3.14 \(\int \frac{\sec ^{-1}(\frac{a}{x})}{x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac{\tanh ^{-1}\left (\sqrt{1-\frac{x^2}{a^2}}\right )}{a}-\frac{\cos ^{-1}\left (\frac{x}{a}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0296163, antiderivative size = 31, 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 = {5264, 4628, 266, 63, 208} \[ \frac{\tanh ^{-1}\left (\sqrt{1-\frac{x^2}{a^2}}\right )}{a}-\frac{\cos ^{-1}\left (\frac{x}{a}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a/x]/x^2,x]

[Out]

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

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]

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 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 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]

Rubi steps

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

Mathematica [B]  time = 0.13649, size = 93, normalized size = 3. \[ \frac{x \sqrt{\frac{a^2}{x^2}-1} \left (\log \left (\frac{a}{x \sqrt{\frac{a^2}{x^2}-1}}+1\right )-\log \left (1-\frac{a}{x \sqrt{\frac{a^2}{x^2}-1}}\right )\right )}{2 a^2 \sqrt{1-\frac{x^2}{a^2}}}-\frac{\sec ^{-1}\left (\frac{a}{x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a/x]/x^2,x]

[Out]

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

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Maple [A]  time = 0.145, size = 41, normalized size = 1.3 \begin{align*} -{\frac{1}{x}{\rm arcsec} \left ({\frac{a}{x}}\right )}+{\frac{1}{a}\ln \left ({\frac{a}{x}}+{\frac{a}{x}\sqrt{1-{\frac{{x}^{2}}{{a}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(a/x)/x^2,x)

[Out]

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

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Maxima [A]  time = 0.961596, size = 70, normalized size = 2.26 \begin{align*} -\frac{\frac{2 \, a \operatorname{arcsec}\left (\frac{a}{x}\right )}{x} - \log \left (\sqrt{-\frac{x^{2}}{a^{2}} + 1} + 1\right ) + \log \left (-\sqrt{-\frac{x^{2}}{a^{2}} + 1} + 1\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a/x)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.41124, size = 225, normalized size = 7.26 \begin{align*} -\frac{2 \, a x \arctan \left (-\frac{x^{2} \sqrt{\frac{a^{2} - x^{2}}{x^{2}}}}{a^{2} - x^{2}}\right ) - 2 \,{\left (a x - a\right )} \operatorname{arcsec}\left (\frac{a}{x}\right ) - x \log \left (x \sqrt{\frac{a^{2} - x^{2}}{x^{2}}} + a\right ) + x \log \left (x \sqrt{\frac{a^{2} - x^{2}}{x^{2}}} - a\right )}{2 \, a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a/x)/x^2,x, algorithm="fricas")

[Out]

-1/2*(2*a*x*arctan(-x^2*sqrt((a^2 - x^2)/x^2)/(a^2 - x^2)) - 2*(a*x - a)*arcsec(a/x) - x*log(x*sqrt((a^2 - x^2
)/x^2) + a) + x*log(x*sqrt((a^2 - x^2)/x^2) - a))/(a*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}{\left (\frac{a}{x} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(a/x)/x**2,x)

[Out]

Integral(asec(a/x)/x**2, x)

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Giac [B]  time = 1.108, size = 82, normalized size = 2.65 \begin{align*} \frac{a{\left (\frac{\log \left ({\left | a + \sqrt{a^{2} - x^{2}} \right |}\right )}{a} - \frac{\log \left ({\left | -a + \sqrt{a^{2} - x^{2}} \right |}\right )}{a}\right )}}{2 \,{\left | a \right |}} - \frac{\arccos \left (\frac{x}{a}\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a/x)/x^2,x, algorithm="giac")

[Out]

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