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

Optimal. Leaf size=1089 \[ \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 {Li}_2\left (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 {Li}_2\left (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 {Li}_2\left (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 {Li}_2\left (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 {Li}_3\left (1-\frac {2}{-c-d x+1}\right ) b^3}{4 f (d e-c f+f)}+\frac {3 d \text {Li}_3\left (1-\frac {2}{c+d x+1}\right ) b^3}{4 f (d e-c f-f)}-\frac {3 d \text {Li}_3\left (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 {Li}_3\left (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 {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right ) b^2}{2 f (d e-c f+f)}+\frac {3 a d \text {Li}_2\left (1-\frac {2}{c+d x+1}\right ) b^2}{2 f (d e-c f-f)}-\frac {3 a d \text {Li}_2\left (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 {Li}_2\left (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}{f^2-(d e-c f)^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)} \]

[Out]

-(a+b*arctanh(d*x+c))^3/f/(f*x+e)+3*a*b^2*d*arctanh(d*x+c)*ln(2/(-d*x-c+1))/f/(-c*f+d*e+f)+3/2*b^3*d*arctanh(d
*x+c)^2*ln(2/(-d*x-c+1))/f/(-c*f+d*e+f)-3/2*a^2*b*d*ln(-d*x-c+1)/f/(-c*f+d*e+f)-3*a*b^2*d*arctanh(d*x+c)*ln(2/
(d*x+c+1))/f/(-c*f+d*e-f)+6*a*b^2*d*arctanh(d*x+c)*ln(2/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)-3/2*b^3*d*arctanh
(d*x+c)^2*ln(2/(d*x+c+1))/f/(-c*f+d*e-f)+3*b^3*d*arctanh(d*x+c)^2*ln(2/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)+3/
2*a^2*b*d*ln(d*x+c+1)/f/(-c*f+d*e-f)+3*a^2*b*d*ln(f*x+e)/(f^2-(-c*f+d*e)^2)-6*a*b^2*d*arctanh(d*x+c)*ln(2*d*(f
*x+e)/(-c*f+d*e+f)/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)-3*b^3*d*arctanh(d*x+c)^2*ln(2*d*(f*x+e)/(-c*f+d*e+f)/(
d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)+3/2*a*b^2*d*polylog(2,(-d*x-c-1)/(-d*x-c+1))/f/(-c*f+d*e+f)+3/2*b^3*d*arct
anh(d*x+c)*polylog(2,1-2/(-d*x-c+1))/f/(-c*f+d*e+f)+3/2*a*b^2*d*polylog(2,1-2/(d*x+c+1))/f/(-c*f+d*e-f)-3*a*b^
2*d*polylog(2,1-2/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)+3/2*b^3*d*arctanh(d*x+c)*polylog(2,1-2/(d*x+c+1))/f/(-c
*f+d*e-f)-3*b^3*d*arctanh(d*x+c)*polylog(2,1-2/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)+3*a*b^2*d*polylog(2,1-2*d*
(f*x+e)/(-c*f+d*e+f)/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)+3*b^3*d*arctanh(d*x+c)*polylog(2,1-2*d*(f*x+e)/(-c*f
+d*e+f)/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)-3/4*b^3*d*polylog(3,1-2/(-d*x-c+1))/f/(-c*f+d*e+f)+3/4*b^3*d*poly
log(3,1-2/(d*x+c+1))/f/(-c*f+d*e-f)-3/2*b^3*d*polylog(3,1-2/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)+3/2*b^3*d*pol
ylog(3,1-2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c+1))/(-c*f+d*e-f)/(-c*f+d*e+f)

________________________________________________________________________________________

Rubi [A]  time = 2.83, antiderivative size = 1094, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 18, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, 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 12

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

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 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 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 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 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 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 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]

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 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 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 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 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 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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 6741

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

Rule 6742

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

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 = 23.21, size = 1946, normalized size = 1.79 \[ \text {result too large to display} \]

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])) - (d*(-3*d*e*ArcTanh[c + d*x]^3 + f*ArcTanh[c + d*x]^3 + 3*c*f*ArcTanh[c + d*x]^3 - (2*S
qrt[1 - c^2 - (d^2*e^2)/f^2 + (2*c*d*e)/f]*f*ArcTanh[c + d*x]^3)/E^ArcTanh[(d*e - c*f)/f] - (3*I)*d*e*Pi*ArcTa
nh[c + d*x]*Log[(1 + E^(2*ArcTanh[c + d*x]))/(2*E^ArcTanh[c + d*x])] + (3*I)*c*f*Pi*ArcTanh[c + d*x]*Log[(1 +
E^(2*ArcTanh[c + d*x]))/(2*E^ArcTanh[c + d*x])] + 3*d*e*ArcTanh[c + d*x]^2*Log[1 - E^(ArcTanh[(d*e - c*f)/f] +
ArcTanh[c + d*x])] - 3*c*f*ArcTanh[c + d*x]^2*Log[1 - E^(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])] + 3*d*e*
ArcTanh[c + d*x]^2*Log[1 + E^(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])] - 3*c*f*ArcTanh[c + d*x]^2*Log[1 + E
^(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])] + 6*d*e*ArcTanh[(d*e - c*f)/f]*ArcTanh[c + d*x]*Log[(I/2)*E^(-Ar
cTanh[(d*e - c*f)/f] - ArcTanh[c + d*x])*(-1 + E^(2*(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])))] - 6*c*f*Arc
Tanh[(d*e - c*f)/f]*ArcTanh[c + d*x]*Log[(I/2)*E^(-ArcTanh[(d*e - c*f)/f] - ArcTanh[c + d*x])*(-1 + E^(2*(ArcT
anh[(d*e - c*f)/f] + ArcTanh[c + d*x])))] + 3*d*e*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)/(2*E^ArcTanh[c + d*x])] - 3*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)/(2*E
^ArcTanh[c + d*x])] + (3*I)*d*e*Pi*ArcTanh[c + d*x]*Log[1/Sqrt[1 - (c + d*x)^2]] - (3*I)*c*f*Pi*ArcTanh[c + d*
x]*Log[1/Sqrt[1 - (c + d*x)^2]] - 3*d*e*ArcTanh[c + d*x]^2*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]] + 3*c*f*ArcTanh[c + d*x]^2*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]] - 6*d*e*ArcTanh[(d*e - c*f)/f]*ArcTanh[c +
d*x]*Log[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]] + 6*c*f*ArcTanh[(d*e - c*f)/f]*ArcTanh[c + d*x]*L
og[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]] + 6*(d*e - c*f)*ArcTanh[c + d*x]*PolyLog[2, -E^(ArcTanh[
(d*e - c*f)/f] + ArcTanh[c + d*x])] + 6*(d*e - c*f)*ArcTanh[c + d*x]*PolyLog[2, E^(ArcTanh[(d*e - c*f)/f] + Ar
cTanh[c + d*x])] - 6*d*e*PolyLog[3, -E^(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])] + 6*c*f*PolyLog[3, -E^(Arc
Tanh[(d*e - c*f)/f] + ArcTanh[c + d*x])] - 6*d*e*PolyLog[3, E^(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])] + 6
*c*f*PolyLog[3, E^(ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x])]))/((d*e - c*f)*(d^2*e^2 - 2*c*d*e*f + (-1 + c^2
)*f^2))))/(d^2*(e + f*x)^2)

________________________________________________________________________________________

fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (d x + c\right ) + a^{3}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (f x + e\right )}^{2}}\,{d x} \]

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)

________________________________________________________________________________________

maple [C]  time = 1.00, size = 5728, normalized size = 5.26 \[ \text {output too large to display} \]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3}{2} \, {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d e f - {\left (c + 1\right )} f^{2}} - \frac {\log \left (d x + c - 1\right )}{d e f - {\left (c - 1\right )} f^{2}} - \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} - 1\right )} f^{2}}\right )} - \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{f^{2} x + e f}\right )} a^{2} b - \frac {a^{3}}{f^{2} x + e f} - \frac {{\left ({\left (d^{2} e f - c d f^{2} - d f^{2}\right )} b^{3} x + {\left (c d e f - c^{2} f^{2} - d e f + f^{2}\right )} b^{3}\right )} \log \left (-d x - c + 1\right )^{3} + 3 \, {\left (2 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2} - f^{2}\right )} a b^{2} - {\left ({\left (d^{2} e f - c d f^{2} + d f^{2}\right )} b^{3} x + {\left (c d e f - c^{2} f^{2} + d e f + f^{2}\right )} b^{3}\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )^{2}}{8 \, {\left (d^{2} e^{3} f - 2 \, c d e^{2} f^{2} + c^{2} e f^{3} - e f^{3} + {\left (d^{2} e^{2} f^{2} - 2 \, c d e f^{3} + c^{2} f^{4} - f^{4}\right )} x\right )}} - \int -\frac {{\left ({\left (d^{2} e f - c d f^{2} - d f^{2}\right )} b^{3} x + {\left (c d e f - c^{2} f^{2} - d e f + f^{2}\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{3} + 6 \, {\left ({\left (d^{2} e f - c d f^{2} - d f^{2}\right )} a b^{2} x + {\left (c d e f - c^{2} f^{2} - d e f + f^{2}\right )} a b^{2}\right )} \log \left (d x + c + 1\right )^{2} + 3 \, {\left (4 \, {\left (d^{2} e f - c d f^{2} - d f^{2}\right )} a b^{2} x + 4 \, {\left (d^{2} e^{2} - c d e f - d e f\right )} a b^{2} - {\left ({\left (d^{2} e f - c d f^{2} - d f^{2}\right )} b^{3} x + {\left (c d e f - c^{2} f^{2} - d e f + f^{2}\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{2} - 2 \, {\left (b^{3} d^{2} f^{2} x^{2} + 2 \, {\left (c d e f - c^{2} f^{2} - d e f + f^{2}\right )} a b^{2} + {\left (c d e f + d e f\right )} b^{3} + {\left (2 \, {\left (d^{2} e f - c d f^{2} - d f^{2}\right )} a b^{2} + {\left (d^{2} e f + c d f^{2} + d f^{2}\right )} b^{3}\right )} x\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{8 \, {\left (c d e^{3} f - c^{2} e^{2} f^{2} - d e^{3} f + e^{2} f^{2} + {\left (d^{2} e f^{3} - c d f^{4} - d f^{4}\right )} x^{3} + {\left (2 \, d^{2} e^{2} f^{2} - c d e f^{3} - c^{2} f^{4} - 3 \, d e f^{3} + f^{4}\right )} x^{2} + {\left (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}\right )} x\right )}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{{\left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{3}}{\left (e + f x\right )^{2}}\, dx \]

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]

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

________________________________________________________________________________________