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

Optimal. Leaf size=448 \[ \frac{3 b c \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{3 b c \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac{3 b c \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d^3}-\frac{3 b^2 c \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d^3}+\frac{3 b^2 c \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d^3}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (c x+1)}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{6 c \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}+\frac{2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{3 c \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}-\frac{19 b^2 c}{16 d^3 (c x+1)}-\frac{b^2 c}{16 d^3 (c x+1)^2}+\frac{19 b^2 c \tanh ^{-1}(c x)}{16 d^3} \]

[Out]

-(b^2*c)/(16*d^3*(1 + c*x)^2) - (19*b^2*c)/(16*d^3*(1 + c*x)) + (19*b^2*c*ArcTanh[c*x])/(16*d^3) - (b*c*(a + b
*ArcTanh[c*x]))/(4*d^3*(1 + c*x)^2) - (9*b*c*(a + b*ArcTanh[c*x]))/(4*d^3*(1 + c*x)) + (17*c*(a + b*ArcTanh[c*
x])^2)/(8*d^3) - (a + b*ArcTanh[c*x])^2/(d^3*x) - (c*(a + b*ArcTanh[c*x])^2)/(2*d^3*(1 + c*x)^2) - (2*c*(a + b
*ArcTanh[c*x])^2)/(d^3*(1 + c*x)) - (6*c*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c*x)])/d^3 - (3*c*(a + b*Ar
cTanh[c*x])^2*Log[2/(1 + c*x)])/d^3 + (2*b*c*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/d^3 + (3*b*c*(a + b*Ar
cTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)])/d^3 - (3*b*c*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d^3 +
(3*b*c*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^3 - (b^2*c*PolyLog[2, -1 + 2/(1 + c*x)])/d^3 - (3*
b^2*c*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^3) + (3*b^2*c*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d^3) + (3*b^2*c*PolyLog
[3, 1 - 2/(1 + c*x)])/(2*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.982703, antiderivative size = 448, normalized size of antiderivative = 1., number of steps used = 36, number of rules used = 17, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.773, Rules used = {5940, 5916, 5988, 5932, 2447, 5914, 6052, 5948, 6058, 6610, 5928, 5926, 627, 44, 207, 5918, 6056} \[ \frac{3 b c \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{3 b c \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac{3 b c \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{b^2 c \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )}{d^3}-\frac{3 b^2 c \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d^3}+\frac{3 b^2 c \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d^3}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (c x+1)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (c x+1)}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{6 c \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}+\frac{2 b c \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}-\frac{3 c \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}-\frac{19 b^2 c}{16 d^3 (c x+1)}-\frac{b^2 c}{16 d^3 (c x+1)^2}+\frac{19 b^2 c \tanh ^{-1}(c x)}{16 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*c)/(16*d^3*(1 + c*x)^2) - (19*b^2*c)/(16*d^3*(1 + c*x)) + (19*b^2*c*ArcTanh[c*x])/(16*d^3) - (b*c*(a + b
*ArcTanh[c*x]))/(4*d^3*(1 + c*x)^2) - (9*b*c*(a + b*ArcTanh[c*x]))/(4*d^3*(1 + c*x)) + (17*c*(a + b*ArcTanh[c*
x])^2)/(8*d^3) - (a + b*ArcTanh[c*x])^2/(d^3*x) - (c*(a + b*ArcTanh[c*x])^2)/(2*d^3*(1 + c*x)^2) - (2*c*(a + b
*ArcTanh[c*x])^2)/(d^3*(1 + c*x)) - (6*c*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c*x)])/d^3 - (3*c*(a + b*Ar
cTanh[c*x])^2*Log[2/(1 + c*x)])/d^3 + (2*b*c*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/d^3 + (3*b*c*(a + b*Ar
cTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)])/d^3 - (3*b*c*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)])/d^3 +
(3*b*c*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^3 - (b^2*c*PolyLog[2, -1 + 2/(1 + c*x)])/d^3 - (3*
b^2*c*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^3) + (3*b^2*c*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d^3) + (3*b^2*c*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 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 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^2 (d+c d x)^3} \, dx &=\int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x^2}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}+\frac{c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)^3}+\frac{2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)^2}+\frac{3 c^2 \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^2} \, dx}{d^3}-\frac{(3 c) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx}{d^3}+\frac{c^2 \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx}{d^3}+\frac{\left (2 c^2\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{d^3}+\frac{\left (3 c^2\right ) \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}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{(2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d^3}+\frac{\left (b c^2\right ) \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{\left (4 b c^2\right ) \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{\left (6 b c^2\right ) \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{\left (12 b c^2\right ) \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{c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}+\frac{(2 b c) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{d^3}+\frac{\left (b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 d^3}-\frac{\left (b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 d^3}+\frac{\left (b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 d^3}+\frac{\left (2 b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{d^3}-\frac{\left (2 b c^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{d^3}-\frac{\left (6 b c^2\right ) \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{\left (6 b c^2\right ) \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 (3 b^2 c^2\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^3}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}+\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}+\frac{\left (b^2 c^2\right ) \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 d^3}+\frac{\left (b^2 c^2\right ) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 d^3}+\frac{\left (2 b^2 c^2\right ) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{d^3}-\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^3}-\frac{\left (3 b^2 c^2\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d^3}+\frac{\left (3 b^2 c^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d^3}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}-\frac{b^2 c \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d^3}-\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}+\frac{\left (b^2 c^2\right ) \int \frac{1}{(1-c x) (1+c x)^3} \, dx}{4 d^3}+\frac{\left (b^2 c^2\right ) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{4 d^3}+\frac{\left (2 b^2 c^2\right ) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{d^3}\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}-\frac{b^2 c \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d^3}-\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}+\frac{\left (b^2 c^2\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^2\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 (2 b^2 c^2\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 c}{16 d^3 (1+c x)^2}-\frac{19 b^2 c}{16 d^3 (1+c x)}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}-\frac{b^2 c \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d^3}-\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}-\frac{\left (b^2 c^2\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{16 d^3}-\frac{\left (b^2 c^2\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{8 d^3}-\frac{\left (b^2 c^2\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{d^3}\\ &=-\frac{b^2 c}{16 d^3 (1+c x)^2}-\frac{19 b^2 c}{16 d^3 (1+c x)}+\frac{19 b^2 c \tanh ^{-1}(c x)}{16 d^3}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)^2}-\frac{9 b c \left (a+b \tanh ^{-1}(c x)\right )}{4 d^3 (1+c x)}+\frac{17 c \left (a+b \tanh ^{-1}(c x)\right )^2}{8 d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 x}-\frac{c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^3 (1+c x)}-\frac{6 c \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{d^3}+\frac{2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d^3}-\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d^3}+\frac{3 b c \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d^3}-\frac{b^2 c \text{Li}_2\left (-1+\frac{2}{1+c x}\right )}{d^3}-\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d^3}+\frac{3 b^2 c \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d^3}\\ \end{align*}

Mathematica [C]  time = 2.26037, size = 479, normalized size = 1.07 \[ \frac{\frac{4 a b \left (48 c x \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+c x \left (32 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+20 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )-20 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )\right )-4 \tanh ^{-1}(c x) \left (24 c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-10 c x \sinh \left (2 \tanh ^{-1}(c x)\right )-c x \sinh \left (4 \tanh ^{-1}(c x)\right )+10 c x \cosh \left (2 \tanh ^{-1}(c x)\right )+c x \cosh \left (4 \tanh ^{-1}(c x)\right )+8\right )\right )}{x}+b^2 c \left (-192 \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-64 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+96 \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )+128 \tanh ^{-1}(c x)^3-\frac{64 \tanh ^{-1}(c x)^2}{c x}+64 \tanh ^{-1}(c x)^2-192 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+128 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+80 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )+8 \tanh ^{-1}(c x)^2 \sinh \left (4 \tanh ^{-1}(c x)\right )+80 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \sinh \left (4 \tanh ^{-1}(c x)\right )+40 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )-80 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )-8 \tanh ^{-1}(c x)^2 \cosh \left (4 \tanh ^{-1}(c x)\right )-80 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )-4 \tanh ^{-1}(c x) \cosh \left (4 \tanh ^{-1}(c x)\right )-40 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )-8 i \pi ^3\right )-\frac{128 a^2 c}{c x+1}-\frac{32 a^2 c}{(c x+1)^2}-192 a^2 c \log (x)+192 a^2 c \log (c x+1)-\frac{64 a^2}{x}}{64 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-64*a^2)/x - (32*a^2*c)/(1 + c*x)^2 - (128*a^2*c)/(1 + c*x) - 192*a^2*c*Log[x] + 192*a^2*c*Log[1 + c*x] + b^
2*c*((-8*I)*Pi^3 + 64*ArcTanh[c*x]^2 - (64*ArcTanh[c*x]^2)/(c*x) + 128*ArcTanh[c*x]^3 - 40*Cosh[2*ArcTanh[c*x]
] - 80*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] - 80*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 4*A
rcTanh[c*x]*Cosh[4*ArcTanh[c*x]] - 8*ArcTanh[c*x]^2*Cosh[4*ArcTanh[c*x]] + 128*ArcTanh[c*x]*Log[1 - E^(-2*ArcT
anh[c*x])] - 192*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - 64*PolyLog[2, E^(-2*ArcTanh[c*x])] - 192*ArcTanh
[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] + 96*PolyLog[3, E^(2*ArcTanh[c*x])] + 40*Sinh[2*ArcTanh[c*x]] + 80*ArcTan
h[c*x]*Sinh[2*ArcTanh[c*x]] + 80*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*S
inh[4*ArcTanh[c*x]] + 8*ArcTanh[c*x]^2*Sinh[4*ArcTanh[c*x]]) + (4*a*b*(48*c*x*PolyLog[2, E^(-2*ArcTanh[c*x])]
+ c*x*(-20*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 32*Log[(c*x)/Sqrt[1 - c^2*x^2]] + 20*Sinh[2*ArcTanh[c
*x]] + Sinh[4*ArcTanh[c*x]]) - 4*ArcTanh[c*x]*(8 + 10*c*x*Cosh[2*ArcTanh[c*x]] + c*x*Cosh[4*ArcTanh[c*x]] + 24
*c*x*Log[1 - E^(-2*ArcTanh[c*x])] - 10*c*x*Sinh[2*ArcTanh[c*x]] - c*x*Sinh[4*ArcTanh[c*x]])))/x)/(64*d^3)

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Maple [C]  time = 0.723, size = 7593, normalized size = 17. \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/(c*d*x+d)^3,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a^{2}{\left (\frac{6 \, c^{2} x^{2} + 9 \, c x + 2}{c^{2} d^{3} x^{3} + 2 \, c d^{3} x^{2} + d^{3} x} - \frac{6 \, c \log \left (c x + 1\right )}{d^{3}} + \frac{6 \, c \log \left (x\right )}{d^{3}}\right )} - \frac{{\left (6 \, b^{2} c^{2} x^{2} + 9 \, b^{2} c x + 2 \, b^{2} - 6 \,{\left (b^{2} c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} + b^{2} c x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \,{\left (c^{2} d^{3} x^{3} + 2 \, c d^{3} x^{2} + d^{3} x\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 (6 \, b^{2} c^{4} x^{4} + 15 \, b^{2} c^{3} x^{3} + 11 \, b^{2} c^{2} x^{2} + 4 \, a b - 2 \,{\left (2 \, a b c - b^{2} c\right )} x - 2 \,{\left (3 \, b^{2} c^{5} x^{5} + 9 \, b^{2} c^{4} x^{4} + 9 \, b^{2} c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \,{\left (c^{4} d^{3} x^{6} + 2 \, c^{3} d^{3} x^{5} - 2 \, c d^{3} x^{3} - d^{3} x^{2}\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/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/2*a^2*((6*c^2*x^2 + 9*c*x + 2)/(c^2*d^3*x^3 + 2*c*d^3*x^2 + d^3*x) - 6*c*log(c*x + 1)/d^3 + 6*c*log(x)/d^3)
- 1/8*(6*b^2*c^2*x^2 + 9*b^2*c*x + 2*b^2 - 6*(b^2*c^3*x^3 + 2*b^2*c^2*x^2 + b^2*c*x)*log(c*x + 1))*log(-c*x +
1)^2/(c^2*d^3*x^3 + 2*c*d^3*x^2 + d^3*x) - integrate(-1/4*((b^2*c*x - b^2)*log(c*x + 1)^2 + 4*(a*b*c*x - a*b)
*log(c*x + 1) + (6*b^2*c^4*x^4 + 15*b^2*c^3*x^3 + 11*b^2*c^2*x^2 + 4*a*b - 2*(2*a*b*c - b^2*c)*x - 2*(3*b^2*c^
5*x^5 + 9*b^2*c^4*x^4 + 9*b^2*c^3*x^3 + 3*b^2*c^2*x^2 + b^2*c*x - b^2)*log(c*x + 1))*log(-c*x + 1))/(c^4*d^3*x
^6 + 2*c^3*d^3*x^5 - 2*c*d^3*x^3 - d^3*x^2), 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^{5} + 3 \, c^{2} d^{3} x^{4} + 3 \, c d^{3} x^{3} + d^{3} 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/(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^5 + 3*c^2*d^3*x^4 + 3*c*d^3*x^3 + d^3*x^2)
, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{3} x^{5} + 3 c^{2} x^{4} + 3 c x^{3} + x^{2}}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{c^{3} x^{5} + 3 c^{2} x^{4} + 3 c x^{3} + x^{2}}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{c^{3} x^{5} + 3 c^{2} x^{4} + 3 c x^{3} + x^{2}}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**3*x**5 + 3*c**2*x**4 + 3*c*x**3 + x**2), x) + Integral(b**2*atanh(c*x)**2/(c**3*x**5 + 3*c*
*2*x**4 + 3*c*x**3 + x**2), x) + Integral(2*a*b*atanh(c*x)/(c**3*x**5 + 3*c**2*x**4 + 3*c*x**3 + x**2), 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^{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/(c*d*x+d)^3,x, algorithm="giac")

[Out]

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