3.74 \(\int \frac{\coth ^{-1}(a+b x)^2}{x^2} \, dx\)

Optimal. Leaf size=251 \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )}{1-a^2}-\frac{b \text{PolyLog}\left (2,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )}{1-a^2}+\frac{b \text{PolyLog}\left (2,-\frac{a+b x+1}{-a-b x+1}\right )}{2 (1-a)}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )}{2 (a+1)}-\frac{2 b \log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{1-a^2}+\frac{2 b \log \left (\frac{2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x)}{1-a^2}-\frac{\coth ^{-1}(a+b x)^2}{x}+\frac{b \log \left (\frac{2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{1-a}+\frac{b \log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{a+1} \]

[Out]

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

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Rubi [A]  time = 0.719278, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 15, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.25, Rules used = {6110, 371, 706, 31, 633, 6741, 6122, 6688, 12, 6725, 5921, 2402, 2315, 2447, 5919} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )}{1-a^2}-\frac{b \text{PolyLog}\left (2,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )}{1-a^2}+\frac{b \text{PolyLog}\left (2,-\frac{a+b x+1}{-a-b x+1}\right )}{2 (1-a)}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )}{2 (a+1)}-\frac{2 b \log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{1-a^2}+\frac{2 b \log \left (\frac{2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x)}{1-a^2}-\frac{\coth ^{-1}(a+b x)^2}{x}+\frac{b \log \left (\frac{2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{1-a}+\frac{b \log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)}{a+1} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]^2/x^2,x]

[Out]

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

Rule 6110

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcCoth[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcCoth[c + d*x])^(p - 1))/(1 - (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -
1]

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 706

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

Rule 31

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

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6122

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x
_)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + (f*x)/d)^m*(-(C/d^2) + (C*x^2)/d^2)^q*(a + b*Ar
cCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 - c^2) + 2*A*c*
d, 0] && EqQ[2*c*C - B*d, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 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]]

Rule 5919

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

Rubi steps

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

Mathematica [C]  time = 1.04006, size = 206, normalized size = 0.82 \[ \frac{b x \text{PolyLog}\left (2,e^{2 \tanh ^{-1}\left (\frac{1}{a}\right )-2 \coth ^{-1}(a+b x)}\right )-\left (\sqrt{1-\frac{1}{a^2}} a b x e^{\tanh ^{-1}\left (\frac{1}{a}\right )}+a^2-1\right ) \coth ^{-1}(a+b x)^2+b x \coth ^{-1}(a+b x) \left (-2 \log \left (1-e^{2 \tanh ^{-1}\left (\frac{1}{a}\right )-2 \coth ^{-1}(a+b x)}\right )+2 \tanh ^{-1}\left (\frac{1}{a}\right )-i \pi \right )+b x \left (i \pi \left (\log \left (e^{2 \coth ^{-1}(a+b x)}+1\right )-\log \left (\frac{1}{\sqrt{1-\frac{1}{(a+b x)^2}}}\right )\right )+2 \tanh ^{-1}\left (\frac{1}{a}\right ) \left (\log \left (1-e^{2 \tanh ^{-1}\left (\frac{1}{a}\right )-2 \coth ^{-1}(a+b x)}\right )-\log \left (i \sinh \left (\coth ^{-1}(a+b x)-\tanh ^{-1}\left (\frac{1}{a}\right )\right )\right )\right )\right )}{\left (a^2-1\right ) x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a + b*x]^2/x^2,x]

[Out]

(-((-1 + a^2 + Sqrt[1 - a^(-2)]*a*b*E^ArcTanh[a^(-1)]*x)*ArcCoth[a + b*x]^2) + b*x*ArcCoth[a + b*x]*((-I)*Pi +
2*ArcTanh[a^(-1)] - 2*Log[1 - E^(-2*ArcCoth[a + b*x] + 2*ArcTanh[a^(-1)])]) + b*x*(I*Pi*(Log[1 + E^(2*ArcCoth
[a + b*x])] - Log[1/Sqrt[1 - (a + b*x)^(-2)]]) + 2*ArcTanh[a^(-1)]*(Log[1 - E^(-2*ArcCoth[a + b*x] + 2*ArcTanh
[a^(-1)])] - Log[I*Sinh[ArcCoth[a + b*x] - ArcTanh[a^(-1)]]])) + b*x*PolyLog[2, E^(-2*ArcCoth[a + b*x] + 2*Arc
Tanh[a^(-1)])])/((-1 + a^2)*x)

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Maple [A]  time = 0.096, size = 342, normalized size = 1.4 \begin{align*} -{\frac{ \left ({\rm arccoth} \left (bx+a\right ) \right ) ^{2}}{x}}+2\,{\frac{b{\rm arccoth} \left (bx+a\right )\ln \left ( bx+a-1 \right ) }{2\,a-2}}-2\,{\frac{b{\rm arccoth} \left (bx+a\right )\ln \left ( bx \right ) }{ \left ( 1+a \right ) \left ( a-1 \right ) }}-2\,{\frac{b{\rm arccoth} \left (bx+a\right )\ln \left ( bx+a+1 \right ) }{2+2\,a}}+{\frac{b}{ \left ( 1+a \right ) \left ( a-1 \right ) }{\it dilog} \left ({\frac{bx+a+1}{1+a}} \right ) }+{\frac{b\ln \left ( bx \right ) }{ \left ( 1+a \right ) \left ( a-1 \right ) }\ln \left ({\frac{bx+a+1}{1+a}} \right ) }-{\frac{b}{ \left ( 1+a \right ) \left ( a-1 \right ) }{\it dilog} \left ({\frac{bx+a-1}{a-1}} \right ) }-{\frac{b\ln \left ( bx \right ) }{ \left ( 1+a \right ) \left ( a-1 \right ) }\ln \left ({\frac{bx+a-1}{a-1}} \right ) }+{\frac{b}{2+2\,a}\ln \left ( -{\frac{bx}{2}}-{\frac{a}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) }-{\frac{b\ln \left ( bx+a+1 \right ) }{2+2\,a}\ln \left ( -{\frac{bx}{2}}-{\frac{a}{2}}+{\frac{1}{2}} \right ) }+{\frac{b}{2+2\,a}{\it dilog} \left ({\frac{1}{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) }+{\frac{b \left ( \ln \left ( bx+a+1 \right ) \right ) ^{2}}{4+4\,a}}+{\frac{b \left ( \ln \left ( bx+a-1 \right ) \right ) ^{2}}{4\,a-4}}-{\frac{b}{2\,a-2}{\it dilog} \left ({\frac{1}{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) }-{\frac{b\ln \left ( bx+a-1 \right ) }{2\,a-2}\ln \left ({\frac{1}{2}}+{\frac{bx}{2}}+{\frac{a}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(b*x+a)^2/x^2,x)

[Out]

-arccoth(b*x+a)^2/x+2*b*arccoth(b*x+a)/(2*a-2)*ln(b*x+a-1)-2*b*arccoth(b*x+a)/(a-1)/(1+a)*ln(b*x)-2*b*arccoth(
b*x+a)/(2+2*a)*ln(b*x+a+1)+b/(a-1)/(1+a)*dilog((b*x+a+1)/(1+a))+b/(a-1)/(1+a)*ln(b*x)*ln((b*x+a+1)/(1+a))-b/(a
-1)/(1+a)*dilog((b*x+a-1)/(a-1))-b/(a-1)/(1+a)*ln(b*x)*ln((b*x+a-1)/(a-1))+1/2*b/(1+a)*ln(-1/2*b*x-1/2*a+1/2)*
ln(1/2+1/2*b*x+1/2*a)-1/2*b/(1+a)*ln(-1/2*b*x-1/2*a+1/2)*ln(b*x+a+1)+1/2*b/(1+a)*dilog(1/2+1/2*b*x+1/2*a)+1/4*
b/(1+a)*ln(b*x+a+1)^2+1/4*b/(a-1)*ln(b*x+a-1)^2-1/2*b/(a-1)*dilog(1/2+1/2*b*x+1/2*a)-1/2*b/(a-1)*ln(b*x+a-1)*l
n(1/2+1/2*b*x+1/2*a)

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Maxima [A]  time = 1.02835, size = 329, normalized size = 1.31 \begin{align*} \frac{1}{4} \, b^{2}{\left (\frac{{\left (a - 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \,{\left (a - 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) +{\left (a + 1\right )} \log \left (b x + a - 1\right )^{2}}{a^{2} b - b} - \frac{4 \,{\left (\log \left (b x + a - 1\right ) \log \left (\frac{1}{2} \, b x + \frac{1}{2} \, a + \frac{1}{2}\right ) +{\rm Li}_2\left (-\frac{1}{2} \, b x - \frac{1}{2} \, a + \frac{1}{2}\right )\right )}}{a^{2} b - b} + \frac{4 \,{\left (\log \left (\frac{b x}{a + 1} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a + 1}\right )\right )}}{a^{2} b - b} - \frac{4 \,{\left (\log \left (\frac{b x}{a - 1} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a - 1}\right )\right )}}{a^{2} b - b}\right )} - b{\left (\frac{\log \left (b x + a + 1\right )}{a + 1} - \frac{\log \left (b x + a - 1\right )}{a - 1} + \frac{2 \, \log \left (x\right )}{a^{2} - 1}\right )} \operatorname{arcoth}\left (b x + a\right ) - \frac{\operatorname{arcoth}\left (b x + a\right )^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2/x^2,x, algorithm="maxima")

[Out]

1/4*b^2*(((a - 1)*log(b*x + a + 1)^2 - 2*(a - 1)*log(b*x + a + 1)*log(b*x + a - 1) + (a + 1)*log(b*x + a - 1)^
2)/(a^2*b - b) - 4*(log(b*x + a - 1)*log(1/2*b*x + 1/2*a + 1/2) + dilog(-1/2*b*x - 1/2*a + 1/2))/(a^2*b - b) +
4*(log(b*x/(a + 1) + 1)*log(x) + dilog(-b*x/(a + 1)))/(a^2*b - b) - 4*(log(b*x/(a - 1) + 1)*log(x) + dilog(-b
*x/(a - 1)))/(a^2*b - b)) - b*(log(b*x + a + 1)/(a + 1) - log(b*x + a - 1)/(a - 1) + 2*log(x)/(a^2 - 1))*arcco
th(b*x + a) - arccoth(b*x + a)^2/x

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcoth}\left (b x + a\right )^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2/x^2,x, algorithm="fricas")

[Out]

integral(arccoth(b*x + a)^2/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(b*x+a)**2/x**2,x)

[Out]

Integral(acoth(a + b*x)**2/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (b x + a\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(b*x+a)^2/x^2,x, algorithm="giac")

[Out]

integrate(arccoth(b*x + a)^2/x^2, x)