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

Optimal. Leaf size=27 \[ x \sec ^{-1}\left (\frac{x}{a}\right )-a \tanh ^{-1}\left (\sqrt{1-\frac{a^2}{x^2}}\right ) \]

[Out]

x*ArcSec[x/a] - a*ArcTanh[Sqrt[1 - a^2/x^2]]

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Rubi [A]  time = 0.0169255, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.833, Rules used = {4833, 5214, 266, 63, 208} \[ x \sec ^{-1}\left (\frac{x}{a}\right )-a \tanh ^{-1}\left (\sqrt{1-\frac{a^2}{x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a/x],x]

[Out]

x*ArcSec[x/a] - a*ArcTanh[Sqrt[1 - a^2/x^2]]

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 5214

Int[ArcSec[(c_.)*(x_)], x_Symbol] :> Simp[x*ArcSec[c*x], x] - Dist[1/c, Int[1/(x*Sqrt[1 - 1/(c^2*x^2)]), x], x
] /; FreeQ[c, x]

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

Mathematica [B]  time = 0.1011, size = 84, normalized size = 3.11 \[ x \cos ^{-1}\left (\frac{a}{x}\right )-\frac{a \sqrt{x^2-a^2} \left (\log \left (\frac{x}{\sqrt{x^2-a^2}}+1\right )-\log \left (1-\frac{x}{\sqrt{x^2-a^2}}\right )\right )}{2 x \sqrt{1-\frac{a^2}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a/x],x]

[Out]

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

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Maple [A]  time = 0.006, size = 30, normalized size = 1.1 \begin{align*} -a \left ( -{\frac{x}{a}\arccos \left ({\frac{a}{x}} \right ) }+{\it Artanh} \left ({\frac{1}{\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}}}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a/x),x)

[Out]

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

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Maxima [A]  time = 1.45402, size = 61, normalized size = 2.26 \begin{align*} -\frac{1}{2} \, a{\left (\log \left (\sqrt{-\frac{a^{2}}{x^{2}} + 1} + 1\right ) - \log \left (\sqrt{-\frac{a^{2}}{x^{2}} + 1} - 1\right )\right )} + x \arccos \left (\frac{a}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.52093, size = 140, normalized size = 5.19 \begin{align*}{\left (x - 1\right )} \arccos \left (\frac{a}{x}\right ) + a \log \left (x \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}} - x\right ) + 2 \, \arctan \left (\frac{x \sqrt{-\frac{a^{2} - x^{2}}{x^{2}}} - x}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.27124, size = 29, normalized size = 1.07 \begin{align*} - a \left (\begin{cases} \operatorname{acosh}{\left (\frac{x}{a} \right )} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{a^{2}}\right |} > 1 \\- i \operatorname{asin}{\left (\frac{x}{a} \right )} & \text{otherwise} \end{cases}\right ) + x \operatorname{acos}{\left (\frac{a}{x} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a/x),x)

[Out]

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

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Giac [A]  time = 1.26154, size = 58, normalized size = 2.15 \begin{align*} -\frac{1}{2} \,{\left (\log \left (a^{2}\right ) \mathrm{sgn}\left (x\right ) - \frac{2 \, \log \left ({\left | -x + \sqrt{-a^{2} + x^{2}} \right |}\right )}{\mathrm{sgn}\left (x\right )}\right )} a + x \arccos \left (\frac{a}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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