3.59 \(\int \frac{\cos ^{-1}(\frac{a}{x})}{x^4} \, dx\)

Optimal. Leaf size=56 \[ -\frac{\left (1-\frac{a^2}{x^2}\right )^{3/2}}{9 a^3}+\frac{\sqrt{1-\frac{a^2}{x^2}}}{3 a^3}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{3 x^3} \]

[Out]

Sqrt[1 - a^2/x^2]/(3*a^3) - (1 - a^2/x^2)^(3/2)/(9*a^3) - ArcSec[x/a]/(3*x^3)

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Rubi [A]  time = 0.0383783, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4833, 5220, 266, 43} \[ -\frac{\left (1-\frac{a^2}{x^2}\right )^{3/2}}{9 a^3}+\frac{\sqrt{1-\frac{a^2}{x^2}}}{3 a^3}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a/x]/x^4,x]

[Out]

Sqrt[1 - a^2/x^2]/(3*a^3) - (1 - a^2/x^2)^(3/2)/(9*a^3) - ArcSec[x/a]/(3*x^3)

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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}\left (\frac{a}{x}\right )}{x^4} \, dx &=\int \frac{\sec ^{-1}\left (\frac{x}{a}\right )}{x^4} \, dx\\ &=-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{3 x^3}+\frac{1}{3} a \int \frac{1}{\sqrt{1-\frac{a^2}{x^2}} x^5} \, dx\\ &=-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{3 x^3}-\frac{1}{6} a \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-a^2 x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{3 x^3}-\frac{1}{6} a \operatorname{Subst}\left (\int \left (\frac{1}{a^2 \sqrt{1-a^2 x}}-\frac{\sqrt{1-a^2 x}}{a^2}\right ) \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{\sqrt{1-\frac{a^2}{x^2}}}{3 a^3}-\frac{\left (1-\frac{a^2}{x^2}\right )^{3/2}}{9 a^3}-\frac{\sec ^{-1}\left (\frac{x}{a}\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0308231, size = 47, normalized size = 0.84 \[ \frac{x \sqrt{1-\frac{a^2}{x^2}} \left (a^2+2 x^2\right )-3 a^3 \cos ^{-1}\left (\frac{a}{x}\right )}{9 a^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a/x]/x^4,x]

[Out]

(Sqrt[1 - a^2/x^2]*x*(a^2 + 2*x^2) - 3*a^3*ArcCos[a/x])/(9*a^3*x^3)

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Maple [A]  time = 0.006, size = 55, normalized size = 1. \begin{align*} -{\frac{1}{{a}^{3}} \left ({\frac{{a}^{3}}{3\,{x}^{3}}\arccos \left ({\frac{a}{x}} \right ) }-{\frac{{a}^{2}}{9\,{x}^{2}}\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}}}-{\frac{2}{9}\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a/x)/x^4,x)

[Out]

-1/a^3*(1/3*arccos(a/x)*a^3/x^3-1/9*a^2/x^2*(1-a^2/x^2)^(1/2)-2/9*(1-a^2/x^2)^(1/2))

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Maxima [A]  time = 1.44467, size = 66, normalized size = 1.18 \begin{align*} -\frac{1}{9} \, a{\left (\frac{{\left (-\frac{a^{2}}{x^{2}} + 1\right )}^{\frac{3}{2}}}{a^{4}} - \frac{3 \, \sqrt{-\frac{a^{2}}{x^{2}} + 1}}{a^{4}}\right )} - \frac{\arccos \left (\frac{a}{x}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a/x)/x^4,x, algorithm="maxima")

[Out]

-1/9*a*((-a^2/x^2 + 1)^(3/2)/a^4 - 3*sqrt(-a^2/x^2 + 1)/a^4) - 1/3*arccos(a/x)/x^3

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Fricas [A]  time = 2.46169, size = 104, normalized size = 1.86 \begin{align*} -\frac{3 \, a^{3} \arccos \left (\frac{a}{x}\right ) -{\left (a^{2} x + 2 \, x^{3}\right )} \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}}}{9 \, a^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a/x)/x^4,x, algorithm="fricas")

[Out]

-1/9*(3*a^3*arccos(a/x) - (a^2*x + 2*x^3)*sqrt(-(a^2 - x^2)/x^2))/(a^3*x^3)

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Sympy [A]  time = 6.3656, size = 102, normalized size = 1.82 \begin{align*} \frac{a \left (\begin{cases} \frac{\sqrt{-1 + \frac{x^{2}}{a^{2}}}}{3 a x^{3}} + \frac{2 \sqrt{-1 + \frac{x^{2}}{a^{2}}}}{3 a^{3} x} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{a^{2}}\right |} > 1 \\\frac{i \sqrt{1 - \frac{x^{2}}{a^{2}}}}{3 a x^{3}} + \frac{2 i \sqrt{1 - \frac{x^{2}}{a^{2}}}}{3 a^{3} x} & \text{otherwise} \end{cases}\right )}{3} - \frac{\operatorname{acos}{\left (\frac{a}{x} \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a/x)/x**4,x)

[Out]

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

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Giac [A]  time = 1.31389, size = 88, normalized size = 1.57 \begin{align*} -\frac{\arccos \left (\frac{a}{x}\right )}{3 \, x^{3}} + \frac{4 \,{\left (a^{2} + 3 \,{\left (x - \sqrt{-a^{2} + x^{2}}\right )}^{2}\right )} a}{9 \,{\left (a^{2} +{\left (x - \sqrt{-a^{2} + x^{2}}\right )}^{2}\right )}^{3} \mathrm{sgn}\left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a/x)/x^4,x, algorithm="giac")

[Out]

-1/3*arccos(a/x)/x^3 + 4/9*(a^2 + 3*(x - sqrt(-a^2 + x^2))^2)*a/((a^2 + (x - sqrt(-a^2 + x^2))^2)^3*sgn(x))