3.58 \(\int \frac{\cos ^{-1}(\frac{a}{x})}{x^3} \, dx\)

Optimal. Leaf size=51 \[ \frac{\sqrt{1-\frac{a^2}{x^2}}}{4 a x}-\frac{\csc ^{-1}\left (\frac{x}{a}\right )}{4 a^2}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{2 x^2} \]

[Out]

Sqrt[1 - a^2/x^2]/(4*a*x) - ArcCsc[x/a]/(4*a^2) - ArcSec[x/a]/(2*x^2)

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Rubi [A]  time = 0.0325045, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4833, 5220, 335, 321, 216} \[ \frac{\sqrt{1-\frac{a^2}{x^2}}}{4 a x}-\frac{\csc ^{-1}\left (\frac{x}{a}\right )}{4 a^2}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a/x]/x^3,x]

[Out]

Sqrt[1 - a^2/x^2]/(4*a*x) - ArcCsc[x/a]/(4*a^2) - ArcSec[x/a]/(2*x^2)

Rule 4833

Int[ArcCos[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcSec[a/c + (b*x^n)/c]^m, x] /;
FreeQ[{a, b, c, n, m}, x]

Rule 5220

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSec[c*x]
))/(d*(m + 1)), x] - Dist[(b*d)/(c*(m + 1)), Int[(d*x)^(m - 1)/Sqrt[1 - 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c
, d, m}, x] && NeQ[m, -1]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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]

Rule 216

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

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}\left (\frac{a}{x}\right )}{x^3} \, dx &=\int \frac{\sec ^{-1}\left (\frac{x}{a}\right )}{x^3} \, dx\\ &=-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{2 x^2}+\frac{1}{2} a \int \frac{1}{\sqrt{1-\frac{a^2}{x^2}} x^4} \, dx\\ &=-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{2 x^2}-\frac{1}{2} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-a^2 x^2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{1-\frac{a^2}{x^2}}}{4 a x}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{2 x^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx,x,\frac{1}{x}\right )}{4 a}\\ &=\frac{\sqrt{1-\frac{a^2}{x^2}}}{4 a x}-\frac{\csc ^{-1}\left (\frac{x}{a}\right )}{4 a^2}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0244137, size = 50, normalized size = 0.98 \[ \frac{a x \sqrt{1-\frac{a^2}{x^2}}-2 a^2 \cos ^{-1}\left (\frac{a}{x}\right )-x^2 \sin ^{-1}\left (\frac{a}{x}\right )}{4 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a/x]/x^3,x]

[Out]

(a*Sqrt[1 - a^2/x^2]*x - 2*a^2*ArcCos[a/x] - x^2*ArcSin[a/x])/(4*a^2*x^2)

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Maple [A]  time = 0.006, size = 47, normalized size = 0.9 \begin{align*} -{\frac{1}{{a}^{2}} \left ({\frac{{a}^{2}}{2\,{x}^{2}}\arccos \left ({\frac{a}{x}} \right ) }-{\frac{a}{4\,x}\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}}}+{\frac{1}{4}\arcsin \left ({\frac{a}{x}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a/x)/x^3,x)

[Out]

-1/a^2*(1/2*arccos(a/x)*a^2/x^2-1/4*a/x*(1-a^2/x^2)^(1/2)+1/4*arcsin(a/x))

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Maxima [A]  time = 1.46544, size = 104, normalized size = 2.04 \begin{align*} -\frac{1}{4} \, a{\left (\frac{x \sqrt{-\frac{a^{2}}{x^{2}} + 1}}{a^{2} x^{2}{\left (\frac{a^{2}}{x^{2}} - 1\right )} - a^{4}} - \frac{\arctan \left (\frac{x \sqrt{-\frac{a^{2}}{x^{2}} + 1}}{a}\right )}{a^{3}}\right )} - \frac{\arccos \left (\frac{a}{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a/x)/x^3,x, algorithm="maxima")

[Out]

-1/4*a*(x*sqrt(-a^2/x^2 + 1)/(a^2*x^2*(a^2/x^2 - 1) - a^4) - arctan(x*sqrt(-a^2/x^2 + 1)/a)/a^3) - 1/2*arccos(
a/x)/x^2

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Fricas [A]  time = 2.47003, size = 97, normalized size = 1.9 \begin{align*} \frac{a x \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}} -{\left (2 \, a^{2} - x^{2}\right )} \arccos \left (\frac{a}{x}\right )}{4 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a/x)/x^3,x, algorithm="fricas")

[Out]

1/4*(a*x*sqrt(-(a^2 - x^2)/x^2) - (2*a^2 - x^2)*arccos(a/x))/(a^2*x^2)

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Sympy [C]  time = 5.30487, size = 102, normalized size = 2. \begin{align*} \frac{a \left (\begin{cases} \frac{i \sqrt{\frac{a^{2}}{x^{2}} - 1}}{2 a^{2} x} + \frac{i \operatorname{acosh}{\left (\frac{a}{x} \right )}}{2 a^{3}} & \text{for}\: \frac{\left |{a^{2}}\right |}{\left |{x^{2}}\right |} > 1 \\- \frac{1}{2 x^{3} \sqrt{- \frac{a^{2}}{x^{2}} + 1}} + \frac{1}{2 a^{2} x \sqrt{- \frac{a^{2}}{x^{2}} + 1}} - \frac{\operatorname{asin}{\left (\frac{a}{x} \right )}}{2 a^{3}} & \text{otherwise} \end{cases}\right )}{2} - \frac{\operatorname{acos}{\left (\frac{a}{x} \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a/x)/x**3,x)

[Out]

a*Piecewise((I*sqrt(a**2/x**2 - 1)/(2*a**2*x) + I*acosh(a/x)/(2*a**3), Abs(a**2)/Abs(x**2) > 1), (-1/(2*x**3*s
qrt(-a**2/x**2 + 1)) + 1/(2*a**2*x*sqrt(-a**2/x**2 + 1)) - asin(a/x)/(2*a**3), True))/2 - acos(a/x)/(2*x**2)

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Giac [A]  time = 1.28026, size = 84, normalized size = 1.65 \begin{align*} \frac{1}{4} \, a{\left (\frac{\arctan \left (\frac{\sqrt{-a^{2} + x^{2}}}{a}\right )}{a^{3} \mathrm{sgn}\left (x\right )} + \frac{\sqrt{-a^{2} + x^{2}}}{a^{2} x^{2} \mathrm{sgn}\left (x\right )}\right )} - \frac{\arccos \left (\frac{a}{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a/x)/x^3,x, algorithm="giac")

[Out]

1/4*a*(arctan(sqrt(-a^2 + x^2)/a)/(a^3*sgn(x)) + sqrt(-a^2 + x^2)/(a^2*x^2*sgn(x))) - 1/2*arccos(a/x)/x^2