\(\int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx\) [308]

Optimal result

Integrand size = 19, antiderivative size = 137 \[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=\frac {b x^4}{12}+\frac {1}{3} x^3 \text {arctanh}(1-d-d \coth (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \operatorname {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,(1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,(1-d) e^{2 a+2 b x}\right )}{8 b^3} \] Output:

1/12*b*x^4-1/3*x^3*arctanh(-1+d+d*coth(b*x+a))-1/6*x^3*ln(1-(1-d)*exp(2*b* 
x+2*a))-1/4*x^2*polylog(2,(1-d)*exp(2*b*x+2*a))/b+1/4*x*polylog(3,(1-d)*ex 
p(2*b*x+2*a))/b^2-1/8*polylog(4,(1-d)*exp(2*b*x+2*a))/b^3
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=\frac {8 b^3 x^3 \text {arctanh}(1-d-d \coth (a+b x))-4 b^3 x^3 \log \left (1+\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{-1+d}\right )+3 \operatorname {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{-1+d}\right )}{24 b^3} \] Input:

Integrate[x^2*ArcTanh[1 - d - d*Coth[a + b*x]],x]
 

Output:

(8*b^3*x^3*ArcTanh[1 - d - d*Coth[a + b*x]] - 4*b^3*x^3*Log[1 + 1/((-1 + d 
)*E^(2*(a + b*x)))] + 6*b^2*x^2*PolyLog[2, -(1/((-1 + d)*E^(2*(a + b*x)))) 
] + 6*b*x*PolyLog[3, -(1/((-1 + d)*E^(2*(a + b*x))))] + 3*PolyLog[4, -(1/( 
(-1 + d)*E^(2*(a + b*x))))])/(24*b^3)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6795, 2615, 2620, 3011, 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.

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

\(\Big \downarrow \) 6795

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2*ArcTanh[1 - d - d*Coth[a + b*x]],x]
 

Output:

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

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 = 3.26 (sec) , antiderivative size = 1694, normalized size of antiderivative = 12.36

method	result	size
risch	\(\text {Expression too large to display}\)	\(1694\)

Input:

int(-x^2*arctanh(-1+d+d*coth(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/12*b*x^4-1/3*x^3*ln(exp(b*x+a))+1/8/b^3/(d-1)*polylog(4,-(d-1)*exp(2*b*x 
+2*a))+1/6/(d-1)*ln(1+(d-1)*exp(2*b*x+2*a))*x^3-1/6*d/(d-1)*ln(1+(d-1)*exp 
(2*b*x+2*a))*x^3-1/6/b^3*a^3/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)+1)+1 
/2/b^3*a^3/(d-1)*ln(1+exp(b*x+a)*(1-d)^(1/2))+1/2/b^3*a^3/(d-1)*ln(1-exp(b 
*x+a)*(1-d)^(1/2))+1/2/b^3*a^2/(d-1)*dilog(1+exp(b*x+a)*(1-d)^(1/2))+1/2/b 
^3*a^2/(d-1)*dilog(1-exp(b*x+a)*(1-d)^(1/2))-1/8/b^3*d/(d-1)*polylog(4,-(d 
-1)*exp(2*b*x+2*a))+1/4/b/(d-1)*polylog(2,-(d-1)*exp(2*b*x+2*a))*x^2-1/3/b 
^3/(d-1)*ln(1+(d-1)*exp(2*b*x+2*a))*a^3-1/4/b^3/(d-1)*polylog(2,-(d-1)*exp 
(2*b*x+2*a))*a^2-1/4/b^2/(d-1)*polylog(3,-(d-1)*exp(2*b*x+2*a))*x+1/2/b^2* 
d/(d-1)*ln(1+(d-1)*exp(2*b*x+2*a))*a^2*x-1/2/b^2*d*a^2/(d-1)*x*ln(1+exp(b* 
x+a)*(1-d)^(1/2))-1/2/b^2*d*a^2/(d-1)*x*ln(1-exp(b*x+a)*(1-d)^(1/2))-1/4/b 
*d/(d-1)*polylog(2,-(d-1)*exp(2*b*x+2*a))*x^2+1/3/b^3*d/(d-1)*ln(1+(d-1)*e 
xp(2*b*x+2*a))*a^3+1/4/b^3*d/(d-1)*polylog(2,-(d-1)*exp(2*b*x+2*a))*a^2+1/ 
4/b^2*d/(d-1)*polylog(3,-(d-1)*exp(2*b*x+2*a))*x-1/2/b^2/(d-1)*ln(1+(d-1)* 
exp(2*b*x+2*a))*a^2*x+1/2/b^2*a^2/(d-1)*x*ln(1+exp(b*x+a)*(1-d)^(1/2))+1/2 
/b^2*a^2/(d-1)*x*ln(1-exp(b*x+a)*(1-d)^(1/2))+1/6/b^3*d*a^3/(d-1)*ln(d*exp 
(2*b*x+2*a)-exp(2*b*x+2*a)+1)-1/2/b^3*d*a^3/(d-1)*ln(1+exp(b*x+a)*(1-d)^(1 
/2))-1/2/b^3*d*a^3/(d-1)*ln(1-exp(b*x+a)*(1-d)^(1/2))-1/2/b^3*d*a^2/(d-1)* 
dilog(1+exp(b*x+a)*(1-d)^(1/2))-1/2/b^3*d*a^2/(d-1)*dilog(1-exp(b*x+a)*(1- 
d)^(1/2))+1/6*x^3*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)+1)-1/12*(-I*Pi*csg...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (111) = 222\).

Time = 0.10 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.79 \[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=\frac {b^{4} x^{4} - 2 \, b^{3} x^{3} \log \left (-\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + {\left (d - 2\right )} \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d + 4}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d + 4}\right ) + 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{3}} \] Input:

integrate(-x^2*arctanh(-1+d+d*coth(b*x+a)),x, algorithm="fricas")
 

Output:

1/12*(b^4*x^4 - 2*b^3*x^3*log(-(d*cosh(b*x + a) + d*sinh(b*x + a))/(d*cosh 
(b*x + a) + (d - 2)*sinh(b*x + a))) - 6*b^2*x^2*dilog(1/2*sqrt(-4*d + 4)*( 
cosh(b*x + a) + sinh(b*x + a))) - 6*b^2*x^2*dilog(-1/2*sqrt(-4*d + 4)*(cos 
h(b*x + a) + sinh(b*x + a))) + 2*a^3*log(2*(d - 1)*cosh(b*x + a) + 2*(d - 
1)*sinh(b*x + a) + sqrt(-4*d + 4)) + 2*a^3*log(2*(d - 1)*cosh(b*x + a) + 2 
*(d - 1)*sinh(b*x + a) - sqrt(-4*d + 4)) + 12*b*x*polylog(3, 1/2*sqrt(-4*d 
 + 4)*(cosh(b*x + a) + sinh(b*x + a))) + 12*b*x*polylog(3, -1/2*sqrt(-4*d 
+ 4)*(cosh(b*x + a) + sinh(b*x + a))) - 2*(b^3*x^3 + a^3)*log(1/2*sqrt(-4* 
d + 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 2*(b^3*x^3 + a^3)*log(-1/2*s 
qrt(-4*d + 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 12*polylog(4, 1/2*sqr 
t(-4*d + 4)*(cosh(b*x + a) + sinh(b*x + a))) - 12*polylog(4, -1/2*sqrt(-4* 
d + 4)*(cosh(b*x + a) + sinh(b*x + a))))/b^3
 

Sympy [F]

\[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=- \int x^{2} \operatorname {atanh}{\left (d \coth {\left (a + b x \right )} + d - 1 \right )}\, dx \] Input:

integrate(-x**2*atanh(-1+d+d*coth(b*x+a)),x)
 

Output:

-Integral(x**2*atanh(d*coth(a + b*x) + d - 1), x)
 

Maxima [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=-\frac {1}{3} \, x^{3} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d - 1\right ) + \frac {1}{36} \, {\left (\frac {3 \, x^{4}}{d} - \frac {2 \, {\left (4 \, b^{3} x^{3} \log \left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{4} d}\right )} b d \] Input:

integrate(-x^2*arctanh(-1+d+d*coth(b*x+a)),x, algorithm="maxima")
 

Output:

-1/3*x^3*arctanh(d*coth(b*x + a) + d - 1) + 1/36*(3*x^4/d - 2*(4*b^3*x^3*l 
og((d - 1)*e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog(-(d - 1)*e^(2*b*x + 2*a) 
) - 6*b*x*polylog(3, -(d - 1)*e^(2*b*x + 2*a)) + 3*polylog(4, -(d - 1)*e^( 
2*b*x + 2*a)))/(b^4*d))*b*d
 

Giac [F]

\[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=\int { -x^{2} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d - 1\right ) \,d x } \] Input:

integrate(-x^2*arctanh(-1+d+d*coth(b*x+a)),x, algorithm="giac")
 

Output:

integrate(-x^2*arctanh(d*coth(b*x + a) + d - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=\int -x^2\,\mathrm {atanh}\left (d+d\,\mathrm {coth}\left (a+b\,x\right )-1\right ) \,d x \] Input:

int(-x^2*atanh(d + d*coth(a + b*x) - 1),x)
 

Output:

int(-x^2*atanh(d + d*coth(a + b*x) - 1), x)
 

Reduce [F]

\[ \int x^2 \text {arctanh}(1-d-d \coth (a+b x)) \, dx=-\left (\int \mathit {atanh} \left (\coth \left (b x +a \right ) d +d -1\right ) x^{2}d x \right ) \] Input:

int(-x^2*atanh(-1+d+d*coth(b*x+a)),x)
 

Output:

 - int(atanh(coth(a + b*x)*d + d - 1)*x**2,x)