Integrand size = 10, antiderivative size = 56 \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=\frac {\sqrt {1-\frac {a^2}{x^2}}}{3 a^3}-\frac {\left (1-\frac {a^2}{x^2}\right )^{3/2}}{9 a^3}-\frac {\sec ^{-1}\left (\frac {x}{a}\right )}{3 x^3} \] Output:
1/3*(1-a^2/x^2)^(1/2)/a^3-1/9*(1-a^2/x^2)^(3/2)/a^3-1/3*arcsec(x/a)/x^3
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84 \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=\frac {\sqrt {1-\frac {a^2}{x^2}} x \left (a^2+2 x^2\right )-3 a^3 \arccos \left (\frac {a}{x}\right )}{9 a^3 x^3} \] Input:
Integrate[ArcCos[a/x]/x^4,x]
Output:
(Sqrt[1 - a^2/x^2]*x*(a^2 + 2*x^2) - 3*a^3*ArcCos[a/x])/(9*a^3*x^3)
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5332, 5743, 798, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 5332 |
\(\displaystyle \int \frac {\sec ^{-1}\left (\frac {x}{a}\right )}{x^4}dx\) |
\(\Big \downarrow \) 5743 |
\(\displaystyle \frac {1}{3} a \int \frac {1}{\sqrt {1-\frac {a^2}{x^2}} x^5}dx-\frac {\sec ^{-1}\left (\frac {x}{a}\right )}{3 x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{6} a \int \frac {1}{\sqrt {1-\frac {a^2}{x^2}} x^2}d\frac {1}{x^2}-\frac {\sec ^{-1}\left (\frac {x}{a}\right )}{3 x^3}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {1}{6} a \int \left (\frac {1}{a^2 \sqrt {1-\frac {a^2}{x^2}}}-\frac {\sqrt {1-\frac {a^2}{x^2}}}{a^2}\right )d\frac {1}{x^2}-\frac {\sec ^{-1}\left (\frac {x}{a}\right )}{3 x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{6} a \left (\frac {2 \left (1-\frac {a^2}{x^2}\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-\frac {a^2}{x^2}}}{a^4}\right )-\frac {\sec ^{-1}\left (\frac {x}{a}\right )}{3 x^3}\) |
Input:
Int[ArcCos[a/x]/x^4,x]
Output:
-1/6*(a*((-2*Sqrt[1 - a^2/x^2])/a^4 + (2*(1 - a^2/x^2)^(3/2))/(3*a^4))) - ArcSec[x/a]/(3*x^3)
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] && 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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
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]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(d*x)^(m + 1)*((a + b*ArcSec[c*x])/(d*(m + 1))), x] - Simp[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]
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(-\frac {\frac {a^{3} \arccos \left (\frac {a}{x}\right )}{3 x^{3}}-\frac {a^{2} \sqrt {1-\frac {a^{2}}{x^{2}}}}{9 x^{2}}-\frac {2 \sqrt {1-\frac {a^{2}}{x^{2}}}}{9}}{a^{3}}\) | \(55\) |
default | \(-\frac {\frac {a^{3} \arccos \left (\frac {a}{x}\right )}{3 x^{3}}-\frac {a^{2} \sqrt {1-\frac {a^{2}}{x^{2}}}}{9 x^{2}}-\frac {2 \sqrt {1-\frac {a^{2}}{x^{2}}}}{9}}{a^{3}}\) | \(55\) |
parts | \(-\frac {\arccos \left (\frac {a}{x}\right )}{3 x^{3}}-\frac {\left (a^{2}-x^{2}\right ) \left (a^{2}+2 x^{2}\right )}{9 a^{3} \sqrt {-\frac {a^{2}-x^{2}}{x^{2}}}\, x^{4}}\) | \(55\) |
orering | \(-\frac {\left (7 a^{4}+4 a^{2} x^{2}-8 x^{4}\right ) \arccos \left (\frac {a}{x}\right )}{9 x^{3} a^{4}}-\frac {\left (a^{2}+2 x^{2}\right ) x^{2} \left (a -x \right ) \left (a +x \right ) \left (\frac {a}{x^{6} \sqrt {1-\frac {a^{2}}{x^{2}}}}-\frac {4 \arccos \left (\frac {a}{x}\right )}{x^{5}}\right )}{9 a^{4}}\) | \(89\) |
Input:
int(arccos(a/x)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/a^3*(1/3*a^3/x^3*arccos(a/x)-1/9*a^2/x^2*(1-a^2/x^2)^(1/2)-2/9*(1-a^2/x ^2)^(1/2))
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=-\frac {3 \, a^{3} \arccos \left (\frac {a}{x}\right ) - {\left (a^{2} x + 2 \, x^{3}\right )} \sqrt {-\frac {a^{2} - x^{2}}{x^{2}}}}{9 \, a^{3} x^{3}} \] Input:
integrate(arccos(a/x)/x^4,x, algorithm="fricas")
Output:
-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 )
Time = 1.59 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.79 \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=\frac {a \left (\begin {cases} \frac {\sqrt {-1 + \frac {x^{2}}{a^{2}}}}{3 a x^{3}} + \frac {2 \sqrt {-1 + \frac {x^{2}}{a^{2}}}}{3 a^{3} x} & \text {for}\: \left |{\frac {x^{2}}{a^{2}}}\right | > 1 \\\frac {i \sqrt {1 - \frac {x^{2}}{a^{2}}}}{3 a x^{3}} + \frac {2 i \sqrt {1 - \frac {x^{2}}{a^{2}}}}{3 a^{3} x} & \text {otherwise} \end {cases}\right )}{3} - \frac {\operatorname {acos}{\left (\frac {a}{x} \right )}}{3 x^{3}} \] Input:
integrate(acos(a/x)/x**4,x)
Output:
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)
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=-\frac {1}{9} \, a {\left (\frac {{\left (-\frac {a^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}}}{a^{4}} - \frac {3 \, \sqrt {-\frac {a^{2}}{x^{2}} + 1}}{a^{4}}\right )} - \frac {\arccos \left (\frac {a}{x}\right )}{3 \, x^{3}} \] Input:
integrate(arccos(a/x)/x^4,x, algorithm="maxima")
Output:
-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
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=-\frac {\frac {3 \, a \arccos \left (\frac {a}{x}\right )}{x^{3}} - \frac {2 \, \sqrt {-\frac {a^{2}}{x^{2}} + 1}}{a^{2}} - \frac {\sqrt {-\frac {a^{2}}{x^{2}} + 1}}{x^{2}}}{9 \, a} \] Input:
integrate(arccos(a/x)/x^4,x, algorithm="giac")
Output:
-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
Timed out. \[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=\int \frac {\mathrm {acos}\left (\frac {a}{x}\right )}{x^4} \,d x \] Input:
int(acos(a/x)/x^4,x)
Output:
int(acos(a/x)/x^4, x)
\[ \int \frac {\arccos \left (\frac {a}{x}\right )}{x^4} \, dx=\int \frac {\mathit {acos} \left (\frac {a}{x}\right )}{x^{4}}d x \] Input:
int(acos(a/x)/x^4,x)
Output:
int(acos(a/x)/x**4,x)