Integrand size = 16, antiderivative size = 155 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{16 b^4} \]
1/20*b*x^5+1/4*x^4*arccoth(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
Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {4 b^4 x^4 \coth ^{-1}(1+d+d \tanh (a+b x))-2 b^4 x^4 \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b x \operatorname {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{1+d}\right )+3 \operatorname {PolyLog}\left (5,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{16 b^4} \]
(4*b^4*x^4*ArcCoth[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 = 1.02 (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 = {6794, 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.
| \(\displaystyle \int x^3 \coth ^{-1}(d \tanh (a+b x)+d+1) \, dx\) |
| \(\Big \downarrow \) 6794 |
| \(\displaystyle \frac {1}{4} b \int \frac {x^4}{e^{2 a+2 b x} (d+1)+1}dx+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \int \frac {e^{2 a+2 b x} x^4}{e^{2 a+2 b x} (d+1)+1}dx\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \int x^3 \log \left (e^{2 a+2 b x} (d+1)+1\right )dx}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \int x^2 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int x \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )dx}{2 b}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )de^{2 a+2 b x}}{4 b^2}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{4} b \left (\frac {x^5}{5}-(d+1) \left (\frac {x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )}{2 b (d+1)}-\frac {2 \left (\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\frac {x \operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\operatorname {PolyLog}\left (5,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b^2}}{b}\right )}{2 b}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{b (d+1)}\right )\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)\) |
(x^4*ArcCoth[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
3.3.7.3.1 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[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCoth[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 = 2.09 (sec) , antiderivative size = 1684, normalized size of antiderivative = 10.86
-1/2/b^3*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x*a^3+1/2/b^3*d*a^3/(1+d)*ln(1 +exp(b*x+a)*(-d-1)^(1/2))*x+1/2/b^3*d*a^3/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/ 2))*x-3/8/b^4*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^4+3/8/b^2*d/(1+d)*polyl og(3,-(1+d)*exp(2*b*x+2*a))*x^2-1/4/b^4*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x +2*a))*a^3-3/8/b^3*d/(1+d)*polylog(4,-(1+d)*exp(2*b*x+2*a))*x+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+1/2/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/ 8/b^4*d*a^4/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)+1/20*b*x^5-1/8/b^4 *a^4/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)-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+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-e xp(b*x+a)*(-d-1)^(1/2))-1/8*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^4-1/8/(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^3/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x*a^3-1/4*x^4*ln(exp(b...
Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (135) = 270\).
Time = 0.27 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.90 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {2 \, b^{5} x^{5} + 5 \, b^{4} x^{4} \log \left (\frac {{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 20 \, b^{3} x^{3} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 ) + \sqrt {-4 \, d - 4}\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 ) - \sqrt {-4 \, d - 4}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 60 \, b^{2} x^{2} {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 120 \, {\rm polylog}\left (5, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \, {\rm polylog}\left (5, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\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 + 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*sq rt(-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*polylog (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
\[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\int x^{3} \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \]
Time = 0.79 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96 \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {1}{4} \, x^{4} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \]
1/4*x^4*arccoth(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
\[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\int { x^{3} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \,d x } \]
Timed out. \[ \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\int x^3\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \]