3.8 \(\int \frac{a+b \csc ^{-1}(c x)}{x} \, dx\)

Optimal. Leaf size=64 \[ \frac{1}{2} i b \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )+\frac{i \left (a+b \csc ^{-1}(c x)\right )^2}{2 b}-\log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \left (a+b \csc ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.0847952, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5219, 4625, 3717, 2190, 2279, 2391} \[ \frac{1}{2} i b \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )+\frac{i \left (a+b \csc ^{-1}(c x)\right )^2}{2 b}-\log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \left (a+b \csc ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsc[c*x])/x,x]

[Out]

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

Rule 5219

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcSin[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, 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{a+b \csc ^{-1}(c x)}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,\frac{1}{x}\right )\\ &=-\operatorname{Subst}\left (\int (a+b x) \cot (x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \csc ^{-1}(c x)\right )^2}{2 b}+2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \csc ^{-1}(c x)\right )^2}{2 b}-\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )+b \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac{i \left (a+b \csc ^{-1}(c x)\right )^2}{2 b}-\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-\frac{1}{2} (i b) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \csc ^{-1}(c x)}\right )\\ &=\frac{i \left (a+b \csc ^{-1}(c x)\right )^2}{2 b}-\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )+\frac{1}{2} i b \text{Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0299711, size = 53, normalized size = 0.83 \[ \frac{1}{2} i b \left (\csc ^{-1}(c x)^2+\text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )\right )+a \log (x)-b \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCsc[c*x])/x,x]

[Out]

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

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Maple [A]  time = 0.257, size = 140, normalized size = 2.2 \begin{align*} a\ln \left ( cx \right ) +{\frac{i}{2}}b \left ({\rm arccsc} \left (cx\right ) \right ) ^{2}-b{\rm arccsc} \left (cx\right )\ln \left ( 1-{\frac{i}{cx}}-\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) -b{\rm arccsc} \left (cx\right )\ln \left ( 1+{\frac{i}{cx}}+\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +ib{\it polylog} \left ( 2,{\frac{-i}{cx}}-\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +ib{\it polylog} \left ( 2,{\frac{i}{cx}}+\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsc(c*x))/x,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left (c^{2} \int \frac{\sqrt{c x + 1} \sqrt{c x - 1} \log \left (x\right )}{c^{4} x^{3} - c^{2} x}\,{d x} + \arctan \left (1, \sqrt{c x + 1} \sqrt{c x - 1}\right ) \log \left (x\right )\right )} b + a \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))/x,x, algorithm="maxima")

[Out]

(c^2*integrate(sqrt(c*x + 1)*sqrt(c*x - 1)*log(x)/(c^4*x^3 - c^2*x), x) + arctan2(1, sqrt(c*x + 1)*sqrt(c*x -
1))*log(x))*b + a*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))/x,x, algorithm="fricas")

[Out]

integral((b*arccsc(c*x) + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsc(c*x))/x,x)

[Out]

Integral((a + b*acsc(c*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))/x,x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)/x, x)