\(\int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx\) [288]

Optimal result

Integrand size = 16, antiderivative size = 155 \[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \text {arctanh}(1+d+d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{16 b^4} \] Output:

1/20*b*x^5+1/4*x^4*arctanh(1+d+d*tanh(b*x+a))-1/8*x^4*ln(1+(1+d)*exp(2*b*x 
+2*a))-1/4*x^3*polylog(2,-(1+d)*exp(2*b*x+2*a))/b+3/8*x^2*polylog(3,-(1+d) 
*exp(2*b*x+2*a))/b^2-3/8*x*polylog(4,-(1+d)*exp(2*b*x+2*a))/b^3+3/16*polyl 
og(5,-(1+d)*exp(2*b*x+2*a))/b^4
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {4 b^4 x^4 \text {arctanh}(1+d+d \tanh (a+b x))-2 b^4 x^4 \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b x \operatorname {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{1+d}\right )+3 \operatorname {PolyLog}\left (5,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{16 b^4} \] Input:

Integrate[x^3*ArcTanh[1 + d + d*Tanh[a + b*x]],x]
 

Output:

(4*b^4*x^4*ArcTanh[1 + d + d*Tanh[a + b*x]] - 2*b^4*x^4*Log[1 + 1/((1 + d) 
*E^(2*(a + b*x)))] + 4*b^3*x^3*PolyLog[2, -(1/((1 + d)*E^(2*(a + b*x))))] 
+ 6*b^2*x^2*PolyLog[3, -(1/((1 + d)*E^(2*(a + b*x))))] + 6*b*x*PolyLog[4, 
-(1/((1 + d)*E^(2*(a + b*x))))] + 3*PolyLog[5, -(1/((1 + d)*E^(2*(a + b*x) 
)))])/(16*b^4)
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6793, 2615, 2620, 3011, 7163, 7163, 2720, 7143}

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.

\(\displaystyle \int x^3 \text {arctanh}(d \tanh (a+b x)+d+1) \, dx\)

\(\Big \downarrow \) 6793

\(\displaystyle \frac {1}{4} b \int \frac {x^4}{e^{2 a+2 b x} (d+1)+1}dx+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \int \frac {e^{2 a+2 b x} x^4}{e^{2 a+2 b x} (d+1)+1}dx\right )+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \int x^3 \log \left (e^{2 a+2 b x} (d+1)+1\right )dx}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \int x^2 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int x \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{2 b}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )de^{2 a+2 b x}}{4 b^2}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {1}{4} x^4 \text {arctanh}(d \tanh (a+b x)+d+1)+\frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\operatorname {PolyLog}\left (5,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b^2}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )\)

Input:

Int[x^3*ArcTanh[1 + d + d*Tanh[a + b*x]],x]
 

Output:

(x^4*ArcTanh[1 + d + d*Tanh[a + b*x]])/4 + (b*(x^5/5 - (1 + d)*((x^4*Log[1 
 + (1 + d)*E^(2*a + 2*b*x)])/(2*b*(1 + d)) - (2*(-1/2*(x^3*PolyLog[2, -((1 
 + d)*E^(2*a + 2*b*x))])/b + (3*((x^2*PolyLog[3, -((1 + d)*E^(2*a + 2*b*x) 
)])/(2*b) - ((x*PolyLog[4, -((1 + d)*E^(2*a + 2*b*x))])/(2*b) - PolyLog[5, 
 -((1 + d)*E^(2*a + 2*b*x))]/(4*b^2))/b))/(2*b)))/(b*(1 + d)))))/4
 

Defintions of rubi rules used

rule 2615

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x 
_))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ 
b/a   Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] 
, x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 6793

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tanh[a + b*x]]/(f*( 
m + 1))), x] + Simp[b/(f*(m + 1))   Int[(e + f*x)^(m + 1)/(c - d + c*E^(2*a 
 + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c 
- d)^2, 1]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [C] (warning: unable to verify)

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

Time = 3.17 (sec) , antiderivative size = 1721, normalized size of antiderivative = 11.10

method	result	size
risch	\(\text {Expression too large to display}\)	\(1721\)

Input:

int(x^3*arctanh(1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

-1/8*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^4+3/16/b^4*d/(1+d)*polylog(5,-(1 
+d)*exp(2*b*x+2*a))+1/2/b^4*a^4/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b^ 
4*a^4/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))+1/2/b^4*a^3/(1+d)*dilog(1+exp(b* 
x+a)*(-d-1)^(1/2))+1/2/b^4*a^3/(1+d)*dilog(1-exp(b*x+a)*(-d-1)^(1/2))-1/4/ 
b/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*x^3-3/8/b^4/(1+d)*ln(1+(1+d)*exp( 
2*b*x+2*a))*a^4+3/8/b^2/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))*x^2-1/4/b^4 
/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*a^3-3/8/b^3/(1+d)*polylog(4,-(1+d) 
*exp(2*b*x+2*a))*x+1/8*x^4*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)-1/8/b^4*a 
^4/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)+1/20*b*x^5-1/4*x^4*ln(exp(b 
*x+a))-1/8/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^4+3/16/b^4/(1+d)*polylog(5,- 
(1+d)*exp(2*b*x+2*a))+1/2/b^4*d*a^4/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/ 
2/b^4*d*a^4/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))+1/2/b^4*d*a^3/(1+d)*dilog( 
1+exp(b*x+a)*(-d-1)^(1/2))+1/2/b^4*d*a^3/(1+d)*dilog(1-exp(b*x+a)*(-d-1)^( 
1/2))+1/2/b^3*a^3/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))*x+1/2/b^3*a^3/(1+d)* 
ln(1-exp(b*x+a)*(-d-1)^(1/2))*x-1/4/b*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2 
*a))*x^3-1/16*(2*I*Pi+2*ln(d)-I*Pi*csgn(I*(d*exp(2*b*x+2*a)+exp(2*b*x+2*a) 
+1))*csgn(I/(exp(2*b*x+2*a)+1)*(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1))^2+I*Pi 
*csgn(I/(exp(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))^2*csgn(I*d)-2*I*Pi*csgn(I/(ex 
p(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))^2+I*Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp 
(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-I*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (135) = 270\).

Time = 0.10 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.91 \[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx =\text {Too large to display} \] Input:

integrate(x^3*arctanh(1+d+d*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/40*(2*b^5*x^5 + 5*b^4*x^4*log(-((d + 2)*cosh(b*x + a) + d*sinh(b*x + a)) 
/(d*cosh(b*x + a) + d*sinh(b*x + a))) - 20*b^3*x^3*dilog(1/2*sqrt(-4*d - 4 
)*(cosh(b*x + a) + sinh(b*x + a))) - 20*b^3*x^3*dilog(-1/2*sqrt(-4*d - 4)* 
(cosh(b*x + a) + sinh(b*x + a))) - 5*a^4*log(2*(d + 1)*cosh(b*x + a) + 2*( 
d + 1)*sinh(b*x + a) + sqrt(-4*d - 4)) - 5*a^4*log(2*(d + 1)*cosh(b*x + a) 
 + 2*(d + 1)*sinh(b*x + a) - sqrt(-4*d - 4)) + 60*b^2*x^2*polylog(3, 1/2*s 
qrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) + 60*b^2*x^2*polylog(3, -1/ 
2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) - 120*b*x*polylog(4, 1/2 
*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) - 120*b*x*polylog(4, -1/2 
*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) - 5*(b^4*x^4 - a^4)*log(1 
/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 5*(b^4*x^4 - a^4) 
*log(-1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + 120*polylo 
g(5, 1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) + 120*polylog(5, 
-1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))))/b^4
 

Sympy [F]

\[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int x^{3} \operatorname {atanh}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \] Input:

integrate(x**3*atanh(1+d+d*tanh(b*x+a)),x)
 

Output:

Integral(x**3*atanh(d*tanh(a + b*x) + d + 1), x)
 

Maxima [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96 \[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\frac {1}{4} \, x^{4} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \] Input:

integrate(x^3*arctanh(1+d+d*tanh(b*x+a)),x, algorithm="maxima")
 

Output:

1/4*x^4*arctanh(d*tanh(b*x + a) + d + 1) + 1/40*(2*x^5/d - 5*(2*b^4*x^4*lo 
g((d + 1)*e^(2*b*x + 2*a) + 1) + 4*b^3*x^3*dilog(-(d + 1)*e^(2*b*x + 2*a)) 
 - 6*b^2*x^2*polylog(3, -(d + 1)*e^(2*b*x + 2*a)) + 6*b*x*polylog(4, -(d + 
 1)*e^(2*b*x + 2*a)) - 3*polylog(5, -(d + 1)*e^(2*b*x + 2*a)))/(b^5*d))*b* 
d
 

Giac [F]

\[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int { x^{3} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \,d x } \] Input:

integrate(x^3*arctanh(1+d+d*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x^3*arctanh(d*tanh(b*x + a) + d + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int x^3\,\mathrm {atanh}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \] Input:

int(x^3*atanh(d + d*tanh(a + b*x) + 1),x)
 

Output:

int(x^3*atanh(d + d*tanh(a + b*x) + 1), x)
 

Reduce [F]

\[ \int x^3 \text {arctanh}(1+d+d \tanh (a+b x)) \, dx=\int \mathit {atanh} \left (\tanh \left (b x +a \right ) d +d +1\right ) x^{3}d x \] Input:

int(x^3*atanh(1+d+d*tanh(b*x+a)),x)
 

Output:

int(atanh(tanh(a + b*x)*d + d + 1)*x**3,x)