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

Optimal. Leaf size=60 \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )+\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-\cos ^{-1}\left (\frac{a}{x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right ) \]

[Out]

(I/2)*ArcCos[a/x]^2 - ArcCos[a/x]*Log[1 + E^((2*I)*ArcCos[a/x])] + (I/2)*PolyLog[2, -E^((2*I)*ArcCos[a/x])]

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Rubi [A]  time = 0.0546249, antiderivative size = 60, 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 = {4831, 3719, 2190, 2279, 2391} \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )+\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-\cos ^{-1}\left (\frac{a}{x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a/x]/x,x]

[Out]

(I/2)*ArcCos[a/x]^2 - ArcCos[a/x]*Log[1 + E^((2*I)*ArcCos[a/x])] + (I/2)*PolyLog[2, -E^((2*I)*ArcCos[a/x])]

Rule 4831

Int[ArcCos[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> -Dist[p^(-1), Subst[Int[x^n*Tan[x], x], x, ArcCos[a*x^p]]
, x] /; FreeQ[{a, p}, x] && IGtQ[n, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}\left (\frac{a}{x}\right )}{x} \, dx &=\operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac{a}{x}\right )\right )\\ &=\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac{a}{x}\right )\right )\\ &=\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-\cos ^{-1}\left (\frac{a}{x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )+\operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac{a}{x}\right )\right )\\ &=\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-\cos ^{-1}\left (\frac{a}{x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )\\ &=\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-\cos ^{-1}\left (\frac{a}{x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )+\frac{1}{2} i \text{Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0208701, size = 60, normalized size = 1. \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right )+\frac{1}{2} i \cos ^{-1}\left (\frac{a}{x}\right )^2-\cos ^{-1}\left (\frac{a}{x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{a}{x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a/x]/x,x]

[Out]

(I/2)*ArcCos[a/x]^2 - ArcCos[a/x]*Log[1 + E^((2*I)*ArcCos[a/x])] + (I/2)*PolyLog[2, -E^((2*I)*ArcCos[a/x])]

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Maple [A]  time = 0.055, size = 77, normalized size = 1.3 \begin{align*}{\frac{i}{2}} \left ( \arccos \left ({\frac{a}{x}} \right ) \right ) ^{2}-\arccos \left ({\frac{a}{x}} \right ) \ln \left ( 1+ \left ({\frac{a}{x}}+i\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}} \right ) ^{2} \right ) +{\frac{i}{2}}{\it polylog} \left ( 2,- \left ({\frac{a}{x}}+i\sqrt{1-{\frac{{a}^{2}}{{x}^{2}}}} \right ) ^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a/x)/x,x)

[Out]

1/2*I*arccos(a/x)^2-arccos(a/x)*ln(1+(a/x+I*(1-a^2/x^2)^(1/2))^2)+1/2*I*polylog(2,-(a/x+I*(1-a^2/x^2)^(1/2))^2
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -i \, a^{2} \int -\frac{\log \left (x\right )}{a^{2} x - x^{3}}\,{d x} - a \int -\frac{\sqrt{a + x} \sqrt{-a + x} \log \left (x\right )}{a^{2} x - x^{3}}\,{d x} + \arctan \left (\sqrt{a + x} \sqrt{-a + x}, a\right ) \log \left (x\right ) - \frac{1}{2} i \, \log \left (x^{2}\right ) \log \left (x\right ) + \frac{1}{2} i \, \log \left (x\right )^{2} + \frac{1}{2} i \, \log \left (x\right ) \log \left (\frac{a + x}{a}\right ) + \frac{1}{2} i \, \log \left (x\right ) \log \left (\frac{a - x}{a}\right ) + \frac{1}{2} i \,{\rm Li}_2\left (\frac{x}{a}\right ) + \frac{1}{2} i \,{\rm Li}_2\left (-\frac{x}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a^2*integrate(-log(x)/(a^2*x - x^3), x) - a*integrate(-sqrt(a + x)*sqrt(-a + x)*log(x)/(a^2*x - x^3), x) +
arctan2(sqrt(a + x)*sqrt(-a + x), a)*log(x) - 1/2*I*log(x^2)*log(x) + 1/2*I*log(x)^2 + 1/2*I*log(x)*log((a + x
)/a) + 1/2*I*log(x)*log((a - x)/a) + 1/2*I*dilog(x/a) + 1/2*I*dilog(-x/a)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arccos \left (\frac{a}{x}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccos(a/x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a/x)/x,x)

[Out]

Integral(acos(a/x)/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arccos \left (\frac{a}{x}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(a/x)/x, x)