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

Optimal. Leaf size=20 \[ -\frac{1}{8} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right ) \]

[Out]

-ArcTanh[Sqrt[16 - x^4]/4]/8

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Rubi [A]  time = 0.01006, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {266, 63, 206} \[ -\frac{1}{8} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[16 - x^4]),x]

[Out]

-ArcTanh[Sqrt[16 - x^4]/4]/8

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 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 \sqrt{16-x^4}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{16-x} x} \, dx,x,x^4\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{16-x^2} \, dx,x,\sqrt{16-x^4}\right )\right )\\ &=-\frac{1}{8} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right )\\ \end{align*}

Mathematica [A]  time = 0.002583, size = 20, normalized size = 1. \[ -\frac{1}{8} \tanh ^{-1}\left (\frac{\sqrt{16-x^4}}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[16 - x^4]),x]

[Out]

-ArcTanh[Sqrt[16 - x^4]/4]/8

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Maple [A]  time = 0.008, size = 15, normalized size = 0.8 \begin{align*} -{\frac{1}{8}{\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/(-x^4+16)^(1/2),x)

[Out]

-1/8*arctanh(4/(-x^4+16)^(1/2))

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Maxima [B]  time = 0.964051, size = 39, normalized size = 1.95 \begin{align*} -\frac{1}{16} \, \log \left (\sqrt{-x^{4} + 16} + 4\right ) + \frac{1}{16} \, \log \left (\sqrt{-x^{4} + 16} - 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.49476, size = 86, normalized size = 4.3 \begin{align*} -\frac{1}{16} \, \log \left (\sqrt{-x^{4} + 16} + 4\right ) + \frac{1}{16} \, \log \left (\sqrt{-x^{4} + 16} - 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.30896, size = 24, normalized size = 1.2 \begin{align*} \begin{cases} - \frac{\operatorname{acosh}{\left (\frac{4}{x^{2}} \right )}}{8} & \text{for}\: \frac{16}{\left |{x^{4}}\right |} > 1 \\\frac{i \operatorname{asin}{\left (\frac{4}{x^{2}} \right )}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-acosh(4/x**2)/8, 16/Abs(x**4) > 1), (I*asin(4/x**2)/8, True))

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Giac [B]  time = 1.23874, size = 42, normalized size = 2.1 \begin{align*} -\frac{1}{16} \, \log \left (\sqrt{-x^{4} + 16} + 4\right ) + \frac{1}{16} \, \log \left (-\sqrt{-x^{4} + 16} + 4\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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