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

Optimal. Leaf size=1089 \[ \text{result too large to display} \]

[Out]

-((a + b*ArcTanh[c + d*x])^3/(f*(e + f*x))) + (3*a*b^2*d*ArcTanh[c + d*x]*Log[2/(1 - c - d*x)])/(f*(d*e + f -
c*f)) + (3*b^3*d*ArcTanh[c + d*x]^2*Log[2/(1 - c - d*x)])/(2*f*(d*e + f - c*f)) - (3*a^2*b*d*Log[1 - c - d*x])
/(2*f*(d*e + f - c*f)) - (3*a*b^2*d*ArcTanh[c + d*x]*Log[2/(1 + c + d*x)])/(f*(d*e - f - c*f)) + (6*a*b^2*d*Ar
cTanh[c + d*x]*Log[2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1 + c)*f)) - (3*b^3*d*ArcTanh[c + d*x]^2*Log[2/(
1 + c + d*x)])/(2*f*(d*e - f - c*f)) + (3*b^3*d*ArcTanh[c + d*x]^2*Log[2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e
- (1 + c)*f)) + (3*a^2*b*d*Log[1 + c + d*x])/(2*f*(d*e - f - c*f)) + (3*a^2*b*d*Log[e + f*x])/(f^2 - (d*e - c
*f)^2) - (6*a*b^2*d*ArcTanh[c + d*x]*Log[(2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d
*e - (1 + c)*f)) - (3*b^3*d*ArcTanh[c + d*x]^2*Log[(2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f
- c*f)*(d*e - (1 + c)*f)) + (3*a*b^2*d*PolyLog[2, -((1 + c + d*x)/(1 - c - d*x))])/(2*f*(d*e + f - c*f)) + (3
*b^3*d*ArcTanh[c + d*x]*PolyLog[2, 1 - 2/(1 - c - d*x)])/(2*f*(d*e + f - c*f)) + (3*a*b^2*d*PolyLog[2, 1 - 2/(
1 + c + d*x)])/(2*f*(d*e - f - c*f)) - (3*a*b^2*d*PolyLog[2, 1 - 2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1
+ c)*f)) + (3*b^3*d*ArcTanh[c + d*x]*PolyLog[2, 1 - 2/(1 + c + d*x)])/(2*f*(d*e - f - c*f)) - (3*b^3*d*ArcTanh
[c + d*x]*PolyLog[2, 1 - 2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (3*a*b^2*d*PolyLog[2, 1 - (2*
d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (3*b^3*d*ArcTanh[c + d*x]
*PolyLog[2, 1 - (2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d*e - (1 + c)*f)) - (3*b^3
*d*PolyLog[3, 1 - 2/(1 - c - d*x)])/(4*f*(d*e + f - c*f)) + (3*b^3*d*PolyLog[3, 1 - 2/(1 + c + d*x)])/(4*f*(d*
e - f - c*f)) - (3*b^3*d*PolyLog[3, 1 - 2/(1 + c + d*x)])/(2*(d*e + f - c*f)*(d*e - (1 + c)*f)) + (3*b^3*d*Pol
yLog[3, 1 - (2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/(2*(d*e + f - c*f)*(d*e - (1 + c)*f))

________________________________________________________________________________________

Rubi [A]  time = 2.82718, antiderivative size = 1094, normalized size of antiderivative = 1., number of steps used = 30, number of rules used = 18, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6109, 6741, 6121, 6688, 12, 6725, 72, 6742, 5918, 2402, 2315, 5920, 2447, 5948, 6058, 6610, 6056, 5922} \[ \frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{-c-d x+1}\right ) b^3}{2 f (d e-c f+f)}-\frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{c+d x+1}\right ) b^3}{2 f (d e-c f-f)}+\frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{c+d x+1}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}-\frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) b^3}{2 f (d e-c f+f)}+\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^3}{2 f (d e-c f-f)}-\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}-\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right ) b^3}{4 f (d e-c f+f)}+\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{c+d x+1}\right ) b^3}{4 f (d e-c f-f)}-\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{c+d x+1}\right ) b^3}{2 (d e-c f+f) (d e-(c+1) f)}+\frac{3 d \text{PolyLog}\left (3,1-\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^3}{2 (d e-c f+f) (d e-(c+1) f)}+\frac{3 a d \tanh ^{-1}(c+d x) \log \left (\frac{2}{-c-d x+1}\right ) b^2}{f (d e-c f+f)}-\frac{3 a d \tanh ^{-1}(c+d x) \log \left (\frac{2}{c+d x+1}\right ) b^2}{f (d e-c f-f)}+\frac{6 a d \tanh ^{-1}(c+d x) \log \left (\frac{2}{c+d x+1}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}-\frac{6 a d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 a d \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right ) b^2}{2 f (d e-c f+f)}+\frac{3 a d \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^2}{2 f (d e-c f-f)}-\frac{3 a d \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 a d \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}-\frac{3 a^2 d \log (-c-d x+1) b}{2 f (d e-c f+f)}+\frac{3 a^2 d \log (c+d x+1) b}{2 f (d e-c f-f)}-\frac{3 a^2 d \log (e+f x) b}{(d e-c f+f) (d e-(c+1) f)}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcTanh[c + d*x])^3/(f*(e + f*x))) + (3*a*b^2*d*ArcTanh[c + d*x]*Log[2/(1 - c - d*x)])/(f*(d*e + f -
c*f)) + (3*b^3*d*ArcTanh[c + d*x]^2*Log[2/(1 - c - d*x)])/(2*f*(d*e + f - c*f)) - (3*a^2*b*d*Log[1 - c - d*x])
/(2*f*(d*e + f - c*f)) - (3*a*b^2*d*ArcTanh[c + d*x]*Log[2/(1 + c + d*x)])/(f*(d*e - f - c*f)) + (6*a*b^2*d*Ar
cTanh[c + d*x]*Log[2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1 + c)*f)) - (3*b^3*d*ArcTanh[c + d*x]^2*Log[2/(
1 + c + d*x)])/(2*f*(d*e - f - c*f)) + (3*b^3*d*ArcTanh[c + d*x]^2*Log[2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e
- (1 + c)*f)) + (3*a^2*b*d*Log[1 + c + d*x])/(2*f*(d*e - f - c*f)) - (3*a^2*b*d*Log[e + f*x])/((d*e + f - c*f
)*(d*e - (1 + c)*f)) - (6*a*b^2*d*ArcTanh[c + d*x]*Log[(2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e
+ f - c*f)*(d*e - (1 + c)*f)) - (3*b^3*d*ArcTanh[c + d*x]^2*Log[(2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x
))])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (3*a*b^2*d*PolyLog[2, -((1 + c + d*x)/(1 - c - d*x))])/(2*f*(d*e +
f - c*f)) + (3*b^3*d*ArcTanh[c + d*x]*PolyLog[2, 1 - 2/(1 - c - d*x)])/(2*f*(d*e + f - c*f)) + (3*a*b^2*d*Poly
Log[2, 1 - 2/(1 + c + d*x)])/(2*f*(d*e - f - c*f)) - (3*a*b^2*d*PolyLog[2, 1 - 2/(1 + c + d*x)])/((d*e + f - c
*f)*(d*e - (1 + c)*f)) + (3*b^3*d*ArcTanh[c + d*x]*PolyLog[2, 1 - 2/(1 + c + d*x)])/(2*f*(d*e - f - c*f)) - (3
*b^3*d*ArcTanh[c + d*x]*PolyLog[2, 1 - 2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (3*a*b^2*d*Poly
Log[2, 1 - (2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (3*b^3*d*Ar
cTanh[c + d*x]*PolyLog[2, 1 - (2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d*e - (1 + c
)*f)) - (3*b^3*d*PolyLog[3, 1 - 2/(1 - c - d*x)])/(4*f*(d*e + f - c*f)) + (3*b^3*d*PolyLog[3, 1 - 2/(1 + c + d
*x)])/(4*f*(d*e - f - c*f)) - (3*b^3*d*PolyLog[3, 1 - 2/(1 + c + d*x)])/(2*(d*e + f - c*f)*(d*e - (1 + c)*f))
+ (3*b^3*d*PolyLog[3, 1 - (2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/(2*(d*e + f - c*f)*(d*e - (1 + c)*
f))

Rule 6109

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcTanh[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcTanh[c + d*x])^(p - 1))/(1 - (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -
1]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6121

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x
_)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + (f*x)/d)^m*(-(C/d^2) + (C*x^2)/d^2)^q*(a + b*Ar
cTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 - c^2) + 2*A*c*
d, 0] && EqQ[2*c*C - B*d, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 5920

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

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+d x)\right )^3}{(e+f x)^2} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{(e+f x) \left (1-(c+d x)^2\right )} \, dx}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{(e+f x) \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right ) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{d \left (a+b \tanh ^{-1}(x)\right )^2}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \operatorname{Subst}\left (\int \left (-\frac{a^2}{(-1+x) (1+x) (d e-c f+f x)}-\frac{2 a b \tanh ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)}-\frac{b^2 \tanh ^{-1}(x)^2}{(-1+x) (1+x) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}-\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}-\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2 (d e+f-c f) (-1+x)}+\frac{1}{2 (-d e+(1+c) f) (1+x)}+\frac{f^2}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \left (\frac{\tanh ^{-1}(x)}{2 (d e+f-c f) (-1+x)}+\frac{\tanh ^{-1}(x)}{2 (-d e+(1+c) f) (1+x)}+\frac{f^2 \tanh ^{-1}(x)}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \left (\frac{\tanh ^{-1}(x)^2}{2 (d e+f-c f) (-1+x)}+\frac{\tanh ^{-1}(x)^2}{2 (-d e+(1+c) f) (1+x)}+\frac{f^2 \tanh ^{-1}(x)^2}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{1+x} \, dx,x,c+d x\right )}{f (d e-f-c f)}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{1+x} \, dx,x,c+d x\right )}{2 f (d e-f-c f)}-\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{-1+x} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{-1+x} \, dx,x,c+d x\right )}{2 f (d e+f-c f)}-\frac{\left (6 a b^2 d f\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{\left (3 b^3 d f\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1-c-d x}\right )}{f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}-\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e-f-c f)}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x) \log \left (\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e-f-c f)}-\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2 (d e-c f+f x)}{(d e+f-c f) (1+x)}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1-c-d x}\right )}{f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}-\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c+d x}\right )}{f (d e-f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 f (d e-f-c f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c-d x}\right )}{f (d e+f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 f (d e+f-c f)}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1-c-d x}\right )}{f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}-\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{4 f (d e+f-c f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{4 f (d e-f-c f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 (d e+f-c f) (d e-(1+c) f)}\\ \end{align*}

Mathematica [C]  time = 23.1682, size = 2683, normalized size = 2.46 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a^3/(f*(e + f*x))) - (3*a^2*b*ArcTanh[c + d*x])/(f*(e + f*x)) + (3*a^2*b*d*Log[1 - c - d*x])/(2*f*(-(d*e) -
f + c*f)) - (3*a^2*b*d*Log[1 + c + d*x])/(2*f*(-(d*e) + f + c*f)) - (3*a^2*b*d*Log[e + f*x])/(d^2*e^2 - 2*c*d*
e*f - f^2 + c^2*f^2) + (3*a*b^2*(1 - (c + d*x)^2)*((d*e - c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 -
(c + d*x)^2])^2*(-(ArcTanh[c + d*x]^2/(E^ArcTanh[(d*e - c*f)/f]*f*Sqrt[1 - (d*e - c*f)^2/f^2])) + ((c + d*x)*A
rcTanh[c + d*x]^2)/(Sqrt[1 - (c + d*x)^2]*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c +
d*x))/Sqrt[1 - (c + d*x)^2])) + ((d*e - c*f)*(I*Pi*Log[1 + E^(2*ArcTanh[c + d*x])] - 2*ArcTanh[c + d*x]*Log[1
- E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))] - I*Pi*(ArcTanh[c + d*x] + Log[1/Sqrt[1 - (c + d*x)^2]]
) - 2*ArcTanh[(d*e - c*f)/f]*(ArcTanh[c + d*x] + Log[1 - E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))] -
Log[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]]) + PolyLog[2, E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[
c + d*x]))]))/(d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2)))/(d*(d*e - c*f)*(e + f*x)^2) + (b^3*(1 - (c + d*x)^2)*((
d*e - c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2])^2*((d*(c + d*x)*ArcTanh[c + d*x]^3)/((
d*e - c*f)*Sqrt[1 - (c + d*x)^2]*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sq
rt[1 - (c + d*x)^2])) - (3*d*((ArcTanh[c + d*x]^2*(-(f*ArcTanh[c + d*x]) + (d*e - c*f)*Log[(d*e)/Sqrt[1 - (c +
d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2]]))/((d*e + f - c*f)*(d*e - (1 + c
)*f)) - (ArcTanh[c + d*x]*((-I)*d*e*Pi*ArcTanh[c + d*x] + I*c*f*Pi*ArcTanh[c + d*x] - 2*f*ArcTanh[c + d*x]^2 +
(Sqrt[1 - c^2 - (d^2*e^2)/f^2 + (2*c*d*e)/f]*f*ArcTanh[c + d*x]^2)/E^ArcTanh[(d*e - c*f)/f] + I*d*e*Pi*Log[1
+ E^(2*ArcTanh[c + d*x])] - I*c*f*Pi*Log[1 + E^(2*ArcTanh[c + d*x])] - 2*d*e*ArcTanh[c + d*x]*Log[1 - E^(-2*(A
rcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))] + 2*c*f*ArcTanh[c + d*x]*Log[1 - E^(-2*(ArcTanh[(d*e - c*f)/f] + A
rcTanh[c + d*x]))] - I*d*e*Pi*Log[1/Sqrt[1 - (c + d*x)^2]] + I*c*f*Pi*Log[1/Sqrt[1 - (c + d*x)^2]] + 2*d*e*Arc
Tanh[c + d*x]*Log[(d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)
^2]] - 2*c*f*ArcTanh[c + d*x]*Log[(d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sq
rt[1 - (c + d*x)^2]] - 2*(d*e - c*f)*ArcTanh[(d*e - c*f)/f]*(ArcTanh[c + d*x] + Log[1 - E^(-2*(ArcTanh[(d*e -
c*f)/f] + ArcTanh[c + d*x]))] - Log[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]]) + (d*e - c*f)*PolyLog[
2, E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))]))/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (((2*d*e + (-2
- 2*c + Sqrt[1 - c^2 - (d^2*e^2)/f^2 + (2*c*d*e)/f]/E^ArcTanh[(d*e - c*f)/f])*f)*ArcTanh[c + d*x]^3)/3 - (d*e
- c*f)*ArcTanh[c + d*x]^2*Log[-1 + E^(2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))] + (d*e - c*f)*ArcTanh[c
+ d*x]*((-I)*Pi*(ArcTanh[c + d*x] - Log[1 + E^(2*ArcTanh[c + d*x])] + Log[(1 + E^(2*ArcTanh[c + d*x]))/(2*E^Ar
cTanh[c + d*x])]) - 2*ArcTanh[(d*e - c*f)/f]*(ArcTanh[c + d*x] + Log[1 - E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTa
nh[c + d*x]))] - Log[(I/2)*E^(-ArcTanh[(d*e - c*f)/f] - ArcTanh[c + d*x])*(-1 + E^(2*(ArcTanh[(d*e - c*f)/f] +
ArcTanh[c + d*x])))])) + (d*e - c*f)*ArcTanh[c + d*x]^2*Log[d*e*(1 + E^(2*ArcTanh[c + d*x])) - (1 + c - E^(2*
ArcTanh[c + d*x]) + c*E^(2*ArcTanh[c + d*x]))*f] - (d*e - c*f)*ArcTanh[c + d*x]^2*(ArcTanh[c + d*x] + Log[1 -
E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))] - Log[-1 + E^(2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]
))] + Log[d*e*(1 + E^(2*ArcTanh[c + d*x])) - (1 + c - E^(2*ArcTanh[c + d*x]) + c*E^(2*ArcTanh[c + d*x]))*f] -
Log[(d*e*(1 + E^(2*ArcTanh[c + d*x])) - (1 + c - E^(2*ArcTanh[c + d*x]) + c*E^(2*ArcTanh[c + d*x]))*f)/(2*E^Ar
cTanh[c + d*x])]) - ((d*e - c*f)*(4*ArcTanh[c + d*x]^3 - 6*ArcTanh[c + d*x]^2*Log[1 - E^(2*(ArcTanh[(d*e - c*f
)/f] + ArcTanh[c + d*x]))] - 6*ArcTanh[c + d*x]*PolyLog[2, E^(2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))]
+ 3*PolyLog[3, E^(2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))]))/6 + ((d*e - c*f)*(4*ArcTanh[c + d*x]^3 - 6
*ArcTanh[c + d*x]^2*Log[1 + (E^(2*ArcTanh[c + d*x])*(d*e + f - c*f))/(d*e - (1 + c)*f)] - 6*ArcTanh[c + d*x]*P
olyLog[2, -((E^(2*ArcTanh[c + d*x])*(d*e + f - c*f))/(d*e - (1 + c)*f))] + 3*PolyLog[3, -((E^(2*ArcTanh[c + d*
x])*(d*e + f - c*f))/(d*e - (1 + c)*f))]))/6 - ((d*e - c*f)*(2*ArcTanh[c + d*x]^3 + 3*ArcTanh[c + d*x]^2*Log[1
- E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]))] - 3*ArcTanh[c + d*x]^2*Log[1 - (E^ArcTanh[c + d*x]*Sqrt
[-(d*e) + (-1 + c)*f])/Sqrt[d*e - (1 + c)*f]] - 3*ArcTanh[c + d*x]^2*Log[1 + (E^ArcTanh[c + d*x]*Sqrt[-(d*e) +
(-1 + c)*f])/Sqrt[d*e - (1 + c)*f]] - 3*ArcTanh[c + d*x]*PolyLog[2, E^(-2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c
+ d*x]))] - 6*ArcTanh[c + d*x]*PolyLog[2, -((E^ArcTanh[c + d*x]*Sqrt[-(d*e) + (-1 + c)*f])/Sqrt[d*e - (1 + c)
*f])] - 6*ArcTanh[c + d*x]*PolyLog[2, (E^ArcTanh[c + d*x]*Sqrt[-(d*e) + (-1 + c)*f])/Sqrt[d*e - (1 + c)*f]] +
6*PolyLog[3, -((E^ArcTanh[c + d*x]*Sqrt[-(d*e) + (-1 + c)*f])/Sqrt[d*e - (1 + c)*f])] + 6*PolyLog[3, (E^ArcTan
h[c + d*x]*Sqrt[-(d*e) + (-1 + c)*f])/Sqrt[d*e - (1 + c)*f]]))/3)/(d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2)))/(d*
e - c*f)))/(d^2*(e + f*x)^2)

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Maple [C]  time = 0.84, size = 5728, normalized size = 5.3 \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(d*x+c))^3/(f*x+e)^2,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(d*(log(d*x + c + 1)/(d*e*f - (c + 1)*f^2) - log(d*x + c - 1)/(d*e*f - (c - 1)*f^2) - 2*log(f*x + e)/(d^2*
e^2 - 2*c*d*e*f + (c^2 - 1)*f^2)) - 2*arctanh(d*x + c)/(f^2*x + e*f))*a^2*b - a^3/(f^2*x + e*f) - 1/8*(((d^2*e
*f - c*d*f^2 - d*f^2)*b^3*x + (c*d*e*f - c^2*f^2 - d*e*f + f^2)*b^3)*log(-d*x - c + 1)^3 + 3*(2*(d^2*e^2 - 2*c
*d*e*f + c^2*f^2 - f^2)*a*b^2 - ((d^2*e*f - c*d*f^2 + d*f^2)*b^3*x + (c*d*e*f - c^2*f^2 + d*e*f + f^2)*b^3)*lo
g(d*x + c + 1))*log(-d*x - c + 1)^2)/(d^2*e^3*f - 2*c*d*e^2*f^2 + c^2*e*f^3 - e*f^3 + (d^2*e^2*f^2 - 2*c*d*e*f
^3 + c^2*f^4 - f^4)*x) - integrate(-1/8*(((d^2*e*f - c*d*f^2 - d*f^2)*b^3*x + (c*d*e*f - c^2*f^2 - d*e*f + f^2
)*b^3)*log(d*x + c + 1)^3 + 6*((d^2*e*f - c*d*f^2 - d*f^2)*a*b^2*x + (c*d*e*f - c^2*f^2 - d*e*f + f^2)*a*b^2)*
log(d*x + c + 1)^2 + 3*(4*(d^2*e*f - c*d*f^2 - d*f^2)*a*b^2*x + 4*(d^2*e^2 - c*d*e*f - d*e*f)*a*b^2 - ((d^2*e*
f - c*d*f^2 - d*f^2)*b^3*x + (c*d*e*f - c^2*f^2 - d*e*f + f^2)*b^3)*log(d*x + c + 1)^2 - 2*(b^3*d^2*f^2*x^2 +
2*(c*d*e*f - c^2*f^2 - d*e*f + f^2)*a*b^2 + (c*d*e*f + d*e*f)*b^3 + (2*(d^2*e*f - c*d*f^2 - d*f^2)*a*b^2 + (d^
2*e*f + c*d*f^2 + d*f^2)*b^3)*x)*log(d*x + c + 1))*log(-d*x - c + 1))/(c*d*e^3*f - c^2*e^2*f^2 - d*e^3*f + e^2
*f^2 + (d^2*e*f^3 - c*d*f^4 - d*f^4)*x^3 + (2*d^2*e^2*f^2 - c*d*e*f^3 - c^2*f^4 - 3*d*e*f^3 + f^4)*x^2 + (d^2*
e^3*f + c*d*e^2*f^2 - 2*c^2*e*f^3 - 3*d*e^2*f^2 + 2*e*f^3)*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (d x + c\right ) + a^{3}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*arctanh(d*x + c)^3 + 3*a*b^2*arctanh(d*x + c)^2 + 3*a^2*b*arctanh(d*x + c) + a^3)/(f^2*x^2 + 2*e
*f*x + e^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(d*x+c))**3/(f*x+e)**2,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 (d x + c\right ) + a\right )}^{3}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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