3.16 \(\int \frac {\sec ^{-1}(\frac {a}{x})}{x^4} \, dx\)

Optimal. Leaf size=60 \[ \frac {\sqrt {1-\frac {x^2}{a^2}}}{6 a x^2}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {x^2}{a^2}}\right )}{6 a^3}-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3} \]

[Out]

-1/3*arccos(x/a)/x^3+1/6*arctanh((1-x^2/a^2)^(1/2))/a^3+1/6*(1-x^2/a^2)^(1/2)/a/x^2

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Rubi [A]  time = 0.04, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5264, 4628, 266, 51, 63, 208} \[ \frac {\sqrt {1-\frac {x^2}{a^2}}}{6 a x^2}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {x^2}{a^2}}\right )}{6 a^3}-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - x^2/a^2]/(6*a*x^2) - ArcCos[x/a]/(3*x^3) + ArcTanh[Sqrt[1 - x^2/a^2]]/(6*a^3)

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 208

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

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 4628

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5264

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

Rubi steps

\begin {align*} \int \frac {\sec ^{-1}\left (\frac {a}{x}\right )}{x^4} \, dx &=\int \frac {\cos ^{-1}\left (\frac {x}{a}\right )}{x^4} \, dx\\ &=-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3}-\frac {\int \frac {1}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx}{3 a}\\ &=-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3}-\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x}{a^2}}} \, dx,x,x^2\right )}{6 a}\\ &=\frac {\sqrt {1-\frac {x^2}{a^2}}}{6 a x^2}-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3}-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,x^2\right )}{12 a^3}\\ &=\frac {\sqrt {1-\frac {x^2}{a^2}}}{6 a x^2}-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3}+\frac {\operatorname {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {x^2}{a^2}}\right )}{6 a}\\ &=\frac {\sqrt {1-\frac {x^2}{a^2}}}{6 a x^2}-\frac {\cos ^{-1}\left (\frac {x}{a}\right )}{3 x^3}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {x^2}{a^2}}\right )}{6 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 69, normalized size = 1.15 \[ \frac {-2 a^3 \sec ^{-1}\left (\frac {a}{x}\right )+a^2 x \sqrt {1-\frac {x^2}{a^2}}+x^3 \log \left (\sqrt {1-\frac {x^2}{a^2}}+1\right )-x^3 \log (x)}{6 a^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x*Sqrt[1 - x^2/a^2] - 2*a^3*ArcSec[a/x] - x^3*Log[x] + x^3*Log[1 + Sqrt[1 - x^2/a^2]])/(6*a^3*x^3)

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fricas [B]  time = 0.94, size = 142, normalized size = 2.37 \[ -\frac {4 \, a^{3} x^{3} \arctan \left (-\frac {x^{2} \sqrt {\frac {a^{2} - x^{2}}{x^{2}}}}{a^{2} - x^{2}}\right ) - x^{3} \log \left (x \sqrt {\frac {a^{2} - x^{2}}{x^{2}}} + a\right ) + x^{3} \log \left (x \sqrt {\frac {a^{2} - x^{2}}{x^{2}}} - a\right ) - 2 \, a x^{2} \sqrt {\frac {a^{2} - x^{2}}{x^{2}}} - 4 \, {\left (a^{3} x^{3} - a^{3}\right )} \operatorname {arcsec}\left (\frac {a}{x}\right )}{12 \, a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*a^3*x^3*arctan(-x^2*sqrt((a^2 - x^2)/x^2)/(a^2 - x^2)) - x^3*log(x*sqrt((a^2 - x^2)/x^2) + a) + x^3*l
og(x*sqrt((a^2 - x^2)/x^2) - a) - 2*a*x^2*sqrt((a^2 - x^2)/x^2) - 4*(a^3*x^3 - a^3)*arcsec(a/x))/(a^3*x^3)

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giac [A]  time = 0.12, size = 80, normalized size = 1.33 \[ \frac {a {\left (\frac {\log \left ({\left | a + \sqrt {a^{2} - x^{2}} \right |}\right )}{a^{3}} - \frac {\log \left ({\left | -a + \sqrt {a^{2} - x^{2}} \right |}\right )}{a^{3}} + \frac {2 \, \sqrt {a^{2} - x^{2}}}{a^{2} x^{2}}\right )}}{12 \, {\left | a \right |}} - \frac {\arccos \left (\frac {x}{a}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*a*(log(abs(a + sqrt(a^2 - x^2)))/a^3 - log(abs(-a + sqrt(a^2 - x^2)))/a^3 + 2*sqrt(a^2 - x^2)/(a^2*x^2))/
abs(a) - 1/3*arccos(x/a)/x^3

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maple [A]  time = 0.06, size = 98, normalized size = 1.63 \[ -\frac {\mathrm {arcsec}\left (\frac {a}{x}\right )}{3 x^{3}}+\frac {-1+\frac {a^{2}}{x^{2}}}{6 a^{3} \sqrt {\frac {\left (-1+\frac {a^{2}}{x^{2}}\right ) x^{2}}{a^{2}}}}+\frac {\sqrt {-1+\frac {a^{2}}{x^{2}}}\, x \ln \left (\frac {a}{x}+\sqrt {-1+\frac {a^{2}}{x^{2}}}\right )}{6 a^{4} \sqrt {\frac {\left (-1+\frac {a^{2}}{x^{2}}\right ) x^{2}}{a^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arcsec(a/x)/x^3+1/6/a^3*(-1+a^2/x^2)/((-1+a^2/x^2)*x^2/a^2)^(1/2)+1/6/a^4*(-1+a^2/x^2)^(1/2)/((-1+a^2/x^2
)*x^2/a^2)^(1/2)*x*ln(a/x+(-1+a^2/x^2)^(1/2))

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maxima [A]  time = 0.42, size = 64, normalized size = 1.07 \[ \frac {\frac {\log \left (\frac {2 \, \sqrt {-\frac {x^{2}}{a^{2}} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a^{2}} + \frac {\sqrt {-\frac {x^{2}}{a^{2}} + 1}}{x^{2}}}{6 \, a} - \frac {\operatorname {arcsec}\left (\frac {a}{x}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(log(2*sqrt(-x^2/a^2 + 1)/abs(x) + 2/abs(x))/a^2 + sqrt(-x^2/a^2 + 1)/x^2)/a - 1/3*arcsec(a/x)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(x/a)/x^4, x)

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sympy [C]  time = 2.52, size = 100, normalized size = 1.67 \[ - \frac {\operatorname {asec}{\left (\frac {a}{x} \right )}}{3 x^{3}} - \frac {\begin {cases} - \frac {\sqrt {\frac {a^{2}}{x^{2}} - 1}}{2 a x} - \frac {\operatorname {acosh}{\left (\frac {a}{x} \right )}}{2 a^{2}} & \text {for}\: \left |{\frac {a^{2}}{x^{2}}}\right | > 1 \\\frac {i a}{2 x^{3} \sqrt {- \frac {a^{2}}{x^{2}} + 1}} - \frac {i}{2 a x \sqrt {- \frac {a^{2}}{x^{2}} + 1}} + \frac {i \operatorname {asin}{\left (\frac {a}{x} \right )}}{2 a^{2}} & \text {otherwise} \end {cases}}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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