2} \left (1-\sqrt {3}\right )}{x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )}\right )|-7-4 \sqrt {3}\right )}{7 \sqrt {\frac {x^2+\sqrt [3]{2}}{\left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )^2}} \sqrt {x^6+2}}+\frac {1}{7} \sqrt {x^6+2} x^4-\frac {8 \sqrt {x^6+2}}{7 \left (x^2+\sqrt [3]{2} \left (1+\sqrt {3}\right )\right )} \]

[In]

Int[x^9/Sqrt[2 + x^6],x]

[Out]

(x^4*Sqrt[2 + x^6])/7 - (8*Sqrt[2 + x^6])/(7*(2^(1/3)*(1 + Sqrt[3]) + x^2)) + (4*2^(1/6)*3^(1/4)*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]*EllipticE[ArcSin[(2^(
1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(7*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1
+ Sqrt[3]) + x^2)^2]*Sqrt[2 + x^6]) - (8*2^(2/3)*(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]*EllipticF[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)], -7 - 4*
Sqrt[3]])/(7*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]

Rule 309

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

Rule 327

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 1891

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[(1 - Sqrt[3])*(d/c)]]
, s = Denom[Simplify[(1 - Sqrt[3])*(d/c)]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x
] - Simp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(r^2
*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1
+ Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{\sqrt {2+x^3}} \, dx,x,x^2\right ) \\ & = \frac {1}{7} x^4 \sqrt {2+x^6}-\frac {4}{7} \text {Subst}\left (\int \frac {x}{\sqrt {2+x^3}} \, dx,x,x^2\right ) \\ & = \frac {1}{7} x^4 \sqrt {2+x^6}-\frac {4}{7} \text {Subst}\left (\int \frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x}{\sqrt {2+x^3}} \, dx,x,x^2\right )+\frac {1}{7} \left (4 \sqrt [3]{2} \left (1-\sqrt {3}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2+x^3}} \, dx,x,x^2\right ) \\ & = \frac {1}{7} x^4 \sqrt {2+x^6}-\frac {8 \sqrt {2+x^6}}{7 \left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )}+\frac {4 \sqrt [6]{2} \sqrt [4]{3} \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}} E\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 )}{7 \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}}-\frac {8\ 2^{2/3} \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 )}{7 \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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.11 \[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\frac {1}{7} x^4 \left (\sqrt {2+x^6}-\sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\frac {x^6}{2}\right )\right ) \]

[In]

Integrate[x^9/Sqrt[2 + x^6],x]

[Out]

(x^4*(Sqrt[2 + x^6] - Sqrt[2]*Hypergeometric2F1[1/2, 2/3, 5/3, -1/2*x^6]))/7

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 6.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.05

method	result	size
meijerg	\(\frac {\sqrt {2}\, x^{10} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};-\frac {x^{6}}{2}\right )}{20}\)	\(20\)
risch	\(\frac {x^{4} \sqrt {x^{6}+2}}{7}-\frac {\sqrt {2}\, x^{4} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};-\frac {x^{6}}{2}\right )}{7}\)	\(33\)

[In]

int(x^9/(x^6+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/20*2^(1/2)*x^10*hypergeom([1/2,5/3],[8/3],-1/2*x^6)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.06 \[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\frac {1}{7} \, \sqrt {x^{6} + 2} x^{4} + \frac {8}{7} \, {\rm weierstrassZeta}\left (0, -8, {\rm weierstrassPInverse}\left (0, -8, x^{2}\right )\right ) \]

[In]

integrate(x^9/(x^6+2)^(1/2),x, algorithm="fricas")

[Out]

1/7*sqrt(x^6 + 2)*x^4 + 8/7*weierstrassZeta(0, -8, weierstrassPInverse(0, -8, x^2))

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.10 \[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\frac {\sqrt {2} x^{10} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{12 \Gamma \left (\frac {8}{3}\right )} \]

[In]

integrate(x**9/(x**6+2)**(1/2),x)

[Out]

sqrt(2)*x**10*gamma(5/3)*hyper((1/2, 5/3), (8/3,), x**6*exp_polar(I*pi)/2)/(12*gamma(8/3))

Maxima [F]

\[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\int { \frac {x^{9}}{\sqrt {x^{6} + 2}} \,d x } \]

[In]

integrate(x^9/(x^6+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^9/sqrt(x^6 + 2), x)

Giac [F]

\[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\int { \frac {x^{9}}{\sqrt {x^{6} + 2}} \,d x } \]

[In]

integrate(x^9/(x^6+2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^9/sqrt(x^6 + 2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^9}{\sqrt {2+x^6}} \, dx=\int \frac {x^9}{\sqrt {x^6+2}} \,d x \]

[In]

int(x^9/(x^6 + 2)^(1/2),x)

[Out]

int(x^9/(x^6 + 2)^(1/2), x)