\(\int x^3 \text {arctanh}(1+d+d \coth (a+b x)) \, dx\) [302]

Optimal result

Integrand size = 16, antiderivative size = 152 \[ \int x^3 \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {b x^5}{20}+\frac {1}{4} x^4 \text {arctanh}(1+d+d \coth (a+b x))-\frac {1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {3 x^2 \operatorname {PolyLog}\left (3,(1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac {3 x \operatorname {PolyLog}\left (4,(1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac {3 \operatorname {PolyLog}\left (5,(1+d) e^{2 a+2 b x}\right )}{16 b^4} \] Output:

1/20*b*x^5+1/4*x^4*arctanh(1+d+d*coth(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)*e 
xp(2*b*x+2*a))/b^2-3/8*x*polylog(4,(1+d)*exp(2*b*x+2*a))/b^3+3/16*polylog( 
5,(1+d)*exp(2*b*x+2*a))/b^4
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int x^3 \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {4 b^4 x^4 \text {arctanh}(1+d+d \coth (a+b x))-2 b^4 x^4 \log \left (1-\frac {e^{-2 (a+b x)}}{1+d}\right )+4 b^3 x^3 \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b x \operatorname {PolyLog}\left (4,\frac {e^{-2 (a+b x)}}{1+d}\right )+3 \operatorname {PolyLog}\left (5,\frac {e^{-2 (a+b x)}}{1+d}\right )}{16 b^4} \] Input:

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

Output:

(4*b^4*x^4*ArcTanh[1 + d + d*Coth[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 = 1.50 (sec) , antiderivative size = 195, 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 = {6795, 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.

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

\(\Big \downarrow \) 6795

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(x^4*ArcTanh[1 + d + d*Coth[a + b*x]])/4 + (b*(x^5/5 + (1 + d)*(-1/2*(x^4* 
Log[1 - (1 + d)*E^(2*a + 2*b*x)])/(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 6795

Int[ArcTanh[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Coth[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 = 3.67 (sec) , antiderivative size = 1693, normalized size of antiderivative = 11.14

method	result	size
risch	\(\text {Expression too large to display}\)	\(1693\)

Input:

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

Output:

-1/16*(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+I*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)-1))^2*csgn(I*exp(2*b*x 
+2*a))-2*I*Pi*csgn(I/(exp(2*b*x+2*a)-1)*d*exp(2*b*x+2*a))^2+2*I*Pi+2*ln(d) 
+I*Pi*csgn(I/(exp(2*b*x+2*a)-1)*d*exp(2*b*x+2*a))^2*csgn(I*d)-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))*csgn(I*exp(2 
*b*x+2*a))-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(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)-1)*d*exp(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)-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)-1))*c 
sgn(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*ex 
p(2*b*x+2*a)+exp(2*b*x+2*a)-1))-I*Pi*csgn(I*exp(2*b*x+2*a))^3+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))*csgn(I/(exp(2*b*x+2*a)-1)*d*exp(2*b*x+2*a))*csgn(I*d)-I*Pi*csg 
n(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)-1))^3)*x^4+1/2/b^4*a^4/(1+d)*ln(1+exp(b 
*x+a)*(1+d)^(1/2))+1/2/b^4*a^3/(1+d)*dilog(1-exp(b*x+a)*(1+d)^(1/2))+1/2/b 
^4*a^3/(1+d)*dilog(1+exp(b*x+a)*(1+d)^(1/2))-1/8*d/(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))-1/8/(1+d)*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (132) = 264\).

Time = 0.11 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.79 \[ \int x^3 \text {arctanh}(1+d+d \coth (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 ) + {\left (d + 2\right )} \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 (\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}} \] Input:

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

Output:

1/40*(2*b^5*x^5 + 5*b^4*x^4*log(-(d*cosh(b*x + a) + (d + 2)*sinh(b*x + a)) 
/(d*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
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.96 \[ \int x^3 \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {1}{4} \, x^{4} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d + 1\right ) + \frac {1}{40} \, {\left (\frac {2 \, x^{5}}{d} - \frac {5 \, {\left (2 \, b^{4} x^{4} \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3} {\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_{3}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x {\rm Li}_{4}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \, {\rm Li}_{5}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \] Input:

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

Output:

1/4*x^4*arctanh(d*coth(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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