3.966 \(\int \frac{1}{x^5 \sqrt{16-x^4}} \, dx\)

Optimal. Leaf size=39 \[ -\frac{\sqrt{16-x^4}}{64 x^4}-\frac{1}{256} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right ) \]

[Out]

-Sqrt[16 - x^4]/(64*x^4) - ArcTanh[Sqrt[16 - x^4]/4]/256

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Rubi [A]  time = 0.0158811, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 206} \[ -\frac{\sqrt{16-x^4}}{64 x^4}-\frac{1}{256} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[16 - x^4]),x]

[Out]

-Sqrt[16 - x^4]/(64*x^4) - ArcTanh[Sqrt[16 - x^4]/4]/256

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^5 \sqrt{16-x^4}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{16-x} x^2} \, dx,x,x^4\right )\\ &=-\frac{\sqrt{16-x^4}}{64 x^4}+\frac{1}{128} \operatorname{Subst}\left (\int \frac{1}{\sqrt{16-x} x} \, dx,x,x^4\right )\\ &=-\frac{\sqrt{16-x^4}}{64 x^4}-\frac{1}{64} \operatorname{Subst}\left (\int \frac{1}{16-x^2} \, dx,x,\sqrt{16-x^4}\right )\\ &=-\frac{\sqrt{16-x^4}}{64 x^4}-\frac{1}{256} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right )\\ \end{align*}

Mathematica [A]  time = 0.0116212, size = 37, normalized size = 0.95 \[ \frac{1}{256} \left (-\frac{4 \sqrt{16-x^4}}{x^4}-\tanh ^{-1}\left (\sqrt{1-\frac{x^4}{16}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[16 - x^4]),x]

[Out]

((-4*Sqrt[16 - x^4])/x^4 - ArcTanh[Sqrt[1 - x^4/16]])/256

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Maple [A]  time = 0.01, size = 30, normalized size = 0.8 \begin{align*} -{\frac{1}{64\,{x}^{4}}\sqrt{-{x}^{4}+16}}-{\frac{1}{256}{\it Artanh} \left ( 4\,{\frac{1}{\sqrt{-{x}^{4}+16}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-x^4+16)^(1/2),x)

[Out]

-1/64*(-x^4+16)^(1/2)/x^4-1/256*arctanh(4/(-x^4+16)^(1/2))

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Maxima [A]  time = 0.970023, size = 58, normalized size = 1.49 \begin{align*} -\frac{\sqrt{-x^{4} + 16}}{64 \, x^{4}} - \frac{1}{512} \, \log \left (\sqrt{-x^{4} + 16} + 4\right ) + \frac{1}{512} \, \log \left (\sqrt{-x^{4} + 16} - 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^4+16)^(1/2),x, algorithm="maxima")

[Out]

-1/64*sqrt(-x^4 + 16)/x^4 - 1/512*log(sqrt(-x^4 + 16) + 4) + 1/512*log(sqrt(-x^4 + 16) - 4)

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Fricas [A]  time = 1.46552, size = 127, normalized size = 3.26 \begin{align*} -\frac{x^{4} \log \left (\sqrt{-x^{4} + 16} + 4\right ) - x^{4} \log \left (\sqrt{-x^{4} + 16} - 4\right ) + 8 \, \sqrt{-x^{4} + 16}}{512 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^4+16)^(1/2),x, algorithm="fricas")

[Out]

-1/512*(x^4*log(sqrt(-x^4 + 16) + 4) - x^4*log(sqrt(-x^4 + 16) - 4) + 8*sqrt(-x^4 + 16))/x^4

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Sympy [A]  time = 2.53298, size = 73, normalized size = 1.87 \begin{align*} \begin{cases} - \frac{\operatorname{acosh}{\left (\frac{4}{x^{2}} \right )}}{256} - \frac{\sqrt{-1 + \frac{16}{x^{4}}}}{64 x^{2}} & \text{for}\: \frac{16}{\left |{x^{4}}\right |} > 1 \\\frac{i \operatorname{asin}{\left (\frac{4}{x^{2}} \right )}}{256} - \frac{i}{64 x^{2} \sqrt{1 - \frac{16}{x^{4}}}} + \frac{i}{4 x^{6} \sqrt{1 - \frac{16}{x^{4}}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-x**4+16)**(1/2),x)

[Out]

Piecewise((-acosh(4/x**2)/256 - sqrt(-1 + 16/x**4)/(64*x**2), 16/Abs(x**4) > 1), (I*asin(4/x**2)/256 - I/(64*x
**2*sqrt(1 - 16/x**4)) + I/(4*x**6*sqrt(1 - 16/x**4)), True))

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Giac [A]  time = 1.14318, size = 61, normalized size = 1.56 \begin{align*} -\frac{\sqrt{-x^{4} + 16}}{64 \, x^{4}} - \frac{1}{512} \, \log \left (\sqrt{-x^{4} + 16} + 4\right ) + \frac{1}{512} \, \log \left (-\sqrt{-x^{4} + 16} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^4+16)^(1/2),x, algorithm="giac")

[Out]

-1/64*sqrt(-x^4 + 16)/x^4 - 1/512*log(sqrt(-x^4 + 16) + 4) + 1/512*log(-sqrt(-x^4 + 16) + 4)