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

Optimal. Leaf size=69 \[ \frac{i \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )}{2 d}+\frac{i \csc ^{-1}(a+b x)^2}{2 d}-\frac{\csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0954275, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {5257, 12, 5219, 4625, 3717, 2190, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )}{2 d}+\frac{i \csc ^{-1}(a+b x)^2}{2 d}-\frac{\csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcCsc[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

Rule 5257

Int[((a_.) + ArcCsc[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((f*x)/d)^m*(a + b*ArcCsc[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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{\csc ^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \csc ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\csc ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{x} \, dx,x,\frac{1}{a+b x}\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{i \sin ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{i \sin ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\sin ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}+\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{i \sin ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\sin ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{2 d}\\ &=\frac{i \sin ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\sin ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}+\frac{i \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0563056, size = 59, normalized size = 0.86 \[ \frac{\frac{1}{2} i \left (\csc ^{-1}(a+b x)^2+\text{PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )\right )-\csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCsc[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

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Maple [A]  time = 0.332, size = 158, normalized size = 2.3 \begin{align*}{\frac{{\frac{i}{2}} \left ({\rm arccsc} \left (bx+a\right ) \right ) ^{2}}{d}}-{\frac{{\rm arccsc} \left (bx+a\right )}{d}\ln \left ( 1+{\frac{i}{bx+a}}+\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) }+{\frac{i}{d}{\it polylog} \left ( 2,{\frac{-i}{bx+a}}-\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) }-{\frac{{\rm arccsc} \left (bx+a\right )}{d}\ln \left ( 1-{\frac{i}{bx+a}}-\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) }+{\frac{i}{d}{\it polylog} \left ( 2,{\frac{i}{bx+a}}+\sqrt{1- \left ( bx+a \right ) ^{-2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arccsc(b*x + a)/(b*d*x + a*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Integral(acsc(a + b*x)/(a + b*x), x)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsc(b*x + a)/(d*x + a*d/b), x)