Integrand size = 19, antiderivative size = 168 \[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \text {arctanh}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{16 b^4} \]
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*poly log(5,-(1-d)*exp(2*b*x+2*a))/b^4
Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.88 \[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {4 b^4 x^4 \text {arctanh}(1-d-d \tanh (a+b x))-2 b^4 x^4 \log \left (1-\frac {e^{-2 (a+b x)}}{-1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b x \operatorname {PolyLog}\left (4,\frac {e^{-2 (a+b x)}}{-1+d}\right )+3 \operatorname {PolyLog}\left (5,\frac {e^{-2 (a+b x)}}{-1+d}\right )}{16 b^4} \]
(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)
Time = 1.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.30, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, 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 |
| \(\displaystyle \frac {1}{4} b \int \frac {x^4}{e^{2 a+2 b x} (1-d)+1}dx+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \int \frac {e^{2 a+2 b x} x^4}{e^{2 a+2 b x} (1-d)+1}dx\right )+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \left (\frac {x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {2 \int x^3 \log \left (e^{2 a+2 b x} (1-d)+1\right )dx}{b (1-d)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \left (\frac {x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {2 \left (\frac {3 \int x^2 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )dx}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (1-d)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \left (\frac {x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int x \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )dx}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (1-d)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \left (\frac {x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )dx}{2 b}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (1-d)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \left (\frac {x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )de^{2 a+2 b x}}{4 b^2}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (1-d)}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{4} x^4 \text {arctanh}(d (-\tanh (a+b x))-d+1)+\frac {1}{4} b \left (\frac {x^5}{5}-(1-d) \left (\frac {x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b^2}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (1-d)}\right )\right )\) |
(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
3.3.93.3.1 Defintions of rubi rules used
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]
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]
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]]]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.32 (sec) , antiderivative size = 1725, normalized size of antiderivative = 10.27
1/8*x^4*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1)+1/2/b^3/(d-1)*ln(1-(d-1)*exp (2*b*x+2*a))*x*a^3-1/4/b*d/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*x^3-3/8/b ^4*d/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*a^4+3/8/b^2*d/(d-1)*polylog(3,(d-1)* exp(2*b*x+2*a))*x^2-1/4/b^4*d/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*a^3-3/ 8/b^3*d/(d-1)*polylog(4,(d-1)*exp(2*b*x+2*a))*x-1/2/b^3*a^3/(d-1)*ln(1-exp (b*x+a)*(d-1)^(1/2))*x-1/2/b^3*a^3/(d-1)*ln(1+exp(b*x+a)*(d-1)^(1/2))*x+1/ 2/b^4*d*a^4/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))+1/2/b^4*d*a^4/(d-1)*ln(1+ex p(b*x+a)*(d-1)^(1/2))+1/2/b^4*d*a^3/(d-1)*dilog(1-exp(b*x+a)*(d-1)^(1/2))+ 1/20*b*x^5-1/8/b^4*d*a^4/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1)-1/8*d /(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x^4-1/2/b^4*a^4/(d-1)*ln(1+exp(b*x+a)*(d -1)^(1/2))-1/2/b^4*a^3/(d-1)*dilog(1-exp(b*x+a)*(d-1)^(1/2))-1/2/b^4*a^3/( d-1)*dilog(1+exp(b*x+a)*(d-1)^(1/2))+3/16/b^4*d/(d-1)*polylog(5,(d-1)*exp( 2*b*x+2*a))+1/4/b/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*x^3+3/8/b^4/(d-1)* ln(1-(d-1)*exp(2*b*x+2*a))*a^4-3/8/b^2/(d-1)*polylog(3,(d-1)*exp(2*b*x+2*a ))*x^2+1/4/b^4/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*a^3+3/8/b^3/(d-1)*pol ylog(4,(d-1)*exp(2*b*x+2*a))*x-1/2/b^4*a^4/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/ 2))+1/8/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x^4-3/16/b^4/(d-1)*polylog(5,(d-1 )*exp(2*b*x+2*a))+1/8/b^4*a^4/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1)- 1/16*(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+2*I*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+2*I*Pi*csgn(I/(e...
Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (132) = 264\).
Time = 0.26 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.52 \[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {2 \, b^{5} x^{5} - 5 \, b^{4} x^{4} \log \left (-\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \, a^{4} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) - 5 \, a^{4} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \, {\left (b^{4} x^{4} - a^{4}\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 5 \, {\left (b^{4} x^{4} - a^{4}\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 120 \, {\rm polylog}\left (5, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \, {\rm polylog}\left (5, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{40 \, b^{4}} \]
1/40*(2*b^5*x^5 - 5*b^4*x^4*log(-(d*cosh(b*x + a) + d*sinh(b*x + a))/((d - 2)*cosh(b*x + a) + d*sinh(b*x + a))) - 20*b^3*x^3*dilog(sqrt(d - 1)*(cosh (b*x + a) + sinh(b*x + a))) - 20*b^3*x^3*dilog(-sqrt(d - 1)*(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) + 2*sqrt(d - 1)) - 5*a^4*log(2*(d - 1)*cosh(b*x + a) + 2*(d - 1)*si nh(b*x + a) - 2*sqrt(d - 1)) + 60*b^2*x^2*polylog(3, sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) + 60*b^2*x^2*polylog(3, -sqrt(d - 1)*(cosh(b*x + a ) + sinh(b*x + a))) - 120*b*x*polylog(4, sqrt(d - 1)*(cosh(b*x + a) + sinh (b*x + a))) - 120*b*x*polylog(4, -sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) - 5*(b^4*x^4 - a^4)*log(sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 5*(b^4*x^4 - a^4)*log(-sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + 120*polylog(5, sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) + 120*po lylog(5, -sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))))/b^4
\[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=- \int x^{3} \operatorname {atanh}{\left (d \tanh {\left (a + b x \right )} + d - 1 \right )}\, dx \]
Time = 0.76 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=-\frac {1}{4} \, x^{4} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d - 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \]
-1/4*x^4*arctanh(d*tanh(b*x + a) + d - 1) + 1/40*(2*x^5/d - 5*(2*b^4*x^4*l og(-(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
\[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\int { -x^{3} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d - 1\right ) \,d x } \]
Timed out. \[ \int x^3 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\int -x^3\,\mathrm {atanh}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \]