∫ (a+b tanh−1(cx))2 ___________________________

3.157 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{x (d+e x)} \, dx\)

Optimal. Leaf size=319 \[ \frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d} \]

[Out]

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

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

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

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

g[3, 1 - 2/(1 - c*x)])/(2*d) - (b^2*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2

*d) + (b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d)

________________________________________________________________________________________

Rubi [A] time = 0.430597, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5940, 5914, 6052, 5948, 6058, 6610, 5922} \[ \frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x*(d + e*x)),x]

[Out]

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

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

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

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

g[3, 1 - 2/(1 - c*x)])/(2*d) - (b^2*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2

*d) + (b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d)

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E

xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[

p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5914

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1

- c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTanh[c*x])^(p - 1)*ArcTanh[1 - 2/(1 - c*x)])/(1 - c^2*x^2), x], x]

/; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 6052

Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[

(Log[1 + u]*(a + b*ArcTanh[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTanh[c*x])^p)/(d

+ e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x

))^2, 0]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p

+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT

anh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u]

)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 -

2/(1 - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /

; !FalseQ[w]] /; FreeQ[n, x]

Rule 5922

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[

2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(

b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcTanh[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,

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x (d+e x)} \, dx &=\int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}-\frac{e \left (a+b \tanh ^{-1}(c x)\right )^2}{d (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx}{d}-\frac{e \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx}{d}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{b^2 \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}-\frac{(4 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{b^2 \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}+\frac{(2 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}-\frac{(2 b c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{b^2 \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}+\frac{\left (b^2 c\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}-\frac{\left (b^2 c\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{b^2 \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}\\ \end{align*}

Mathematica [C] time = 12.1933, size = 1034, normalized size = 3.24 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(x*(d + e*x)),x]

[Out]

(a^2*Log[x])/d - (a^2*Log[d + e*x])/d + (a*b*((-I)*c*d*Pi*ArcTanh[c*x] - 2*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x] +

c*d*ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] + 2*c*d

*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*c*d*ArcTanh[(c*d)/e]*Log

[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[

c*x]))] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]]

- c*d*PolyLog[2, E^(-2*ArcTanh[c*x])] + c*d*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(c*d^2) + (

b^2*(I*c*d*Pi^3 - 8*c*d*ArcTanh[c*x]^3 - 8*e*ArcTanh[c*x]^3 + 24*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])

] + 24*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - 12*c*d*PolyLog[3, E^(2*ArcTanh[c*x])] - (24*(c*d - e)

*(c*d + e)*(-6*c*d*ArcTanh[c*x]^3 + 2*e*ArcTanh[c*x]^3 - (4*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^3)/E^ArcTan

h[(c*d)/e] - (6*I)*c*d*Pi*ArcTanh[c*x]*Log[(E^(-ArcTanh[c*x]) + E^ArcTanh[c*x])/2] - 6*c*d*ArcTanh[c*x]^2*Log[

1 + ((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e)] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x

])] + 6*c*d*ArcTanh[c*x]^2*Log[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*(A

rcTanh[(c*d)/e] + ArcTanh[c*x]))] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/2)*E^(-ArcTanh[(c*d)/e] - ArcT

anh[c*x])*(-1 + E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])))] + 6*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*ArcTanh[c*x

])) + c*d*(1 + E^(2*ArcTanh[c*x])))/(2*E^ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]^2*Log[(c*(d + e*x))/Sqrt[1 - c^2*

x^2]] - (3*I)*c*d*Pi*ArcTanh[c*x]*Log[1 - c^2*x^2] - 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*Sinh[ArcTanh[(

c*d)/e] + ArcTanh[c*x]]] - 6*c*d*ArcTanh[c*x]*PolyLog[2, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e))] + 12*c*d

*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 12*c*d*ArcTanh[c*x]*PolyLog[2, E^(ArcTanh[(c*

d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 3*c*d*PolyLo

g[3, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d - e))] - 12*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] -

12*c*d*PolyLog[3, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 3*c*d*PolyLog[3, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x

]))]))/(6*c^2*d^2 - 6*e^2)))/(24*c*d^2)

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Maple [C] time = 0.44, size = 1799, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/x/(e*x+d),x)

[Out]

1/2*I*b^2/d*Pi*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*

x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^2+1/2*I*b^2/d*Pi*arctanh(c*x)^2*csgn(I*(((c*x+1)^2/(-c^2*x

^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1

))/((c*x+1)^2/(-c^2*x^2+1)+1))^2-a^2/d*ln(c*e*x+c*d)-b^2/d*e/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x+1)^2/(-c

^2*x^2+1)/(-c*d+e))-b^2/d*e/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/2*I*b^2/

d*Pi*arctanh(c*x)^2*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+

1)+1))^3+a*b/d*dilog((c*e*x+e)/(-c*d+e))-a*b/d*dilog((c*e*x-e)/(-c*d-e))-b^2*arctanh(c*x)^2/d*ln(c*e*x+c*d)+b^

2*arctanh(c*x)^2/d*ln(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))+1/2*b^2*c/(c*d+e)*polylog(3

,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+a^2/d*ln(c*x)+b^2/d*arctanh(c*x)^2*ln(c*x)+b^2/d*arctanh(c*x)^2*ln(1

+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d*arctanh(c*x)^2*

ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2/d*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))-b^2/d*arctanh(c*x)

^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)-a*b/d*dilog(c*x)-a*b/d*dilog(c*x+1)-2*b^2/d*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/

2))-2*b^2/d*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*I*b^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/((

c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2+1/2*I*b

^2/d*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2-1/2*I*b^2/d*Pi*csgn(I*(

(c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-1/2*

I*b^2/d*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*

arctanh(c*x)^2-1/2*I*b^2/d*Pi*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1

)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/(

(c*x+1)^2/(-c^2*x^2+1)+1))-2*a*b*arctanh(c*x)/d*ln(c*e*x+c*d)+a*b/d*ln(c*e*x+c*d)*ln((c*e*x+e)/(-c*d+e))-a*b/d

*ln(c*e*x+c*d)*ln((c*e*x-e)/(-c*d-e))+1/2*b^2/d*e/(c*d+e)*polylog(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-b

^2*c/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-b^2*c/(c*d+e)*arctanh(c*x)^2*ln(1

-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-a*b/d*ln(c*x)*ln(c*x+1)+2*a*b/d*arctanh(c*x)*ln(c*x)

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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -a^{2}{\left (\frac{\log \left (e x + d\right )}{d} - \frac{\log \left (x\right )}{d}\right )} + \int \frac{b^{2}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}^{2}}{4 \,{\left (e x^{2} + d x\right )}} + \frac{a b{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{e x^{2} + d x}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x/(e*x+d),x, algorithm="maxima")

[Out]

-a^2*(log(e*x + d)/d - log(x)/d) + integrate(1/4*b^2*(log(c*x + 1) - log(-c*x + 1))^2/(e*x^2 + d*x) + a*b*(log

(c*x + 1) - log(-c*x + 1))/(e*x^2 + d*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x/(e*x+d),x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(e*x^2 + d*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x/(e*x+d),x)

[Out]

Integral((a + b*atanh(c*x))**2/(x*(d + e*x)), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/((e*x + d)*x), x)

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