3.116 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{x (d+c d x)^3} \, dx\)

Optimal. Leaf size=362 \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d^3}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d^3}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (c x+1)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}+\frac{11 b^2}{16 d^3 (c x+1)}+\frac{b^2}{16 d^3 (c x+1)^2}-\frac{11 b^2 \tanh ^{-1}(c x)}{16 d^3} \]

[Out]

b^2/(16*d^3*(1 + c*x)^2) + (11*b^2)/(16*d^3*(1 + c*x)) - (11*b^2*ArcTanh[c*x])/(16*d^3) + (b*(a + b*ArcTanh[c*
x]))/(4*d^3*(1 + c*x)^2) + (5*b*(a + b*ArcTanh[c*x]))/(4*d^3*(1 + c*x)) - (5*(a + b*ArcTanh[c*x])^2)/(8*d^3) +
(a + b*ArcTanh[c*x])^2/(2*d^3*(1 + c*x)^2) + (a + b*ArcTanh[c*x])^2/(d^3*(1 + c*x)) + (2*(a + b*ArcTanh[c*x])
^2*ArcTanh[1 - 2/(1 - c*x)])/d^3 + ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/d^3 - (b*(a + b*ArcTanh[c*x])*Pol
yLog[2, 1 - 2/(1 - c*x)])/d^3 + (b*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d^3 - (b*(a + b*ArcTanh[
c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^3 + (b^2*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^3) - (b^2*PolyLog[3, -1 + 2/(1
- c*x)])/(2*d^3) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.804001, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 13, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.591, Rules used = {5940, 5914, 6052, 5948, 6058, 6610, 5928, 5926, 627, 44, 207, 5918, 6056} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d^3}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d^3}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (c x+1)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}+\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}+\frac{11 b^2}{16 d^3 (c x+1)}+\frac{b^2}{16 d^3 (c x+1)^2}-\frac{11 b^2 \tanh ^{-1}(c x)}{16 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x*(d + c*d*x)^3),x]

[Out]

b^2/(16*d^3*(1 + c*x)^2) + (11*b^2)/(16*d^3*(1 + c*x)) - (11*b^2*ArcTanh[c*x])/(16*d^3) + (b*(a + b*ArcTanh[c*
x]))/(4*d^3*(1 + c*x)^2) + (5*b*(a + b*ArcTanh[c*x]))/(4*d^3*(1 + c*x)) - (5*(a + b*ArcTanh[c*x])^2)/(8*d^3) +
(a + b*ArcTanh[c*x])^2/(2*d^3*(1 + c*x)^2) + (a + b*ArcTanh[c*x])^2/(d^3*(1 + c*x)) + (2*(a + b*ArcTanh[c*x])
^2*ArcTanh[1 - 2/(1 - c*x)])/d^3 + ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/d^3 - (b*(a + b*ArcTanh[c*x])*Pol
yLog[2, 1 - 2/(1 - c*x)])/d^3 + (b*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d^3 - (b*(a + b*ArcTanh[
c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^3 + (b^2*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^3) - (b^2*PolyLog[3, -1 + 2/(1
- c*x)])/(2*d^3) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*d^3)

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 5928

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcTanh[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p
- 1), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] &
& NeQ[q, -1]

Rule 5926

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 5918

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

Rule 6056

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x (d+c d x)^3} \, dx &=\int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)^3}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)^2}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}\right ) \, dx\\ &=\frac{\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx}{d^3}-\frac{c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx}{d^3}-\frac{c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{d^3}-\frac{c \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{d^3}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{(b c) \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac{a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^3}-\frac{(2 b c) \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^3}-\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^3}-\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^3}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}-\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 d^3}+\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 d^3}-\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 d^3}-\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{d^3}+\frac{(b c) \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{d^3}+\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^3}-\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^3}+\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^3}\\ &=\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}-\frac{\left (b^2 c\right ) \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 d^3}-\frac{\left (b^2 c\right ) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 d^3}-\frac{\left (b^2 c\right ) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{d^3}+\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^3}-\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^3}\\ &=\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}-\frac{\left (b^2 c\right ) \int \frac{1}{(1-c x) (1+c x)^3} \, dx}{4 d^3}-\frac{\left (b^2 c\right ) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{4 d^3}-\frac{\left (b^2 c\right ) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{d^3}\\ &=\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}-\frac{\left (b^2 c\right ) \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 d^3}-\frac{\left (b^2 c\right ) \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 d^3}-\frac{\left (b^2 c\right ) \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^3}\\ &=\frac{b^2}{16 d^3 (1+c x)^2}+\frac{11 b^2}{16 d^3 (1+c x)}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}+\frac{\left (b^2 c\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{16 d^3}+\frac{\left (b^2 c\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{8 d^3}+\frac{\left (b^2 c\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{2 d^3}\\ &=\frac{b^2}{16 d^3 (1+c x)^2}+\frac{11 b^2}{16 d^3 (1+c x)}-\frac{11 b^2 \tanh ^{-1}(c x)}{16 d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}-\frac{5 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}+\frac{2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}\\ \end{align*}

Mathematica [C]  time = 1.40131, size = 376, normalized size = 1.04 \[ \frac{12 a b \left (-16 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-12 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )+12 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (8 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-6 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )+6 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )+b^2 \left (192 \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-96 \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-128 \tanh ^{-1}(c x)^3+192 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-144 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )-24 \tanh ^{-1}(c x)^2 \sinh \left (4 \tanh ^{-1}(c x)\right )-144 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-12 \tanh ^{-1}(c x) \sinh \left (4 \tanh ^{-1}(c x)\right )-72 \sinh \left (2 \tanh ^{-1}(c x)\right )-3 \sinh \left (4 \tanh ^{-1}(c x)\right )+144 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+24 \tanh ^{-1}(c x)^2 \cosh \left (4 \tanh ^{-1}(c x)\right )+144 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+12 \tanh ^{-1}(c x) \cosh \left (4 \tanh ^{-1}(c x)\right )+72 \cosh \left (2 \tanh ^{-1}(c x)\right )+3 \cosh \left (4 \tanh ^{-1}(c x)\right )+8 i \pi ^3\right )+\frac{192 a^2}{c x+1}+\frac{96 a^2}{(c x+1)^2}+192 a^2 \log (c x)-192 a^2 \log (c x+1)}{192 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(x*(d + c*d*x)^3),x]

[Out]

((96*a^2)/(1 + c*x)^2 + (192*a^2)/(1 + c*x) + 192*a^2*Log[c*x] - 192*a^2*Log[1 + c*x] + 12*a*b*(12*Cosh[2*ArcT
anh[c*x]] + Cosh[4*ArcTanh[c*x]] - 16*PolyLog[2, E^(-2*ArcTanh[c*x])] - 12*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh[c*
x]*(6*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 8*Log[1 - E^(-2*ArcTanh[c*x])] - 6*Sinh[2*ArcTanh[c*x]] -
Sinh[4*ArcTanh[c*x]]) - Sinh[4*ArcTanh[c*x]]) + b^2*((8*I)*Pi^3 - 128*ArcTanh[c*x]^3 + 72*Cosh[2*ArcTanh[c*x]]
+ 144*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] + 144*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x]] + 3*Cosh[4*ArcTanh[c*x]] +
12*ArcTanh[c*x]*Cosh[4*ArcTanh[c*x]] + 24*ArcTanh[c*x]^2*Cosh[4*ArcTanh[c*x]] + 192*ArcTanh[c*x]^2*Log[1 - E^(
2*ArcTanh[c*x])] + 192*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - 96*PolyLog[3, E^(2*ArcTanh[c*x])] - 72*Si
nh[2*ArcTanh[c*x]] - 144*ArcTanh[c*x]*Sinh[2*ArcTanh[c*x]] - 144*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]] - 3*Sinh[
4*ArcTanh[c*x]] - 12*ArcTanh[c*x]*Sinh[4*ArcTanh[c*x]] - 24*ArcTanh[c*x]^2*Sinh[4*ArcTanh[c*x]]))/(192*d^3)

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Maple [C]  time = 0.432, size = 1752, normalized size = 4.8 \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/(c*d*x+d)^3,x)

[Out]

1/2*a^2/d^3/(c*x+1)^2+a^2/d^3/(c*x+1)-a^2/d^3*ln(c*x+1)+a^2/d^3*ln(c*x)-5/8*b^2/d^3*arctanh(c*x)^2-2/3*b^2/d^3
*arctanh(c*x)^3+1/2*I*b^2/d^3*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^3*Pi*csgn(I/((c*x+1)^2/(-c^2
*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x
)^2+1/64*b^2/d^3/(c*x+1)^2+3/8*b^2/d^3/(c*x+1)+1/2*I*b^2/d^3*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c
*x+1)^2/(c^2*x^2-1))*arctanh(c*x)^2-1/2*I*b^2/d^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1
)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2-1/2*I*b^2/d^3*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^3*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^3*Pi*csgn
(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-3/4*b^2/d^3*arctanh(c*x)/(c*x+1)*c*x+1/16*b^2/d^3*arctanh(c*x)/(c*x
+1)^2*c^2*x^2-1/8*b^2/d^3*arctanh(c*x)/(c*x+1)^2*c*x+1/2*I*b^2/d^3*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/
(-c^2*x^2+1)+1))^3*arctanh(c*x)^2+1/2*I*b^2/d^3*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^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^
2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+I*b^2/d^3*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-
1))^2*arctanh(c*x)^2-2*b^2/d^3*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))-2*b^2/d^3*polylog(3,(c*x+1)/(-c^2*x^2+1)
^(1/2))+1/64*b^2/d^3/(c*x+1)^2*c^2*x^2-1/32*b^2/d^3/(c*x+1)^2*c*x-3/8*b^2/d^3/(c*x+1)*c*x+a*b/d^3*ln(-1/2*c*x+
1/2)*ln(1/2+1/2*c*x)+a*b/d^3*arctanh(c*x)/(c*x+1)^2+2*a*b/d^3*arctanh(c*x)/(c*x+1)-2*a*b/d^3*arctanh(c*x)*ln(c
*x+1)-a*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/4*a*b/d^3/(c*x+1)^2+5/4*a*b/d^3/(c*x+1)+a*b/d^3*dilog(1/2+1/2*c*x)+
1/2*a*b/d^3*ln(c*x+1)^2+5/8*a*b/d^3*ln(c*x-1)-5/8*a*b/d^3*ln(c*x+1)+2*b^2/d^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*
x^2+1)^(1/2))-b^2/d^3*arctanh(c*x)^2*ln(c*x+1)+1/2*b^2/d^3*arctanh(c*x)^2/(c*x+1)^2+b^2/d^3*arctanh(c*x)^2/(c*
x+1)+b^2/d^3*arctanh(c*x)^2*ln(2)+3/4*b^2/d^3*arctanh(c*x)/(c*x+1)+1/16*b^2/d^3*arctanh(c*x)/(c*x+1)^2+2*b^2/d
^3*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-b^2/d^3*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+b^2
/d^3*arctanh(c*x)^2*ln(c*x)-a*b/d^3*dilog(c*x)-a*b/d^3*dilog(c*x+1)+b^2/d^3*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*
x^2+1)^(1/2))+2*b^2/d^3*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d^3*arctanh(c*x)^2*ln(1+(c*x+1)
/(-c^2*x^2+1)^(1/2))+2*a*b/d^3*arctanh(c*x)*ln(c*x)-a*b/d^3*ln(c*x)*ln(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{2 \, c x + 3}{c^{2} d^{3} x^{2} + 2 \, c d^{3} x + d^{3}} - \frac{2 \, \log \left (c x + 1\right )}{d^{3}} + \frac{2 \, \log \left (x\right )}{d^{3}}\right )} + \frac{{\left (2 \, b^{2} c x + 3 \, b^{2} - 2 \,{\left (b^{2} c^{2} x^{2} + 2 \, b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \,{\left (c^{2} d^{3} x^{2} + 2 \, c d^{3} x + d^{3}\right )}} + \int \frac{{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )^{2} + 4 \,{\left (a b c x - a b\right )} \log \left (c x + 1\right ) -{\left (2 \, b^{2} c^{3} x^{3} + 5 \, b^{2} c^{2} x^{2} - 4 \, a b +{\left (4 \, a b c + 3 \, b^{2} c\right )} x - 2 \,{\left (b^{2} c^{4} x^{4} + 3 \, b^{2} c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \,{\left (c^{4} d^{3} x^{5} + 2 \, c^{3} d^{3} x^{4} - 2 \, c d^{3} x^{2} - d^{3} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

1/2*a^2*((2*c*x + 3)/(c^2*d^3*x^2 + 2*c*d^3*x + d^3) - 2*log(c*x + 1)/d^3 + 2*log(x)/d^3) + 1/8*(2*b^2*c*x + 3
*b^2 - 2*(b^2*c^2*x^2 + 2*b^2*c*x + b^2)*log(c*x + 1))*log(-c*x + 1)^2/(c^2*d^3*x^2 + 2*c*d^3*x + d^3) + integ
rate(1/4*((b^2*c*x - b^2)*log(c*x + 1)^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) - (2*b^2*c^3*x^3 + 5*b^2*c^2*x^2 - 4
*a*b + (4*a*b*c + 3*b^2*c)*x - 2*(b^2*c^4*x^4 + 3*b^2*c^3*x^3 + 3*b^2*c^2*x^2 + b^2)*log(c*x + 1))*log(-c*x +
1))/(c^4*d^3*x^5 + 2*c^3*d^3*x^4 - 2*c*d^3*x^2 - d^3*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}}{c^{3} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{3} + 3 \, c d^{3} x^{2} + d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(c^3*d^3*x^4 + 3*c^2*d^3*x^3 + 3*c*d^3*x^2 + d^3*x),
x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{3} x^{4} + 3 c^{2} x^{3} + 3 c x^{2} + x}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{c^{3} x^{4} + 3 c^{2} x^{3} + 3 c x^{2} + x}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{c^{3} x^{4} + 3 c^{2} x^{3} + 3 c x^{2} + x}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x/(c*d*x+d)**3,x)

[Out]

(Integral(a**2/(c**3*x**4 + 3*c**2*x**3 + 3*c*x**2 + x), x) + Integral(b**2*atanh(c*x)**2/(c**3*x**4 + 3*c**2*
x**3 + 3*c*x**2 + x), x) + Integral(2*a*b*atanh(c*x)/(c**3*x**4 + 3*c**2*x**3 + 3*c*x**2 + x), x))/d**3

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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 (c d x + d\right )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x/(c*d*x+d)^3,x, algorithm="giac")

[Out]

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