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

Optimal. Leaf size=120 \[ -\frac{\text{PolyLog}\left (2,1-\frac{2 b (c+d x)}{(a+b x+1) (-a d+b c+d)}\right )}{2 d}+\frac{\text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )}{2 d}+\frac{\coth ^{-1}(a+b x) \log \left (\frac{2 b (c+d x)}{(a+b x+1) (-a d+b c+d)}\right )}{d}-\frac{\log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.126501, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6112, 5921, 2402, 2315, 2447} \[ -\frac{\text{PolyLog}\left (2,1-\frac{2 b (c+d x)}{(a+b x+1) (-a d+b c+d)}\right )}{2 d}+\frac{\text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )}{2 d}+\frac{\coth ^{-1}(a+b x) \log \left (\frac{2 b (c+d x)}{(a+b x+1) (-a d+b c+d)}\right )}{d}-\frac{\log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]/(c + d*x),x]

[Out]

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

Rule 6112

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

Rule 5921

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcCoth[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

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

Mathematica [A]  time = 0.0665651, size = 185, normalized size = 1.54 \[ -\frac{\text{PolyLog}\left (2,\frac{b (c+d x)}{-a d+b c-d}\right )}{2 d}+\frac{\text{PolyLog}\left (2,\frac{b (c+d x)}{-a d+b c+d}\right )}{2 d}+\frac{\log (c+d x) \log \left (\frac{d (-a-b x+1)}{-a d+b c+d}\right )}{2 d}-\frac{\log \left (\frac{a+b x-1}{a+b x}\right ) \log (c+d x)}{2 d}-\frac{\log (c+d x) \log \left (-\frac{d (a+b x+1)}{-a d+b c-d}\right )}{2 d}+\frac{\log \left (\frac{a+b x+1}{a+b x}\right ) \log (c+d x)}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.137, size = 176, normalized size = 1.5 \begin{align*}{\frac{\ln \left ( d \left ( bx+a \right ) -ad+bc \right ){\rm arccoth} \left (bx+a\right )}{d}}-{\frac{\ln \left ( d \left ( bx+a \right ) -ad+bc \right ) }{2\,d}\ln \left ({\frac{d \left ( bx+a \right ) +d}{ad-bc+d}} \right ) }-{\frac{1}{2\,d}{\it dilog} \left ({\frac{d \left ( bx+a \right ) +d}{ad-bc+d}} \right ) }+{\frac{\ln \left ( d \left ( bx+a \right ) -ad+bc \right ) }{2\,d}\ln \left ({\frac{d \left ( bx+a \right ) -d}{ad-bc-d}} \right ) }+{\frac{1}{2\,d}{\it dilog} \left ({\frac{d \left ( bx+a \right ) -d}{ad-bc-d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.987377, size = 259, normalized size = 2.16 \begin{align*} -\frac{1}{2} \, b{\left (\frac{\log \left (b x + a - 1\right ) \log \left (\frac{b d x + a d - d}{b c - a d + d} + 1\right ) +{\rm Li}_2\left (-\frac{b d x + a d - d}{b c - a d + d}\right )}{b d} - \frac{\log \left (b x + a + 1\right ) \log \left (\frac{b d x + a d + d}{b c - a d - d} + 1\right ) +{\rm Li}_2\left (-\frac{b d x + a d + d}{b c - a d - d}\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 + c\right )}{2 \, d} + \frac{\operatorname{arcoth}\left (b x + a\right ) \log \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(b*x + a)/(d*x + c), x)

________________________________________________________________________________________

Sympy [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(acoth(b*x+a)/(d*x+c),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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