3.12 \(\int \frac{\csc ^{-1}(\frac{a}{x})}{x} \, dx\)

Optimal. Leaf size=59 \[ -\frac{1}{2} i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )-\frac{1}{2} i \sin ^{-1}\left (\frac{x}{a}\right )^2+\sin ^{-1}\left (\frac{x}{a}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0636623, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5265, 4625, 3717, 2190, 2279, 2391} \[ -\frac{1}{2} i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )-\frac{1}{2} i \sin ^{-1}\left (\frac{x}{a}\right )^2+\sin ^{-1}\left (\frac{x}{a}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCsc[a/x]/x,x]

[Out]

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

Rule 5265

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

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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{\csc ^{-1}\left (\frac{a}{x}\right )}{x} \, dx &=\int \frac{\sin ^{-1}\left (\frac{x}{a}\right )}{x} \, dx\\ &=\operatorname{Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}\left (\frac{x}{a}\right )\right )\\ &=-\frac{1}{2} i \sin ^{-1}\left (\frac{x}{a}\right )^2-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac{x}{a}\right )\right )\\ &=-\frac{1}{2} i \sin ^{-1}\left (\frac{x}{a}\right )^2+\sin ^{-1}\left (\frac{x}{a}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )-\operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{x}{a}\right )\right )\\ &=-\frac{1}{2} i \sin ^{-1}\left (\frac{x}{a}\right )^2+\sin ^{-1}\left (\frac{x}{a}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )+\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )\\ &=-\frac{1}{2} i \sin ^{-1}\left (\frac{x}{a}\right )^2+\sin ^{-1}\left (\frac{x}{a}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )-\frac{1}{2} i \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{x}{a}\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0329512, size = 54, normalized size = 0.92 \[ \csc ^{-1}\left (\frac{a}{x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (\frac{a}{x}\right )}\right )-\frac{1}{2} i \left (\csc ^{-1}\left (\frac{a}{x}\right )^2+\text{PolyLog}\left (2,e^{2 i \csc ^{-1}\left (\frac{a}{x}\right )}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCsc[a/x]/x,x]

[Out]

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

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Maple [A]  time = 0.238, size = 125, normalized size = 2.1 \begin{align*} -{\frac{i}{2}} \left ({\rm arccsc} \left ({\frac{a}{x}}\right ) \right ) ^{2}+{\rm arccsc} \left ({\frac{a}{x}}\right )\ln \left ( 1-{\frac{ix}{a}}-\sqrt{1-{\frac{{x}^{2}}{{a}^{2}}}} \right ) +{\rm arccsc} \left ({\frac{a}{x}}\right )\ln \left ( 1+{\frac{ix}{a}}+\sqrt{1-{\frac{{x}^{2}}{{a}^{2}}}} \right ) -i{\it polylog} \left ( 2,{\frac{-ix}{a}}-\sqrt{1-{\frac{{x}^{2}}{{a}^{2}}}} \right ) -i{\it polylog} \left ( 2,{\frac{ix}{a}}+\sqrt{1-{\frac{{x}^{2}}{{a}^{2}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsc(a/x)/x,x)

[Out]

-1/2*I*arccsc(a/x)^2+arccsc(a/x)*ln(1-I/a*x-(1-x^2/a^2)^(1/2))+arccsc(a/x)*ln(1+I*x/a+(1-x^2/a^2)^(1/2))-I*pol
ylog(2,-I/a*x-(1-x^2/a^2)^(1/2))-I*polylog(2,I*x/a+(1-x^2/a^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccsc(a/x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsc(a/x)/x,x)

[Out]

Integral(acsc(a/x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsc(a/x)/x, x)