3.10.68 \(\int \frac {x^6}{\sqrt {16-x^4}} \, dx\) [968]

Optimal. Leaf size=43 \[ -\frac {1}{5} x^3 \sqrt {16-x^4}+\frac {96}{5} E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \]

[Out]

96/5*EllipticE(1/2*x,I)-96/5*EllipticF(1/2*x,I)-1/5*x^3*(-x^4+16)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {327, 313, 227, 1195, 21, 435} \begin {gather*} -\frac {96}{5} F\left (\left .\text {ArcSin}\left (\frac {x}{2}\right )\right |-1\right )+\frac {96}{5} E\left (\left .\text {ArcSin}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{5} \sqrt {16-x^4} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[16 - x^4],x]

[Out]

-1/5*(x^3*Sqrt[16 - x^4]) + (96*EllipticE[ArcSin[x/2], -1])/5 - (96*EllipticF[ArcSin[x/2], -1])/5

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/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 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {x^6}{\sqrt {16-x^4}} \, dx &=-\frac {1}{5} x^3 \sqrt {16-x^4}+\frac {48}{5} \int \frac {x^2}{\sqrt {16-x^4}} \, dx\\ &=-\frac {1}{5} x^3 \sqrt {16-x^4}-\frac {192}{5} \int \frac {1}{\sqrt {16-x^4}} \, dx+\frac {192}{5} \int \frac {1+\frac {x^2}{4}}{\sqrt {16-x^4}} \, dx\\ &=-\frac {1}{5} x^3 \sqrt {16-x^4}-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\frac {192}{5} \int \frac {1+\frac {x^2}{4}}{\sqrt {4-x^2} \sqrt {4+x^2}} \, dx\\ &=-\frac {1}{5} x^3 \sqrt {16-x^4}-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\frac {48}{5} \int \frac {\sqrt {4+x^2}}{\sqrt {4-x^2}} \, dx\\ &=-\frac {1}{5} x^3 \sqrt {16-x^4}+\frac {96}{5} E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {96}{5} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 38, normalized size = 0.88 \begin {gather*} -\frac {1}{5} x^3 \left (\sqrt {16-x^4}-4 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {x^4}{16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[16 - x^4],x]

[Out]

-1/5*(x^3*(Sqrt[16 - x^4] - 4*Hypergeometric2F1[1/2, 3/4, 7/4, x^4/16]))

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Maple [A]
time = 0.18, size = 58, normalized size = 1.35

method	result	size
meijerg	\(\frac {x^{7} \hypergeom \left (\left [\frac {1}{2}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], \frac {x^{4}}{16}\right )}{28}\)	\(17\)
default	\(-\frac {x^{3} \sqrt {-x^{4}+16}}{5}-\frac {96 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (\frac {x}{2}, i\right )-\EllipticE \left (\frac {x}{2}, i\right )\right )}{5 \sqrt {-x^{4}+16}}\)	\(58\)
elliptic	\(-\frac {x^{3} \sqrt {-x^{4}+16}}{5}-\frac {96 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (\frac {x}{2}, i\right )-\EllipticE \left (\frac {x}{2}, i\right )\right )}{5 \sqrt {-x^{4}+16}}\)	\(58\)
risch	\(\frac {x^{3} \left (x^{4}-16\right )}{5 \sqrt {-x^{4}+16}}-\frac {96 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (\frac {x}{2}, i\right )-\EllipticE \left (\frac {x}{2}, i\right )\right )}{5 \sqrt {-x^{4}+16}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(-x^4+16)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/5*x^3*(-x^4+16)^(1/2)-96/5*(-x^2+4)^(1/2)*(x^2+4)^(1/2)/(-x^4+16)^(1/2)*(EllipticF(1/2*x,I)-EllipticE(1/2*x
,I))

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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^6/(-x^4+16)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(-x^4 + 16), x)

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Fricas [A]
time = 0.08, size = 19, normalized size = 0.44 \begin {gather*} -\frac {{\left (x^{4} + 48\right )} \sqrt {-x^{4} + 16}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^4+16)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(x^4 + 48)*sqrt(-x^4 + 16)/x

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Sympy [A]
time = 0.38, size = 32, normalized size = 0.74 \begin {gather*} \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(-x**4+16)**(1/2),x)

[Out]

x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), x**4*exp_polar(2*I*pi)/16)/(16*gamma(11/4))

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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^6/(-x^4+16)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(-x^4 + 16), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^6}{\sqrt {16-x^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(16 - x^4)^(1/2),x)

[Out]

int(x^6/(16 - x^4)^(1/2), x)

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