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

Optimal. Leaf size=412 \[ \frac{b e \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac{b e \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac{b e \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac{b e \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}-\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d^2}-\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d^2}-\frac{b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d}-\frac{2 e \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac{e \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac{e \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d} \]

[Out]

(c*(a + b*ArcTanh[c*x])^2)/d - (a + b*ArcTanh[c*x])^2/(d*x) - (2*e*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c
*x)])/d^2 - (e*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/d^2 + (e*(a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((
c*d + e)*(1 + c*x))])/d^2 + (2*b*c*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/d + (b*e*(a + b*ArcTanh[c*x])*Po
lyLog[2, 1 - 2/(1 - c*x)])/d^2 - (b*e*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d^2 + (b*e*(a + b*Arc
Tanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^2 - (b^2*c*PolyLog[2, -1 + 2/(1 + c*x)])/d - (b*e*(a + b*ArcTanh[c*x
])*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/d^2 - (b^2*e*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^2) +
(b^2*e*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d^2) + (b^2*e*PolyLog[3, 1 - 2/(1 + c*x)])/(2*d^2) - (b^2*e*PolyLog[3,
1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.603766, antiderivative size = 412, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {5940, 5916, 5988, 5932, 2447, 5914, 6052, 5948, 6058, 6610, 5922} \[ \frac{b e \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac{b e \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac{b e \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac{b e \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}-\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d^2}-\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d^2}-\frac{b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d}-\frac{2 e \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac{e \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac{e \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}+\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(a + b*ArcTanh[c*x])^2)/d - (a + b*ArcTanh[c*x])^2/(d*x) - (2*e*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c
*x)])/d^2 - (e*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/d^2 + (e*(a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((
c*d + e)*(1 + c*x))])/d^2 + (2*b*c*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/d + (b*e*(a + b*ArcTanh[c*x])*Po
lyLog[2, 1 - 2/(1 - c*x)])/d^2 - (b*e*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d^2 + (b*e*(a + b*Arc
Tanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^2 - (b^2*c*PolyLog[2, -1 + 2/(1 + c*x)])/d - (b*e*(a + b*ArcTanh[c*x
])*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/d^2 - (b^2*e*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^2) +
(b^2*e*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d^2) + (b^2*e*PolyLog[3, 1 - 2/(1 + c*x)])/(2*d^2) - (b^2*e*PolyLog[3,
1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*d^2)

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 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5988

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

Rule 5932

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

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 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,
0]

Rubi steps

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

Mathematica [C]  time = 13.593, size = 1188, normalized size = 2.88 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a^2/(d*x)) - (a^2*e*Log[x])/d^2 + (a^2*e*Log[d + e*x])/d^2 + (a*b*(I*c*d*e*Pi*ArcTanh[c*x] - (2*c*d^2*ArcTan
h[c*x])/x + 2*c*d*e*ArcTanh[(c*d)/e]*ArcTanh[c*x] - c*d*e*ArcTanh[c*x]^2 + e^2*ArcTanh[c*x]^2 - (Sqrt[1 - (c^2
*d^2)/e^2]*e^2*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] - 2*c*d*e*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - I*c*d*
e*Pi*Log[1 + E^(2*ArcTanh[c*x])] + 2*c*d*e*ArcTanh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]
+ 2*c*d*e*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 2*c^2*d^2*Log[(c*x)/Sqrt[1 - c^2*x^
2]] - (I/2)*c*d*e*Pi*Log[1 - c^2*x^2] - 2*c*d*e*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]]
+ c*d*e*PolyLog[2, E^(-2*ArcTanh[c*x])] - c*d*e*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(c*d^3)
+ (b^2*((-I)*c*d*e*Pi^3 + 24*c^2*d^2*ArcTanh[c*x]^2 - (24*c*d^2*ArcTanh[c*x]^2)/x + 8*c*d*e*ArcTanh[c*x]^3 +
8*e^2*ArcTanh[c*x]^3 + 48*c^2*d^2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - 24*c*d*e*ArcTanh[c*x]^2*Log[1 -
E^(2*ArcTanh[c*x])] - 24*c^2*d^2*PolyLog[2, E^(-2*ArcTanh[c*x])] - 24*c*d*e*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTa
nh[c*x])] + 12*c*d*e*PolyLog[3, E^(2*ArcTanh[c*x])]))/(24*c*d^3) + (b^2*(c*d - e)*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^ArcTanh[(c*d)/e] - (6*I)*c*d*Pi*A
rcTanh[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*(ArcTanh[(c*d)/e] + ArcTanh[c
*x]))] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/2)*E^(-ArcTanh[(c*d)/e] - ArcTanh[c*x])*(-1 + E^(2*(ArcTa
nh[(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*ArcTan
h[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*PolyLog[3, -(((c*d + e)*E^(2*ArcT
anh[c*x]))/(c*d - e))] - 12*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 12*c*d*PolyLog[3, E^(ArcTan
h[(c*d)/e] + ArcTanh[c*x])] - 3*c*d*PolyLog[3, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(d^3*(6*c^3*d^2 - 6*
c*e^2))

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Maple [C]  time = 2.493, size = 26776, normalized size = 65. \begin{align*} \text{output 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^2/(e*x+d),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2}{\left (\frac{e \log \left (e x + d\right )}{d^{2}} - \frac{e \log \left (x\right )}{d^{2}} - \frac{1}{d x}\right )} - \frac{b^{2} \log \left (-c x + 1\right )^{2}}{4 \, d x} - \int -\frac{{\left (b^{2} c d x - b^{2} d\right )} \log \left (c x + 1\right )^{2} + 4 \,{\left (a b c d x - a b d\right )} \log \left (c x + 1\right ) + 2 \,{\left (b^{2} c e x^{2} + 2 \, a b d -{\left (2 \, a b c d - b^{2} c d\right )} x -{\left (b^{2} c d x - b^{2} d\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \,{\left (c d e x^{4} - d^{2} x^{2} +{\left (c d^{2} - d e\right )} x^{3}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/(d*x)) - 1/4*b^2*log(-c*x + 1)^2/(d*x) - integrate(-1/4*((b^2*c*d*x
- b^2*d)*log(c*x + 1)^2 + 4*(a*b*c*d*x - a*b*d)*log(c*x + 1) + 2*(b^2*c*e*x^2 + 2*a*b*d - (2*a*b*c*d - b^2*c*
d)*x - (b^2*c*d*x - b^2*d)*log(c*x + 1))*log(-c*x + 1))/(c*d*e*x^4 - d^2*x^2 + (c*d^2 - d*e)*x^3), 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^{3} + d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(e*x^3 + d*x^2), x)

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

[Out]

Timed out

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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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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