### 3.132 $$\int \frac{1}{x^3 \sqrt{-16+x^2}} \, dx$$

Optimal. Leaf size=35 $\frac{\sqrt{x^2-16}}{32 x^2}+\frac{1}{128} \tan ^{-1}\left (\frac{\sqrt{x^2-16}}{4}\right )$

[Out]

Sqrt[-16 + x^2]/(32*x^2) + ArcTan[Sqrt[-16 + x^2]/4]/128

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Rubi [A]  time = 0.0132875, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.308, Rules used = {266, 51, 63, 203} $\frac{\sqrt{x^2-16}}{32 x^2}+\frac{1}{128} \tan ^{-1}\left (\frac{\sqrt{x^2-16}}{4}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(x^3*Sqrt[-16 + x^2]),x]

[Out]

Sqrt[-16 + x^2]/(32*x^2) + ArcTan[Sqrt[-16 + x^2]/4]/128

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 203

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

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt{-16+x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-16+x} x^2} \, dx,x,x^2\right )\\ &=\frac{\sqrt{-16+x^2}}{32 x^2}+\frac{1}{64} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-16+x} x} \, dx,x,x^2\right )\\ &=\frac{\sqrt{-16+x^2}}{32 x^2}+\frac{1}{32} \operatorname{Subst}\left (\int \frac{1}{16+x^2} \, dx,x,\sqrt{-16+x^2}\right )\\ &=\frac{\sqrt{-16+x^2}}{32 x^2}+\frac{1}{128} \tan ^{-1}\left (\frac{1}{4} \sqrt{-16+x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0203663, size = 46, normalized size = 1.31 $\frac{1}{256} \sqrt{x^2-16} \left (\frac{8}{x^2}+\frac{2 \tanh ^{-1}\left (\sqrt{1-\frac{x^2}{16}}\right )}{\sqrt{16-x^2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(x^3*Sqrt[-16 + x^2]),x]

[Out]

(Sqrt[-16 + x^2]*(8/x^2 + (2*ArcTanh[Sqrt[1 - x^2/16]])/Sqrt[16 - x^2]))/256

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Maple [A]  time = 0.005, size = 26, normalized size = 0.7 \begin{align*}{\frac{1}{32\,{x}^{2}}\sqrt{{x}^{2}-16}}-{\frac{1}{128}\arctan \left ( 4\,{\frac{1}{\sqrt{{x}^{2}-16}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^2-16)^(1/2),x)

[Out]

1/32*(x^2-16)^(1/2)/x^2-1/128*arctan(4/(x^2-16)^(1/2))

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Maxima [A]  time = 1.40181, size = 30, normalized size = 0.86 \begin{align*} \frac{\sqrt{x^{2} - 16}}{32 \, x^{2}} - \frac{1}{128} \, \arcsin \left (\frac{4}{{\left | x \right |}}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^2-16)^(1/2),x, algorithm="maxima")

[Out]

1/32*sqrt(x^2 - 16)/x^2 - 1/128*arcsin(4/abs(x))

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Fricas [A]  time = 1.87866, size = 96, normalized size = 2.74 \begin{align*} \frac{x^{2} \arctan \left (-\frac{1}{4} \, x + \frac{1}{4} \, \sqrt{x^{2} - 16}\right ) + 2 \, \sqrt{x^{2} - 16}}{64 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^2-16)^(1/2),x, algorithm="fricas")

[Out]

1/64*(x^2*arctan(-1/4*x + 1/4*sqrt(x^2 - 16)) + 2*sqrt(x^2 - 16))/x^2

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Sympy [A]  time = 2.23, size = 66, normalized size = 1.89 \begin{align*} \begin{cases} \frac{i \operatorname{acosh}{\left (\frac{4}{x} \right )}}{128} + \frac{i \sqrt{-1 + \frac{16}{x^{2}}}}{32 x} & \text{for}\: \frac{16}{\left |{x^{2}}\right |} > 1 \\- \frac{\operatorname{asin}{\left (\frac{4}{x} \right )}}{128} + \frac{1}{32 x \sqrt{1 - \frac{16}{x^{2}}}} - \frac{1}{2 x^{3} \sqrt{1 - \frac{16}{x^{2}}}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(x**2-16)**(1/2),x)

[Out]

Piecewise((I*acosh(4/x)/128 + I*sqrt(-1 + 16/x**2)/(32*x), 16/Abs(x**2) > 1), (-asin(4/x)/128 + 1/(32*x*sqrt(1
- 16/x**2)) - 1/(2*x**3*sqrt(1 - 16/x**2)), True))

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Giac [A]  time = 1.05215, size = 34, normalized size = 0.97 \begin{align*} \frac{\sqrt{x^{2} - 16}}{32 \, x^{2}} + \frac{1}{128} \, \arctan \left (\frac{1}{4} \, \sqrt{x^{2} - 16}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^2-16)^(1/2),x, algorithm="giac")

[Out]

1/32*sqrt(x^2 - 16)/x^2 + 1/128*arctan(1/4*sqrt(x^2 - 16))