\(\int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx\) [212]

Optimal result

Integrand size = 19, antiderivative size = 168 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{16 b^4} \]

[Out]

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*pol
ylog(5,-(1-d)*exp(2*b*x+2*a))/b^4

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6375, 2215, 2221, 2611, 6744, 2320, 6724} \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {3 \operatorname {PolyLog}\left (5,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{16 b^4}-\frac {3 x \operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {x^3 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{8} x^4 \log \left ((1-d) e^{2 a+2 b x}+1\right )+\frac {1}{4} x^4 \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^5}{20} \]

[In]

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

[Out]

(b*x^5)/20 + (x^4*ArcCoth[1 - d - d*Tanh[a + b*x]])/4 - (x^4*Log[1 + (1 - d)*E^(2*a + 2*b*x)])/8 - (x^3*PolyLo
g[2, -((1 - d)*E^(2*a + 2*b*x))])/(4*b) + (3*x^2*PolyLog[3, -((1 - d)*E^(2*a + 2*b*x))])/(8*b^2) - (3*x*PolyLo
g[4, -((1 - d)*E^(2*a + 2*b*x))])/(8*b^3) + (3*PolyLog[5, -((1 - d)*E^(2*a + 2*b*x))])/(16*b^4)

Rule 2215

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] - Dist[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 2221

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]
- Dist[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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2611

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] + Dist[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 6375

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] + Dist[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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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 6744

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))+\frac {1}{4} b \int \frac {x^4}{1+(1-d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{4} (b (1-d)) \int \frac {e^{2 a+2 b x} x^4}{1+(1-d) e^{2 a+2 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 {1}{2} \int x^3 \log \left (1+(1-d) e^{2 a+2 b x}\right ) \, 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 \int x^2 \operatorname {PolyLog}\left (2,(-1+d) e^{2 a+2 b x}\right ) \, dx}{4 b} \\ & = \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 \int x \operatorname {PolyLog}\left (3,(-1+d) e^{2 a+2 b x}\right ) \, dx}{4 b^2} \\ & = \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 \int \operatorname {PolyLog}\left (4,(-1+d) e^{2 a+2 b x}\right ) \, dx}{8 b^3} \\ & = \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 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(4,(-1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.88 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\frac {4 b^4 x^4 \coth ^{-1}(1-d-d \tanh (a+b x))-2 b^4 x^4 \log \left (1-\frac {e^{-2 (a+b x)}}{-1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b x \operatorname {PolyLog}\left (4,\frac {e^{-2 (a+b x)}}{-1+d}\right )+3 \operatorname {PolyLog}\left (5,\frac {e^{-2 (a+b x)}}{-1+d}\right )}{16 b^4} \]

[In]

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

[Out]

(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*Poly
Log[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)

Maple [C] (warning: unable to verify)

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

Time = 2.02 (sec) , antiderivative size = 1754, normalized size of antiderivative = 10.44

method	result	size
risch	\(\text {Expression too large to display}\)	\(1754\)

[In]

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

[Out]

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+exp(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))-1/8/
b^4*d*a^4/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1)+1/20*b*x^5-1/8*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)*polylog(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)-1/4*x^4*ln(exp(b*x+a))+1/2/b^4*d*a
^3/(d-1)*dilog(1+exp(b*x+a)*(d-1)^(1/2))+1/2/b^3*d*a^3/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))*x+1/2/b^3*d*a^3/(d-1
)*ln(1+exp(b*x+a)*(d-1)^(1/2))*x-1/2/b^3*d/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x*a^3+1/8*x^4*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)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(d*exp(2*b*x+2*a)-exp(2*b*x
+2*a)-1))^2-2*I*Pi*csgn(I/(exp(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))^2+2*I*Pi*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*d)*csgn(I/(exp(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))^2-I*Pi*csgn(I/(exp(2*b*x
+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+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(2*b*x+2*a))-I*Pi*csgn(I/(exp(2*b*x+2
*a)+1)*(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1))^3-I*Pi*csgn(I*exp(2*b*x+2*a))^3+I*Pi*csgn(I/(exp(2*b*x+2*a)+1)*d*e
xp(2*b*x+2*a))^3+I*Pi*csgn(I/(exp(2*b*x+2*a)+1))*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-I*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+I*Pi*csgn(I/(exp(2*b*x+2*a)
+1))*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)
)-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)/(exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))^2-I*Pi*csgn(I
*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*csgn(I*d)*csgn(I/(exp(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))+2*ln(d))*x^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (135) = 270\).

Time = 0.27 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.52 \[ \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 {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}} \]

[In]

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

[Out]

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)*sinh(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*pol
ylog(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(-sq
rt(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*polylog(5, -sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))))/b^4

Sympy [F]

\[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=- \int x^{3} \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d - 1 \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.79 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=-\frac {1}{4} \, x^{4} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d - 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \]

[In]

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

[Out]

-1/4*x^4*arccoth(d*tanh(b*x + a) + d - 1) + 1/40*(2*x^5/d - 5*(2*b^4*x^4*log(-(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]

\[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\int { x^{3} \operatorname {arcoth}\left (-d \tanh \left (b x + a\right ) - d + 1\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx=\int -x^3\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \]

[In]

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

[Out]

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