3.28 \(\int \frac{(a+b \tanh ^{-1}(c+d x))^3}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=269 \[ -\frac{b^2 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4}-\frac{b^3 \text{PolyLog}\left (3,\frac{2}{c+d x+1}-1\right )}{2 d e^4}-\frac{b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}+\frac{b \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^4}+\frac{b^3 \log (c+d x)}{d e^4}-\frac{b^3 \log \left (1-(c+d x)^2\right )}{2 d e^4} \]

[Out]

-((b^2*(a + b*ArcTanh[c + d*x]))/(d*e^4*(c + d*x))) + (b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^4) - (b*(a + b*Arc
Tanh[c + d*x])^2)/(2*d*e^4*(c + d*x)^2) + (a + b*ArcTanh[c + d*x])^3/(3*d*e^4) - (a + b*ArcTanh[c + d*x])^3/(3
*d*e^4*(c + d*x)^3) + (b^3*Log[c + d*x])/(d*e^4) - (b^3*Log[1 - (c + d*x)^2])/(2*d*e^4) + (b*(a + b*ArcTanh[c
+ d*x])^2*Log[2 - 2/(1 + c + d*x)])/(d*e^4) - (b^2*(a + b*ArcTanh[c + d*x])*PolyLog[2, -1 + 2/(1 + c + d*x)])/
(d*e^4) - (b^3*PolyLog[3, -1 + 2/(1 + c + d*x)])/(2*d*e^4)

________________________________________________________________________________________

Rubi [A]  time = 0.513624, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {6107, 12, 5916, 5982, 266, 36, 31, 29, 5948, 5988, 5932, 6056, 6610} \[ -\frac{b^2 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4}-\frac{b^3 \text{PolyLog}\left (3,\frac{2}{c+d x+1}-1\right )}{2 d e^4}-\frac{b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}+\frac{b \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^4}+\frac{b^3 \log (c+d x)}{d e^4}-\frac{b^3 \log \left (1-(c+d x)^2\right )}{2 d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-((b^2*(a + b*ArcTanh[c + d*x]))/(d*e^4*(c + d*x))) + (b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^4) - (b*(a + b*Arc
Tanh[c + d*x])^2)/(2*d*e^4*(c + d*x)^2) + (a + b*ArcTanh[c + d*x])^3/(3*d*e^4) - (a + b*ArcTanh[c + d*x])^3/(3
*d*e^4*(c + d*x)^3) + (b^3*Log[c + d*x])/(d*e^4) - (b^3*Log[1 - (c + d*x)^2])/(2*d*e^4) + (b*(a + b*ArcTanh[c
+ d*x])^2*Log[2 - 2/(1 + c + d*x)])/(d*e^4) - (b^2*(a + b*ArcTanh[c + d*x])*PolyLog[2, -1 + 2/(1 + c + d*x)])/
(d*e^4) - (b^3*PolyLog[3, -1 + 2/(1 + c + d*x)])/(2*d*e^4)

Rule 6107

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

Rule 12

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

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 5982

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 1.24494, size = 393, normalized size = 1.46 \[ \frac{6 a b^2 \left (-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )-\frac{(c+d x)^2+\tanh ^{-1}(c+d x)^2}{(c+d x)^3}+\tanh ^{-1}(c+d x) \left (-\frac{1-(c+d x)^2}{(c+d x)^2}+\tanh ^{-1}(c+d x)+2 \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )\right )+6 b^3 \left (\tanh ^{-1}(c+d x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c+d x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c+d x)}\right )+\log \left (\frac{c+d x}{\sqrt{1-(c+d x)^2}}\right )-\frac{\left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^3}{3 (c+d x)^3}-\frac{\tanh ^{-1}(c+d x)^3}{3 (c+d x)}-\frac{1}{3} \tanh ^{-1}(c+d x)^3-\frac{\left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac{\tanh ^{-1}(c+d x)}{c+d x}+\tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\frac{i \pi ^3}{24}\right )-3 a^2 b \log \left (-c^2-2 c d x-d^2 x^2+1\right )-\frac{3 a^2 b}{(c+d x)^2}+6 a^2 b \log (c+d x)-\frac{6 a^2 b \tanh ^{-1}(c+d x)}{(c+d x)^3}-\frac{2 a^3}{(c+d x)^3}}{6 d e^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

((-2*a^3)/(c + d*x)^3 - (3*a^2*b)/(c + d*x)^2 - (6*a^2*b*ArcTanh[c + d*x])/(c + d*x)^3 + 6*a^2*b*Log[c + d*x]
- 3*a^2*b*Log[1 - c^2 - 2*c*d*x - d^2*x^2] + 6*a*b^2*(-(((c + d*x)^2 + ArcTanh[c + d*x]^2)/(c + d*x)^3) + ArcT
anh[c + d*x]*(-((1 - (c + d*x)^2)/(c + d*x)^2) + ArcTanh[c + d*x] + 2*Log[1 - E^(-2*ArcTanh[c + d*x])]) - Poly
Log[2, E^(-2*ArcTanh[c + d*x])]) + 6*b^3*((I/24)*Pi^3 - ArcTanh[c + d*x]/(c + d*x) - ((1 - (c + d*x)^2)*ArcTan
h[c + d*x]^2)/(2*(c + d*x)^2) - ArcTanh[c + d*x]^3/3 - ArcTanh[c + d*x]^3/(3*(c + d*x)) - ((1 - (c + d*x)^2)*A
rcTanh[c + d*x]^3)/(3*(c + d*x)^3) + ArcTanh[c + d*x]^2*Log[1 - E^(2*ArcTanh[c + d*x])] + Log[(c + d*x)/Sqrt[1
- (c + d*x)^2]] + ArcTanh[c + d*x]*PolyLog[2, E^(2*ArcTanh[c + d*x])] - PolyLog[3, E^(2*ArcTanh[c + d*x])]/2)
)/(6*d*e^4)

________________________________________________________________________________________

Maple [C]  time = 0.519, size = 2247, normalized size = 8.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^4,x)

[Out]

1/4*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))^3*Pi+1/4*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(
I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3*Pi-1/4*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I/((d*x+
c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c+1)^2/(1-(d
*x+c)^2)+1))*Pi+1/2*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*((d*x+c+1)^2/(1-
(d*x+c)^2)-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*Pi-1/d*a*b^2/e^4*arctanh(d*
x+c)*ln(d*x+c-1)-1/d*a*b^2/e^4*arctanh(d*x+c)/(d*x+c)^2-1/d*a^2*b/e^4/(d*x+c)^3*arctanh(d*x+c)+2/d*a*b^2/e^4*a
rctanh(d*x+c)*ln(d*x+c)-1/d*a*b^2/e^4*arctanh(d*x+c)*ln(d*x+c+1)+1/2/d*a*b^2/e^4*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/
2*c)-1/2/d*a*b^2/e^4*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+1/2/d*a*b^2/e^4*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+
1/2*c)+1/2*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))^2*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/
2))*Pi+1/4*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2)
)^2*Pi-1/4*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d
*x+c+1)^2/(1-(d*x+c)^2)+1))^2*Pi-1/2*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I
*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*Pi+1/4*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I/((d
*x+c+1)^2/(1-(d*x+c)^2)+1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*Pi-1/2*I/d*b^3/e
^4*arctanh(d*x+c)^2*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*csgn(I*((d*x+c+1)^2/
(1-(d*x+c)^2)-1))*Pi-1/3/d*a^3/e^4/(d*x+c)^3-2/d*b^3/e^4*polylog(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/d*b^3/e^4
*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/3/d*b^3/e^4*arctanh(d*x+c)^3+1/2/d*b^3/e^4*arctanh(d*x+c)^2+1/d*b^3/e^4
*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2)-1)-1/d*b^3/e^4*arctanh(d*x+c)-2/d*b^3/e^4*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^
(1/2))-1/2*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^2*Pi+1/2*I/d*b^3/e^4*arctanh(d*x
+c)^2*csgn(I/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3*Pi+1/2*I/d*b^3/e^4*arctanh(d*x+c)^2*csgn(I*((d*x+c+1)^2/(1-(d*x+
c)^2)-1)/((d*x+c+1)^2/(1-(d*x+c)^2)+1))^3*Pi-1/d*a*b^2/e^4*ln(d*x+c)*ln(d*x+c+1)-1/d*a*b^2/e^4/(d*x+c)^3*arcta
nh(d*x+c)^2+1/2*I/d*b^3/e^4*arctanh(d*x+c)^2*Pi-1/2/d*b^3/e^4*arctanh(d*x+c)^2*ln(d*x+c-1)+1/d*b^3/e^4*ln(d*x+
c)*arctanh(d*x+c)^2+1/d*b^3/e^4*arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/d*a^2*b/e^4*ln(d*x+c)-1
/2/d*a^2*b/e^4*ln(d*x+c+1)-1/d*b^3/e^4*arctanh(d*x+c)/(d*x+c)+2/d*b^3/e^4*arctanh(d*x+c)*polylog(2,(d*x+c+1)/(
1-(d*x+c)^2)^(1/2))+1/d*b^3/e^4*arctanh(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/d*b^3/e^4*arctanh(d*x+c)^
2*ln(2)-1/2/d*a^2*b/e^4*ln(d*x+c-1)-1/d*a*b^2/e^4*dilog(d*x+c)-1/d*a*b^2/e^4*dilog(d*x+c+1)+1/d*a*b^2/e^4*dilo
g(1/2+1/2*d*x+1/2*c)+1/d*b^3/e^4*arctanh(d*x+c)^2*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/d*b^3/e^4*arctanh(d*x+
c)^2*ln((d*x+c+1)^2/(1-(d*x+c)^2)-1)+1/4/d*a*b^2/e^4*ln(d*x+c+1)^2-1/2/d*b^3/e^4*arctanh(d*x+c)^2*ln(d*x+c+1)-
1/2/d*b^3/e^4*arctanh(d*x+c)^2/(d*x+c)^2-1/2/d*a*b^2/e^4*ln(d*x+c-1)-1/3/d*b^3/e^4/(d*x+c)^3*arctanh(d*x+c)^3-
1/4/d*a*b^2/e^4*ln(d*x+c-1)^2+1/2/d*a*b^2/e^4*ln(d*x+c+1)-1/2/d*a^2*b/e^4/(d*x+c)^2-1/d*a*b^2/e^4/(d*x+c)+2/d*
b^3/e^4*arctanh(d*x+c)*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))

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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/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/2*(d*(1/(d^4*e^4*x^2 + 2*c*d^3*e^4*x + c^2*d^2*e^4) + log(d*x + c + 1)/(d^2*e^4) - 2*log(d*x + c)/(d^2*e^4)
+ log(d*x + c - 1)/(d^2*e^4)) + 2*arctanh(d*x + c)/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e
^4))*a^2*b - 1/3*a^3/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 1/24*((b^3*d^3*x^3 + 3*b^
3*c*d^2*x^2 + 3*b^3*c^2*d*x + (c^3 - 1)*b^3)*log(-d*x - c + 1)^3 + 3*(b^3*d*x + b^3*c + 2*a*b^2 + (b^3*d^3*x^3
+ 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + (c^3 + 1)*b^3)*log(d*x + c + 1))*log(-d*x - c + 1)^2)/(d^4*e^4*x^3 + 3*c*
d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - integrate(-1/8*((b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^3 + 6*(a
*b^2*d*x + a*b^2*(c - 1))*log(d*x + c + 1)^2 + (2*b^3*d^2*x^2 + 2*b^3*c^2 + 4*a*b^2*c - 3*(b^3*d*x + b^3*(c -
1))*log(d*x + c + 1)^2 + 4*(b^3*c*d + a*b^2*d)*x + 2*(b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + (c^4
+ c)*b^3 - 6*a*b^2*(c - 1) + ((4*c^3*d + d)*b^3 - 6*a*b^2*d)*x)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^5*e^4
*x^5 + c^5*e^4 - c^4*e^4 + (5*c*d^4*e^4 - d^4*e^4)*x^4 + 2*(5*c^2*d^3*e^4 - 2*c*d^3*e^4)*x^3 + 2*(5*c^3*d^2*e^
4 - 3*c^2*d^2*e^4)*x^2 + (5*c^4*d*e^4 - 4*c^3*d*e^4)*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}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^4,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)/(d^4*e^4*x^4 +
4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{b^{3} \operatorname{atanh}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{3 a b^{2} \operatorname{atanh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{3 a^{2} b \operatorname{atanh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**3/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**3*atanh(c
+ d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a*b**2*atanh(c +
d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a**2*b*atanh(c +
d*x)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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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 (d e x + c e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(d*x + c) + a)^3/(d*e*x + c*e)^4, x)