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

Optimal. Leaf size=35 \[ \frac{\text{PolyLog}\left (2,-\frac{1}{a+b x}\right )}{2 d}-\frac{\text{PolyLog}\left (2,\frac{1}{a+b x}\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0301393, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {6108, 12, 5913} \[ \frac{\text{PolyLog}\left (2,-\frac{1}{a+b x}\right )}{2 d}-\frac{\text{PolyLog}\left (2,\frac{1}{a+b x}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6108

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcCoth[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 5913

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[(b*PolyLog[2, -(c*x)^(-1)
])/2, x] - Simp[(b*PolyLog[2, 1/(c*x)])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \coth ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=\frac{\text{Li}_2\left (-\frac{1}{a+b x}\right )}{2 d}-\frac{\text{Li}_2\left (\frac{1}{a+b x}\right )}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.023135, size = 312, normalized size = 8.91 \[ b \left (-\frac{\text{PolyLog}(2,-a-b x)}{2 b d}+\frac{\text{PolyLog}(2,a+b x)}{2 b d}-\frac{\log ^2\left (\frac{a b-(a-1) b}{b (a+b x)}\right )}{4 b d}+\frac{\log ^2\left (\frac{a b-(a+1) b}{b (a+b x)}\right )}{4 b d}-\frac{\log \left (\frac{b (a+b x-1)}{(a-1) b-a b}\right ) \log \left (\frac{a b-(a-1) b}{b (a+b x)}\right )}{2 b d}+\frac{\log \left (\frac{a+b x-1}{a+b x}\right ) \log \left (\frac{a b-(a-1) b}{b (a+b x)}\right )}{2 b d}+\frac{\log \left (\frac{b (-a-b x-1)}{(-a-1) b+a b}\right ) \log \left (\frac{a b-(a+1) b}{b (a+b x)}\right )}{2 b d}-\frac{\log \left (\frac{a b-(a+1) b}{b (a+b x)}\right ) \log \left (\frac{a+b x+1}{a+b x}\right )}{2 b d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

b*(-(Log[(b*(-1 + a + b*x))/((-1 + a)*b - a*b)]*Log[(-((-1 + a)*b) + a*b)/(b*(a + b*x))])/(2*b*d) - Log[(-((-1
+ a)*b) + a*b)/(b*(a + b*x))]^2/(4*b*d) + (Log[(b*(-1 - a - b*x))/((-1 - a)*b + a*b)]*Log[(a*b - (1 + a)*b)/(
b*(a + b*x))])/(2*b*d) + Log[(a*b - (1 + a)*b)/(b*(a + b*x))]^2/(4*b*d) + (Log[(-((-1 + a)*b) + a*b)/(b*(a + b
*x))]*Log[(-1 + a + b*x)/(a + b*x)])/(2*b*d) - (Log[(a*b - (1 + a)*b)/(b*(a + b*x))]*Log[(1 + a + b*x)/(a + b*
x)])/(2*b*d) - PolyLog[2, -a - b*x]/(2*b*d) + PolyLog[2, a + b*x]/(2*b*d))

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Maple [A]  time = 0.044, size = 59, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( bx+a \right ){\rm arccoth} \left (bx+a\right )}{d}}-{\frac{{\it dilog} \left ( bx+a \right ) }{2\,d}}-{\frac{{\it dilog} \left ( bx+a+1 \right ) }{2\,d}}-{\frac{\ln \left ( bx+a \right ) \ln \left ( bx+a+1 \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*ln(b*x+a)*arccoth(b*x+a)-1/2/d*dilog(b*x+a)-1/2/d*dilog(b*x+a+1)-1/2/d*ln(b*x+a)*ln(b*x+a+1)

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Maxima [B]  time = 0.98962, size = 178, normalized size = 5.09 \begin{align*} -\frac{1}{2} \, b{\left (\frac{\log \left (b x + a\right ) \log \left (b x + a - 1\right ) +{\rm Li}_2\left (-b x - a + 1\right )}{b d} - \frac{\log \left (b x + a + 1\right ) \log \left (-b x - a\right ) +{\rm Li}_2\left (b x + a + 1\right )}{b d}\right )} - \frac{b{\left (\frac{\log \left (b x + a + 1\right )}{b} - \frac{\log \left (b x + a - 1\right )}{b}\right )} \log \left (d x + \frac{a d}{b}\right )}{2 \, d} + \frac{\operatorname{arcoth}\left (b x + a\right ) \log \left (d x + \frac{a d}{b}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*((log(b*x + a)*log(b*x + a - 1) + dilog(-b*x - a + 1))/(b*d) - (log(b*x + a + 1)*log(-b*x - a) + dilog(
b*x + a + 1))/(b*d)) - 1/2*b*(log(b*x + a + 1)/b - log(b*x + a - 1)/b)*log(d*x + a*d/b)/d + arccoth(b*x + a)*l
og(d*x + a*d/b)/d

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arcoth}\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(arccoth(b*x+a)/(a*d/b+d*x),x, algorithm="fricas")

[Out]

integral(b*arccoth(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{acoth}{\left (a + b x \right )}}{a + b x}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Integral(acoth(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{arcoth}\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(arccoth(b*x+a)/(a*d/b+d*x),x, algorithm="giac")

[Out]

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