3.969 \(\int \frac {x^4}{\sqrt {16-x^4}} \, dx\)

Optimal. Leaf size=29 \[ \frac {8}{3} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{3} x \sqrt {16-x^4} \]

[Out]

8/3*EllipticF(1/2*x,I)-1/3*x*(-x^4+16)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {321, 221} \[ \frac {8}{3} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{3} x \sqrt {16-x^4} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[16 - x^4],x]

[Out]

-(x*Sqrt[16 - x^4])/3 + (8*EllipticF[ArcSin[x/2], -1])/3

Rule 221

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 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]

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {16-x^4}} \, dx &=-\frac {1}{3} x \sqrt {16-x^4}+\frac {16}{3} \int \frac {1}{\sqrt {16-x^4}} \, dx\\ &=-\frac {1}{3} x \sqrt {16-x^4}+\frac {8}{3} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 36, normalized size = 1.24 \[ -\frac {1}{3} x \left (\sqrt {16-x^4}-4 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {x^4}{16}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[16 - x^4],x]

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + 16} x^{4}}{x^{4} - 16}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + 16)*x^4/(x^4 - 16), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {-x^{4} + 16}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/sqrt(-x^4 + 16), x)

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maple [B]  time = 0.01, size = 47, normalized size = 1.62 \[ -\frac {\sqrt {-x^{4}+16}\, x}{3}+\frac {8 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \EllipticF \left (\frac {x}{2}, i\right )}{3 \sqrt {-x^{4}+16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(-x^4+16)^(1/2),x)

[Out]

-1/3*x*(-x^4+16)^(1/2)+8/3*(-x^2+4)^(1/2)*(x^2+4)^(1/2)/(-x^4+16)^(1/2)*EllipticF(1/2*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {-x^{4} + 16}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/sqrt(-x^4 + 16), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^4}{\sqrt {16-x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(16 - x^4)^(1/2),x)

[Out]

int(x^4/(16 - x^4)^(1/2), x)

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sympy [A]  time = 0.99, size = 32, normalized size = 1.10 \[ \frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), x**4*exp_polar(2*I*pi)/16)/(16*gamma(9/4))

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