\(\int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx\) [27]

Optimal result

Integrand size = 23, antiderivative size = 166 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\frac {3 b (a+b \text {arctanh}(c+d x))^2}{2 d e^3}-\frac {3 b (a+b \text {arctanh}(c+d x))^2}{2 d e^3 (c+d x)}+\frac {(a+b \text {arctanh}(c+d x))^3}{2 d e^3}-\frac {(a+b \text {arctanh}(c+d x))^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 (a+b \text {arctanh}(c+d x)) \log \left (2-\frac {2}{1+c+d x}\right )}{d e^3}-\frac {3 b^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c+d x}\right )}{2 d e^3} \] Output:

3/2*b*(a+b*arctanh(d*x+c))^2/d/e^3-3/2*b*(a+b*arctanh(d*x+c))^2/d/e^3/(d*x 
+c)+1/2*(a+b*arctanh(d*x+c))^3/d/e^3-1/2*(a+b*arctanh(d*x+c))^3/d/e^3/(d*x 
+c)^2+3*b^2*(a+b*arctanh(d*x+c))*ln(2-2/(d*x+c+1))/d/e^3-3/2*b^3*polylog(2 
,-1+2/(d*x+c+1))/d/e^3
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.93 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.02 \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\frac {-4 a^3-12 a^2 b c+i b^3 c^3 \pi ^3-12 a^2 b d x+2 i b^3 c^2 d \pi ^3 x+i b^3 c d^2 \pi ^3 x^2+12 b^2 (-1+c+d x) (b (c+d x)+a (1+c+d x)) \text {arctanh}(c+d x)^2+4 b^3 \left (-1+c^2+2 c d x+d^2 x^2\right ) \text {arctanh}(c+d x)^3+12 b \text {arctanh}(c+d x) \left (a \left (-2 b (c+d x)+a \left (-1+c^2+2 c d x+d^2 x^2\right )\right )+2 b^2 (c+d x)^2 \log \left (1-e^{-2 \text {arctanh}(c+d x)}\right )\right )+24 a b^2 c^2 \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )+48 a b^2 c d x \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )+24 a b^2 d^2 x^2 \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )-12 b^3 (c+d x)^2 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c+d x)}\right )}{8 d e^3 (c+d x)^2} \] Input:

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

Output:

(-4*a^3 - 12*a^2*b*c + I*b^3*c^3*Pi^3 - 12*a^2*b*d*x + (2*I)*b^3*c^2*d*Pi^ 
3*x + I*b^3*c*d^2*Pi^3*x^2 + 12*b^2*(-1 + c + d*x)*(b*(c + d*x) + a*(1 + c 
 + d*x))*ArcTanh[c + d*x]^2 + 4*b^3*(-1 + c^2 + 2*c*d*x + d^2*x^2)*ArcTanh 
[c + d*x]^3 + 12*b*ArcTanh[c + d*x]*(a*(-2*b*(c + d*x) + a*(-1 + c^2 + 2*c 
*d*x + d^2*x^2)) + 2*b^2*(c + d*x)^2*Log[1 - E^(-2*ArcTanh[c + d*x])]) + 2 
4*a*b^2*c^2*Log[(c + d*x)/Sqrt[1 - (c + d*x)^2]] + 48*a*b^2*c*d*x*Log[(c + 
 d*x)/Sqrt[1 - (c + d*x)^2]] + 24*a*b^2*d^2*x^2*Log[(c + d*x)/Sqrt[1 - (c 
+ d*x)^2]] - 12*b^3*(c + d*x)^2*PolyLog[2, E^(-2*ArcTanh[c + d*x])])/(8*d* 
e^3*(c + d*x)^2)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {6657, 27, 6452, 6544, 6452, 6510, 6550, 6494, 2897}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx\)

\(\Big \downarrow \) 6657

\(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c+d x))^3}{e^3 (c+d x)^3}d(c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(a+b \text {arctanh}(c+d x))^3}{(c+d x)^3}d(c+d x)}{d e^3}\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {\frac {3}{2} b \int \frac {(a+b \text {arctanh}(c+d x))^2}{(c+d x)^2 \left (1-(c+d x)^2\right )}d(c+d x)-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

\(\Big \downarrow \) 6544

\(\displaystyle \frac {\frac {3}{2} b \left (\int \frac {(a+b \text {arctanh}(c+d x))^2}{(c+d x)^2}d(c+d x)+\int \frac {(a+b \text {arctanh}(c+d x))^2}{1-(c+d x)^2}d(c+d x)\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {\frac {3}{2} b \left (2 b \int \frac {a+b \text {arctanh}(c+d x)}{(c+d x) \left (1-(c+d x)^2\right )}d(c+d x)+\int \frac {(a+b \text {arctanh}(c+d x))^2}{1-(c+d x)^2}d(c+d x)-\frac {(a+b \text {arctanh}(c+d x))^2}{c+d x}\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

\(\Big \downarrow \) 6510

\(\displaystyle \frac {\frac {3}{2} b \left (2 b \int \frac {a+b \text {arctanh}(c+d x)}{(c+d x) \left (1-(c+d x)^2\right )}d(c+d x)+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}-\frac {(a+b \text {arctanh}(c+d x))^2}{c+d x}\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

\(\Big \downarrow \) 6550

\(\displaystyle \frac {\frac {3}{2} b \left (2 b \left (\int \frac {a+b \text {arctanh}(c+d x)}{(c+d x) (c+d x+1)}d(c+d x)+\frac {(a+b \text {arctanh}(c+d x))^2}{2 b}\right )+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}-\frac {(a+b \text {arctanh}(c+d x))^2}{c+d x}\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

\(\Big \downarrow \) 6494

\(\displaystyle \frac {\frac {3}{2} b \left (2 b \left (-b \int \frac {\log \left (2-\frac {2}{c+d x+1}\right )}{1-(c+d x)^2}d(c+d x)+\frac {(a+b \text {arctanh}(c+d x))^2}{2 b}+\log \left (2-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))\right )+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}-\frac {(a+b \text {arctanh}(c+d x))^2}{c+d x}\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

\(\Big \downarrow \) 2897

\(\displaystyle \frac {\frac {3}{2} b \left (2 b \left (\frac {(a+b \text {arctanh}(c+d x))^2}{2 b}+\log \left (2-\frac {2}{c+d x+1}\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right )\right )+\frac {(a+b \text {arctanh}(c+d x))^3}{3 b}-\frac {(a+b \text {arctanh}(c+d x))^2}{c+d x}\right )-\frac {(a+b \text {arctanh}(c+d x))^3}{2 (c+d x)^2}}{d e^3}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2897

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[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

rule 6494

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] - 
Simp[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 6510

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> 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 6544

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( 
e_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x 
], x] - Simp[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 6550

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] + Simp[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 6657

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_.), x_Symbol] :> Simp[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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 3.48 (sec) , antiderivative size = 4949, normalized size of antiderivative = 29.81

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(4949\)
default	\(\text {Expression too large to display}\)	\(4949\)
parts	\(\text {Expression too large to display}\)	\(4957\)

Input:

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

Output:

1/d*(-1/2*a^3/e^3/(d*x+c)^2+b^3/e^3*(-3/2*arctanh(d*x+c)^2+3/2*polylog(2,- 
(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/2*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2) 
)-3/2/(d*x+c)*arctanh(d*x+c)^2+1/2*arctanh(d*x+c)^3-3/4*arctanh(d*x+c)^2*l 
n(d*x+c-1)+3/4*arctanh(d*x+c)^2*ln(d*x+c+1)-3/2*arctanh(d*x+c)^2*ln((d*x+c 
+1)/(1-(d*x+c)^2)^(1/2))+3/8*I*Pi*csgn(I/(1-(d*x+c+1)^2/((d*x+c)^2-1)))*cs 
gn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c 
+1)^2/((d*x+c)^2-1)))*arctanh(d*x+c)^2+3/4*I*Pi*arctanh(d*x+c)^2-3/2*dilog 
((d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/2*dilog(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2)) 
-3/8*I*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2- 
1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2*dilog((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-3 
/8*I*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1) 
/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2)) 
-3/8*I*Pi*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*csgn(I*(d*x+c+1)^2/((d*x+c)^2- 
1)/(1-(d*x+c+1)^2/((d*x+c)^2-1)))^2*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/ 
2))+3/8*I*Pi*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))^2*csgn(I*(d*x+c+1)^2/(( 
d*x+c)^2-1))*dilog((d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/8*I*Pi*csgn(I/(1-(d*x+ 
c+1)^2/((d*x+c)^2-1)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+1)^2/((d 
*x+c)^2-1)))^2*polylog(2,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/8*I*Pi*csgn(I/(1 
-(d*x+c+1)^2/((d*x+c)^2-1)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1-(d*x+c+1) 
^2/((d*x+c)^2-1)))^2*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+3/4*I*Pi...
 

Fricas [F]

\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\frac {\int \frac {a^{3}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}} \,d x } \] Input:

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

Output:

-3/4*(d*(2/(d^3*e^3*x + c*d^2*e^3) - log(d*x + c + 1)/(d^2*e^3) + log(d*x 
+ c - 1)/(d^2*e^3)) + 2*arctanh(d*x + c)/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^ 
2*d*e^3))*a^2*b - 3/8*(d^2*((log(d*x + c + 1)^2 - 2*log(d*x + c + 1)*log(d 
*x + c - 1) + log(d*x + c - 1)^2 + 4*log(d*x + c - 1))/(d^3*e^3) + 4*log(d 
*x + c + 1)/(d^3*e^3) - 8*log(d*x + c)/(d^3*e^3)) + 4*d*(2/(d^3*e^3*x + c* 
d^2*e^3) - log(d*x + c + 1)/(d^2*e^3) + log(d*x + c - 1)/(d^2*e^3))*arctan 
h(d*x + c))*a*b^2 - 1/16*b^3*(((d^2*x^2 + 2*c*d*x + c^2 - 1)*log(-d*x - c 
+ 1)^3 + 3*(2*d*x - (d^2*x^2 + 2*c*d*x + c^2 - 1)*log(d*x + c + 1) + 2*c)* 
log(-d*x - c + 1)^2)/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) + 2*integra 
te(-((d*x + c - 1)*log(d*x + c + 1)^3 + 3*(2*d^2*x^2 + 4*c*d*x - (d*x + c 
- 1)*log(d*x + c + 1)^2 + 2*c^2 - (d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2*d 
- d)*x - c)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^4*e^3*x^4 + c^4*e^3 - 
c^3*e^3 + (4*c*d^3*e^3 - d^3*e^3)*x^3 + 3*(2*c^2*d^2*e^3 - c*d^2*e^3)*x^2 
+ (4*c^3*d*e^3 - 3*c^2*d*e^3)*x), x)) - 3/2*a*b^2*arctanh(d*x + c)^2/(d^3* 
e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) - 1/2*a^3/(d^3*e^3*x^2 + 2*c*d^2*e^3* 
x + c^2*d*e^3)
 

Giac [F]

\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c+d x))^3}{(c e+d e x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*atanh(c + d*x)**3*b**3*c**3 + 4*atanh(c + d*x)**3*b**3*c**2*d*x + 2*ata 
nh(c + d*x)**3*b**3*c*d**2*x**2 - 2*atanh(c + d*x)**3*b**3*c + 6*atanh(c + 
 d*x)**2*a*b**2*c**3 + 12*atanh(c + d*x)**2*a*b**2*c**2*d*x + 6*atanh(c + 
d*x)**2*a*b**2*c*d**2*x**2 - 6*atanh(c + d*x)**2*a*b**2*c + 6*atanh(c + d* 
x)**2*b**3*c*d*x + 6*atanh(c + d*x)**2*b**3*d**2*x**2 - 6*atanh(c + d*x)*a 
**2*b*c - 6*atanh(c + d*x)*a*b**2*c**2 + 6*atanh(c + d*x)*a*b**2*d**2*x**2 
 - 6*atanh(c + d*x)*b**3*c - 12*atanh(c + d*x)*b**3*d*x + 12*int((atanh(c 
+ d*x)*x)/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 - c**3 + 10*c**2*d**3*x** 
3 - 3*c**2*d*x + 5*c*d**4*x**4 - 3*c*d**2*x**2 + d**5*x**5 - d**3*x**3),x) 
*b**3*c**2*d**2 + 24*int((atanh(c + d*x)*x)/(c**5 + 5*c**4*d*x + 10*c**3*d 
**2*x**2 - c**3 + 10*c**2*d**3*x**3 - 3*c**2*d*x + 5*c*d**4*x**4 - 3*c*d** 
2*x**2 + d**5*x**5 - d**3*x**3),x)*b**3*c*d**3*x + 12*int((atanh(c + d*x)* 
x)/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 - c**3 + 10*c**2*d**3*x**3 - 3*c 
**2*d*x + 5*c*d**4*x**4 - 3*c*d**2*x**2 + d**5*x**5 - d**3*x**3),x)*b**3*d 
**4*x**2 - 3*log(c + d*x - 1)*a**2*b*c**3 - 6*log(c + d*x - 1)*a**2*b*c**2 
*d*x - 3*log(c + d*x - 1)*a**2*b*c*d**2*x**2 - 6*log(c + d*x - 1)*a*b**2*c 
**3 - 12*log(c + d*x - 1)*a*b**2*c**2*d*x + 3*log(c + d*x - 1)*a*b**2*c**2 
 - 6*log(c + d*x - 1)*a*b**2*c*d**2*x**2 + 6*log(c + d*x - 1)*a*b**2*c*d*x 
 + 3*log(c + d*x - 1)*a*b**2*d**2*x**2 + 3*log(c + d*x - 1)*b**3*c**3 + 6* 
log(c + d*x - 1)*b**3*c**2*d*x - 6*log(c + d*x - 1)*b**3*c**2 + 3*log(c...