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]

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

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.43, size = 49, normalized size = 0.88 \[ -\frac {3 \, a^{3} \arccos \left (\frac {a}{x}\right ) - {\left (a^{2} x + 2 \, x^{3}\right )} \sqrt {-\frac {a^{2} - x^{2}}{x^{2}}}}{9 \, a^{3} x^{3}} \]

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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giac [A]  time = 3.99, size = 52, normalized size = 0.93 \[ -\frac {\frac {3 \, a \arccos \left (\frac {a}{x}\right )}{x^{3}} - \frac {2 \, \sqrt {-\frac {a^{2}}{x^{2}} + 1}}{a^{2}} - \frac {\sqrt {-\frac {a^{2}}{x^{2}} + 1}}{x^{2}}}{9 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 55, normalized size = 0.98 \[ -\frac {\frac {\arccos \left (\frac {a}{x}\right ) a^{3}}{3 x^{3}}-\frac {a^{2} \sqrt {1-\frac {a^{2}}{x^{2}}}}{9 x^{2}}-\frac {2 \sqrt {1-\frac {a^{2}}{x^{2}}}}{9}}{a^{3}} \]

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 = 0.40, size = 49, normalized size = 0.88 \[ -\frac {1}{9} \, a {\left (\frac {{\left (-\frac {a^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{a^{4}} - \frac {3 \, \sqrt {-\frac {a^{2}}{x^{2}} + 1}}{a^{4}}\right )} - \frac {\arccos \left (\frac {a}{x}\right )}{3 \, x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (\frac {a}{x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(a/x)/x^4, x)

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sympy [A]  time = 3.03, size = 100, normalized size = 1.79 \[ \frac {a \left (\begin {cases} \frac {\sqrt {-1 + \frac {x^{2}}{a^{2}}}}{3 a x^{3}} + \frac {2 \sqrt {-1 + \frac {x^{2}}{a^{2}}}}{3 a^{3} x} & \text {for}\: \left |{\frac {x^{2}}{a^{2}}}\right | > 1 \\\frac {i \sqrt {1 - \frac {x^{2}}{a^{2}}}}{3 a x^{3}} + \frac {2 i \sqrt {1 - \frac {x^{2}}{a^{2}}}}{3 a^{3} x} & \text {otherwise} \end {cases}\right )}{3} - \frac {\operatorname {acos}{\left (\frac {a}{x} \right )}}{3 x^{3}} \]

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/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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