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

Optimal. Leaf size=148 \[ \frac{1}{2} \text{PolyLog}\left (3,1-\frac{2}{a+b x+1}\right )-\frac{1}{2} \text{PolyLog}\left (3,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )+\coth ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )-\coth ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )-\log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)^2+\log \left (\frac{2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x)^2 \]

[Out]

-(ArcCoth[a + b*x]^2*Log[2/(1 + a + b*x)]) + ArcCoth[a + b*x]^2*Log[(2*b*x)/((1 - a)*(1 + a + b*x))] + ArcCoth
[a + b*x]*PolyLog[2, 1 - 2/(1 + a + b*x)] - ArcCoth[a + b*x]*PolyLog[2, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))] +
PolyLog[3, 1 - 2/(1 + a + b*x)]/2 - PolyLog[3, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))]/2

________________________________________________________________________________________

Rubi [A]  time = 0.0913941, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6112, 5923} \[ \frac{1}{2} \text{PolyLog}\left (3,1-\frac{2}{a+b x+1}\right )-\frac{1}{2} \text{PolyLog}\left (3,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )+\coth ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2}{a+b x+1}\right )-\coth ^{-1}(a+b x) \text{PolyLog}\left (2,1-\frac{2 b x}{(1-a) (a+b x+1)}\right )-\log \left (\frac{2}{a+b x+1}\right ) \coth ^{-1}(a+b x)^2+\log \left (\frac{2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a + b*x]^2/x,x]

[Out]

-(ArcCoth[a + b*x]^2*Log[2/(1 + a + b*x)]) + ArcCoth[a + b*x]^2*Log[(2*b*x)/((1 - a)*(1 + a + b*x))] + ArcCoth
[a + b*x]*PolyLog[2, 1 - 2/(1 + a + b*x)] - ArcCoth[a + b*x]*PolyLog[2, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))] +
PolyLog[3, 1 - 2/(1 + a + b*x)]/2 - PolyLog[3, 1 - (2*b*x)/((1 - a)*(1 + a + b*x))]/2

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 5923

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^2*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcCoth[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
b*(a + b*ArcCoth[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcCoth[c*x])*PolyLog[2, 1 - (2*c*(
d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(b^2*PolyLog
[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2,
0]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a+b x)^2}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)^2}{-\frac{a}{b}+\frac{x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\coth ^{-1}(a+b x)^2 \log \left (\frac{2}{1+a+b x}\right )+\coth ^{-1}(a+b x)^2 \log \left (\frac{2 b x}{(1-a) (1+a+b x)}\right )+\coth ^{-1}(a+b x) \text{Li}_2\left (1-\frac{2}{1+a+b x}\right )-\coth ^{-1}(a+b x) \text{Li}_2\left (1-\frac{2 b x}{(1-a) (1+a+b x)}\right )+\frac{1}{2} \text{Li}_3\left (1-\frac{2}{1+a+b x}\right )-\frac{1}{2} \text{Li}_3\left (1-\frac{2 b x}{(1-a) (1+a+b x)}\right )\\ \end{align*}

Mathematica [C]  time = 3.20549, size = 714, normalized size = 4.82 \[ 2 \coth ^{-1}(a+b x) \text{PolyLog}\left (2,-\sqrt{\frac{a-1}{a+1}} e^{\coth ^{-1}(a+b x)}\right )+2 \coth ^{-1}(a+b x) \text{PolyLog}\left (2,\sqrt{\frac{a-1}{a+1}} e^{\coth ^{-1}(a+b x)}\right )-\coth ^{-1}(a+b x) \text{PolyLog}\left (2,e^{2 \coth ^{-1}(a+b x)}\right )-\coth ^{-1}(a+b x) \text{PolyLog}\left (2,\frac{(a-1) e^{2 \coth ^{-1}(a+b x)}}{a+1}\right )-2 \text{PolyLog}\left (3,-\sqrt{\frac{a-1}{a+1}} e^{\coth ^{-1}(a+b x)}\right )-2 \text{PolyLog}\left (3,\sqrt{\frac{a-1}{a+1}} e^{\coth ^{-1}(a+b x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{2 \coth ^{-1}(a+b x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,\frac{(a-1) e^{2 \coth ^{-1}(a+b x)}}{a+1}\right )+\coth ^{-1}(a+b x) \text{PolyLog}\left (2,e^{2 \coth ^{-1}(a+b x)-2 \tanh ^{-1}\left (\frac{1}{a}\right )}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \coth ^{-1}(a+b x)-2 \tanh ^{-1}\left (\frac{1}{a}\right )}\right )+\frac{2}{3} \sqrt{1-\frac{1}{a^2}} a e^{\tanh ^{-1}\left (\frac{1}{a}\right )} \coth ^{-1}(a+b x)^3-\frac{2}{3} a \coth ^{-1}(a+b x)^3-\frac{2}{3} \coth ^{-1}(a+b x)^3+\coth ^{-1}(a+b x)^2 \log \left (1-\sqrt{\frac{a-1}{a+1}} e^{\coth ^{-1}(a+b x)}\right )+\coth ^{-1}(a+b x)^2 \log \left (\sqrt{\frac{a-1}{a+1}} e^{\coth ^{-1}(a+b x)}+1\right )-\coth ^{-1}(a+b x)^2 \log \left (1-e^{2 \coth ^{-1}(a+b x)}\right )-\coth ^{-1}(a+b x)^2 \log \left (1-\frac{(a-1) e^{2 \coth ^{-1}(a+b x)}}{a+1}\right )+\coth ^{-1}(a+b x)^2 \log \left (\frac{1}{2} e^{-\coth ^{-1}(a+b x)} \left (a \left (e^{2 \coth ^{-1}(a+b x)}-1\right )-e^{2 \coth ^{-1}(a+b x)}-1\right )\right )-\log \left (-\frac{b x}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}\right ) \coth ^{-1}(a+b x)^2-i \pi \coth ^{-1}(a+b x) \log \left (\frac{1}{2} \left (e^{-\coth ^{-1}(a+b x)}+e^{\coth ^{-1}(a+b x)}\right )\right )+i \pi \log \left (\frac{1}{\sqrt{1-\frac{1}{(a+b x)^2}}}\right ) \coth ^{-1}(a+b x)+\coth ^{-1}(a+b x)^2 \log \left (1-e^{2 \coth ^{-1}(a+b x)-2 \tanh ^{-1}\left (\frac{1}{a}\right )}\right )-2 \tanh ^{-1}\left (\frac{1}{a}\right ) \coth ^{-1}(a+b x) \log \left (\frac{1}{2} i \left (e^{\coth ^{-1}(a+b x)-\tanh ^{-1}\left (\frac{1}{a}\right )}-e^{\tanh ^{-1}\left (\frac{1}{a}\right )-\coth ^{-1}(a+b x)}\right )\right )+2 \tanh ^{-1}\left (\frac{1}{a}\right ) \coth ^{-1}(a+b x) \log \left (i \sinh \left (\coth ^{-1}(a+b x)-\tanh ^{-1}\left (\frac{1}{a}\right )\right )\right )-\frac{i \pi ^3}{24} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a + b*x]^2/x,x]

[Out]

(-I/24)*Pi^3 - (2*ArcCoth[a + b*x]^3)/3 - (2*a*ArcCoth[a + b*x]^3)/3 + (2*Sqrt[1 - a^(-2)]*a*E^ArcTanh[a^(-1)]
*ArcCoth[a + b*x]^3)/3 - I*Pi*ArcCoth[a + b*x]*Log[(E^(-ArcCoth[a + b*x]) + E^ArcCoth[a + b*x])/2] + ArcCoth[a
+ b*x]^2*Log[1 - Sqrt[(-1 + a)/(1 + a)]*E^ArcCoth[a + b*x]] + ArcCoth[a + b*x]^2*Log[1 + Sqrt[(-1 + a)/(1 + a
)]*E^ArcCoth[a + b*x]] - ArcCoth[a + b*x]^2*Log[1 - E^(2*ArcCoth[a + b*x])] - ArcCoth[a + b*x]^2*Log[1 - ((-1
+ a)*E^(2*ArcCoth[a + b*x]))/(1 + a)] + ArcCoth[a + b*x]^2*Log[1 - E^(2*ArcCoth[a + b*x] - 2*ArcTanh[a^(-1)])]
- 2*ArcCoth[a + b*x]*ArcTanh[a^(-1)]*Log[(I/2)*(E^(ArcCoth[a + b*x] - ArcTanh[a^(-1)]) - E^(-ArcCoth[a + b*x]
+ ArcTanh[a^(-1)]))] + ArcCoth[a + b*x]^2*Log[(-1 - E^(2*ArcCoth[a + b*x]) + a*(-1 + E^(2*ArcCoth[a + b*x])))
/(2*E^ArcCoth[a + b*x])] + I*Pi*ArcCoth[a + b*x]*Log[1/Sqrt[1 - (a + b*x)^(-2)]] - ArcCoth[a + b*x]^2*Log[-((b
*x)/((a + b*x)*Sqrt[1 - (a + b*x)^(-2)]))] + 2*ArcCoth[a + b*x]*ArcTanh[a^(-1)]*Log[I*Sinh[ArcCoth[a + b*x] -
ArcTanh[a^(-1)]]] + 2*ArcCoth[a + b*x]*PolyLog[2, -(Sqrt[(-1 + a)/(1 + a)]*E^ArcCoth[a + b*x])] + 2*ArcCoth[a
+ b*x]*PolyLog[2, Sqrt[(-1 + a)/(1 + a)]*E^ArcCoth[a + b*x]] - ArcCoth[a + b*x]*PolyLog[2, E^(2*ArcCoth[a + b*
x])] - ArcCoth[a + b*x]*PolyLog[2, ((-1 + a)*E^(2*ArcCoth[a + b*x]))/(1 + a)] + ArcCoth[a + b*x]*PolyLog[2, E^
(2*ArcCoth[a + b*x] - 2*ArcTanh[a^(-1)])] - 2*PolyLog[3, -(Sqrt[(-1 + a)/(1 + a)]*E^ArcCoth[a + b*x])] - 2*Pol
yLog[3, Sqrt[(-1 + a)/(1 + a)]*E^ArcCoth[a + b*x]] + PolyLog[3, E^(2*ArcCoth[a + b*x])]/2 + PolyLog[3, ((-1 +
a)*E^(2*ArcCoth[a + b*x]))/(1 + a)]/2 - PolyLog[3, E^(2*ArcCoth[a + b*x] - 2*ArcTanh[a^(-1)])]/2

________________________________________________________________________________________

Maple [C]  time = 0.543, size = 985, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(b*x)*arccoth(b*x+a)^2-arccoth(b*x+a)^2*ln(a*((b*x+a+1)/(b*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1)+arccoth(b*x+a)^2
*ln((b*x+a+1)/(b*x+a-1)-1)-arccoth(b*x+a)^2*ln(1-1/((b*x+a-1)/(b*x+a+1))^(1/2))-2*arccoth(b*x+a)*polylog(2,1/(
(b*x+a-1)/(b*x+a+1))^(1/2))+2*polylog(3,1/((b*x+a-1)/(b*x+a+1))^(1/2))-arccoth(b*x+a)^2*ln(1+1/((b*x+a-1)/(b*x
+a+1))^(1/2))-2*arccoth(b*x+a)*polylog(2,-1/((b*x+a-1)/(b*x+a+1))^(1/2))+2*polylog(3,-1/((b*x+a-1)/(b*x+a+1))^
(1/2))-1/2*I*Pi*arccoth(b*x+a)^2*csgn(I*(a*((b*x+a+1)/(b*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1))*csgn(I*(a*((b*x+a+1
)/(b*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1)/((b*x+a+1)/(b*x+a-1)-1))^2+I*Pi*arccoth(b*x+a)^2*csgn(I*(a*((b*x+a+1)/(b
*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1)/((b*x+a+1)/(b*x+a-1)-1))^2+1/2*I*Pi*arccoth(b*x+a)^2*csgn(I*(a*((b*x+a+1)/(b
*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1))*csgn(I*(a*((b*x+a+1)/(b*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1)/((b*x+a+1)/(b*x+a-
1)-1))*csgn(I/((b*x+a+1)/(b*x+a-1)-1))-I*Pi*arccoth(b*x+a)^2-1/2*I*Pi*arccoth(b*x+a)^2*csgn(I*(a*((b*x+a+1)/(b
*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1)/((b*x+a+1)/(b*x+a-1)-1))^2*csgn(I/((b*x+a+1)/(b*x+a-1)-1))-1/2*I*Pi*arccoth(
b*x+a)^2*csgn(I*(a*((b*x+a+1)/(b*x+a-1)-1)-(b*x+a+1)/(b*x+a-1)-1)/((b*x+a+1)/(b*x+a-1)-1))^3+a/(a-1)*arccoth(b
*x+a)^2*ln(1-(a-1)*(b*x+a+1)/(b*x+a-1)/(1+a))+a/(a-1)*arccoth(b*x+a)*polylog(2,(a-1)*(b*x+a+1)/(b*x+a-1)/(1+a)
)-1/2*a/(a-1)*polylog(3,(a-1)*(b*x+a+1)/(b*x+a-1)/(1+a))-1/(a-1)*arccoth(b*x+a)^2*ln(1-(a-1)*(b*x+a+1)/(b*x+a-
1)/(1+a))-1/(a-1)*arccoth(b*x+a)*polylog(2,(a-1)*(b*x+a+1)/(b*x+a-1)/(1+a))+1/2/(a-1)*polylog(3,(a-1)*(b*x+a+1
)/(b*x+a-1)/(1+a))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(b*x + a)^2/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(b*x + a)^2/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(a + b*x)**2/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(b*x + a)^2/x, x)