3.973 \(\int \frac {1}{x^4 \sqrt {16-x^4}} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{96} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {\sqrt {16-x^4}}{48 x^3} \]

[Out]

1/96*EllipticF(1/2*x,I)-1/48*(-x^4+16)^(1/2)/x^3

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Rubi [A]  time = 0.01, antiderivative size = 31, 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 = {325, 221} \[ \frac {1}{96} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {\sqrt {16-x^4}}{48 x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[16 - x^4]),x]

[Out]

-Sqrt[16 - x^4]/(48*x^3) + EllipticF[ArcSin[x/2], -1]/96

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [C]  time = 0.00, size = 24, normalized size = 0.77 \[ -\frac {\, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};\frac {x^4}{16}\right )}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[16 - x^4]),x]

[Out]

-1/12*Hypergeometric2F1[-3/4, 1/2, 1/4, x^4/16]/x^3

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + 16}}{x^{8} - 16 \, x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + 16)/(x^8 - 16*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-x^{4} + 16} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 16)*x^4), x)

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maple [B]  time = 0.01, size = 49, normalized size = 1.58 \[ \frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \EllipticF \left (\frac {x}{2}, i\right )}{96 \sqrt {-x^{4}+16}}-\frac {\sqrt {-x^{4}+16}}{48 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(-x^4+16)^(1/2)/x^3+1/96*(-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 \[ \int \frac {1}{\sqrt {-x^{4} + 16} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 16)*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(16 - x^4)^(1/2)),x)

[Out]

int(1/(x^4*(16 - x^4)^(1/2)), x)

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sympy [A]  time = 1.11, size = 36, normalized size = 1.16 \[ \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), x**4*exp_polar(2*I*pi)/16)/(16*x**3*gamma(1/4))

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