3.15.20 \(\int \frac {x^{13}}{(2+x^6)^{3/2}} \, dx\) [1420]

Optimal. Leaf size=202 \[ -\frac {x^8}{3 \sqrt {2+x^6}}+\frac {8}{15} x^2 \sqrt {2+x^6}-\frac {16\ 2^{5/6} \sqrt {2+\sqrt {3}} \left (\sqrt [3]{2}+x^2\right ) \sqrt {\frac {2^{2/3}-\sqrt [3]{2} x^2+x^4}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{2} \left (1-\sqrt {3}\right )+x^2}{\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} \sqrt {\frac {\sqrt [3]{2}+x^2}{\left (\sqrt [3]{2} \left (1+\sqrt {3}\right )+x^2\right )^2}} \sqrt {2+x^6}} \]

[Out]

-1/3*x^8/(x^6+2)^(1/2)+8/15*x^2*(x^6+2)^(1/2)-16/45*2^(5/6)*(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)*(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)*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)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 202, 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 = {281, 294, 327, 224} \begin {gather*} -\frac {16\ 2^{5/6} \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}} F\left (\text {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 )}{15 \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 {x^8}{3 \sqrt {x^6+2}}+\frac {8}{15} \sqrt {x^6+2} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^13/(2 + x^6)^(3/2),x]

[Out]

-1/3*x^8/Sqrt[2 + x^6] + (8*x^2*Sqrt[2 + x^6])/15 - (16*2^(5/6)*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]*EllipticF[ArcSin[(2^(1/3)*(1 - Sqrt[3]) + x^2)/(2^(1/3
)*(1 + Sqrt[3]) + x^2)], -7 - 4*Sqrt[3]])/(15*3^(1/4)*Sqrt[(2^(1/3) + x^2)/(2^(1/3)*(1 + Sqrt[3]) + x^2)^2]*Sq
rt[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 294

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 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]

Rubi steps

\begin {align*} \int \frac {x^{13}}{\left (2+x^6\right )^{3/2}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^6}{\left (2+x^3\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {x^8}{3 \sqrt {2+x^6}}+\frac {4}{3} \text {Subst}\left (\int \frac {x^3}{\sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=-\frac {x^8}{3 \sqrt {2+x^6}}+\frac {8}{15} x^2 \sqrt {2+x^6}-\frac {16}{15} \text {Subst}\left (\int \frac {1}{\sqrt {2+x^3}} \, dx,x,x^2\right )\\ &=-\frac {x^8}{3 \sqrt {2+x^6}}+\frac {8}{15} x^2 \sqrt {2+x^6}-\frac {16\ 2^{5/6} \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}} 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 )}{15 \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.82, size = 56, normalized size = 0.28 \begin {gather*} \frac {x^2 \left (16+3 x^6-8 \sqrt {2} \sqrt {2+x^6} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {x^6}{2}\right )\right )}{15 \sqrt {2+x^6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^13/(2 + x^6)^(3/2),x]

[Out]

(x^2*(16 + 3*x^6 - 8*Sqrt[2]*Sqrt[2 + x^6]*Hypergeometric2F1[1/3, 1/2, 4/3, -1/2*x^6]))/(15*Sqrt[2 + x^6])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.19, size = 20, normalized size = 0.10

method	result	size
meijerg	\(\frac {\sqrt {2}\, x^{14} \hypergeom \left (\left [\frac {3}{2}, \frac {7}{3}\right ], \left [\frac {10}{3}\right ], -\frac {x^{6}}{2}\right )}{56}\)	\(20\)
risch	\(\frac {x^{2} \left (3 x^{6}+16\right )}{15 \sqrt {x^{6}+2}}-\frac {8 \sqrt {2}\, x^{2} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {4}{3}\right ], -\frac {x^{6}}{2}\right )}{15}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(x^6+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/56*2^(1/2)*x^14*hypergeom([3/2,7/3],[10/3],-1/2*x^6)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(x^6+2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^13/(x^6 + 2)^(3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.07, size = 43, normalized size = 0.21 \begin {gather*} -\frac {32 \, {\left (x^{6} + 2\right )} {\rm weierstrassPInverse}\left (0, -8, x^{2}\right ) - {\left (3 \, x^{8} + 16 \, x^{2}\right )} \sqrt {x^{6} + 2}}{15 \, {\left (x^{6} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(x^6+2)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(32*(x^6 + 2)*weierstrassPInverse(0, -8, x^2) - (3*x^8 + 16*x^2)*sqrt(x^6 + 2))/(x^6 + 2)

________________________________________________________________________________________

Sympy [A]
time = 0.65, size = 36, normalized size = 0.18 \begin {gather*} \frac {\sqrt {2} x^{14} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {x^{6} e^{i \pi }}{2}} \right )}}{24 \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13/(x**6+2)**(3/2),x)

[Out]

sqrt(2)*x**14*gamma(7/3)*hyper((3/2, 7/3), (10/3,), x**6*exp_polar(I*pi)/2)/(24*gamma(10/3))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(x^6+2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^13/(x^6 + 2)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{13}}{{\left (x^6+2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(x^6 + 2)^(3/2),x)

[Out]

int(x^13/(x^6 + 2)^(3/2), x)

________________________________________________________________________________________