\(\int x \text {arctanh}(c+d \coth (a+b x)) \, dx\) [299]

Optimal result

Integrand size = 13, antiderivative size = 229 \[ \int x \text {arctanh}(c+d \coth (a+b x)) \, dx=\frac {1}{2} x^2 \text {arctanh}(c+d \coth (a+b x))+\frac {1}{4} x^2 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{4} x^2 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {\operatorname {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^2}+\frac {\operatorname {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.87 \[ \int x \text {arctanh}(c+d \coth (a+b x)) \, dx=\frac {4 b^2 x^2 \text {arctanh}(c+d \coth (a+b x))+2 b^2 x^2 \log \left (1+\frac {(1-c+d) e^{-2 (a+b x)}}{-1+c+d}\right )-2 b^2 x^2 \log \left (1+\frac {(-1-c+d) e^{-2 (a+b x)}}{1+c+d}\right )-2 b x \operatorname {PolyLog}\left (2,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )+2 b x \operatorname {PolyLog}\left (2,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )-\operatorname {PolyLog}\left (3,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )+\operatorname {PolyLog}\left (3,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )}{8 b^2} \] Input:

Integrate[x*ArcTanh[c + d*Coth[a + b*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6799, 2620, 3011, 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.

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

\(\Big \downarrow \) 6799

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x*ArcTanh[c + d*Coth[a + b*x]],x]
 

Output:

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

Defintions of rubi rules used

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 6799

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*((1 - c - d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*( 
E^(2*a + 2*b*x)/(1 - c + d - (1 - c - d)*E^(2*a + 2*b*x))), x], x] + Simp[b 
*((1 + c + d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^(2*a + 2*b*x)/(1 + c 
- d - (1 + c + d)*E^(2*a + 2*b*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x 
] && IGtQ[m, 0] && NeQ[(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]
 

Maple [C] (warning: unable to verify)

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

Time = 5.16 (sec) , antiderivative size = 4944, normalized size of antiderivative = 21.59

method	result	size
risch	\(\text {Expression too large to display}\)	\(4944\)

Input:

int(x*arctanh(c+d*coth(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 730 vs. \(2 (195) = 390\).

Time = 0.12 (sec) , antiderivative size = 730, normalized size of antiderivative = 3.19 \[ \int x \text {arctanh}(c+d \coth (a+b x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*(b^2*x^2*log(-(d*cosh(b*x + a) + (c + 1)*sinh(b*x + a))/(d*cosh(b*x + 
a) + (c - 1)*sinh(b*x + a))) - 2*b*x*dilog(sqrt((c + d + 1)/(c - d + 1))*( 
cosh(b*x + a) + sinh(b*x + a))) - 2*b*x*dilog(-sqrt((c + d + 1)/(c - d + 1 
))*(cosh(b*x + a) + sinh(b*x + a))) + 2*b*x*dilog(sqrt((c + d - 1)/(c - d 
- 1))*(cosh(b*x + a) + sinh(b*x + a))) + 2*b*x*dilog(-sqrt((c + d - 1)/(c 
- d - 1))*(cosh(b*x + a) + sinh(b*x + a))) - a^2*log(2*(c + d + 1)*cosh(b* 
x + a) + 2*(c + d + 1)*sinh(b*x + a) + 2*(c - d + 1)*sqrt((c + d + 1)/(c - 
 d + 1))) - a^2*log(2*(c + d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + 
 a) - 2*(c - d + 1)*sqrt((c + d + 1)/(c - d + 1))) + a^2*log(2*(c + d - 1) 
*cosh(b*x + a) + 2*(c + d - 1)*sinh(b*x + a) + 2*(c - d - 1)*sqrt((c + d - 
 1)/(c - d - 1))) + a^2*log(2*(c + d - 1)*cosh(b*x + a) + 2*(c + d - 1)*si 
nh(b*x + a) - 2*(c - d - 1)*sqrt((c + d - 1)/(c - d - 1))) - (b^2*x^2 - a^ 
2)*log(sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) 
- (b^2*x^2 - a^2)*log(-sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh 
(b*x + a)) + 1) + (b^2*x^2 - a^2)*log(sqrt((c + d - 1)/(c - d - 1))*(cosh( 
b*x + a) + sinh(b*x + a)) + 1) + (b^2*x^2 - a^2)*log(-sqrt((c + d - 1)/(c 
- d - 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + 2*polylog(3, sqrt((c + d 
+ 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 2*polylog(3, -sqrt((c 
 + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) - 2*polylog(3, sqr 
t((c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) - 2*polylog...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.93 \[ \int x \text {arctanh}(c+d \coth (a+b x)) \, dx=-\frac {1}{8} \, b d {\left (\frac {2 \, b^{2} x^{2} \log \left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 2 \, b x {\rm Li}_2\left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - {\rm Li}_{3}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{3} d} - \frac {2 \, b^{2} x^{2} \log \left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 2 \, b x {\rm Li}_2\left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - {\rm Li}_{3}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{3} d}\right )} + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + c\right ) \] Input:

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

Output:

-1/8*b*d*((2*b^2*x^2*log(-(c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1) + 1) + 2 
*b*x*dilog((c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1)) - polylog(3, (c + d + 
1)*e^(2*b*x + 2*a)/(c - d + 1)))/(b^3*d) - (2*b^2*x^2*log(-(c + d - 1)*e^( 
2*b*x + 2*a)/(c - d - 1) + 1) + 2*b*x*dilog((c + d - 1)*e^(2*b*x + 2*a)/(c 
 - d - 1)) - polylog(3, (c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1)))/(b^3*d)) 
 + 1/2*x^2*arctanh(d*coth(b*x + a) + c)
 

Giac [F]

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

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

Output:

integrate(x*arctanh(d*coth(b*x + a) + c), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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