\(\int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx\) [304]

Optimal result

Integrand size = 14, antiderivative size = 100 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {b x^3}{6}+\frac {1}{2} x^2 \text {arctanh}(1+d+d \coth (a+b x))-\frac {1}{4} x^2 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac {x \operatorname {PolyLog}\left (2,(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (3,(1+d) e^{2 a+2 b x}\right )}{8 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {2 b^2 x^2 \left (2 \text {arctanh}(1+d+d \coth (a+b x))-\log \left (1-\frac {e^{-2 (a+b x)}}{1+d}\right )\right )+2 b x \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{1+d}\right )+\operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{1+d}\right )}{8 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6795, 2615, 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.

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

\(\Big \downarrow \) 6795

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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]
 

Maple [C] (warning: unable to verify)

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

Time = 2.76 (sec) , antiderivative size = 1555, normalized size of antiderivative = 15.55

method	result	size
risch	\(\text {Expression too large to display}\)	\(1555\)

Input:

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

Output:

-1/2*x^2*ln(exp(b*x+a))+1/6*b*x^3+1/2/b^2*a/(1+d)*dilog(1+exp(b*x+a)*(1+d) 
^(1/2))-1/4/b^2*a^2/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)-1)-1/4*d/(1+d 
)*ln(1-(1+d)*exp(2*b*x+2*a))*x^2+1/8/b^2/(1+d)*polylog(3,(1+d)*exp(2*b*x+2 
*a))-1/4/b^2/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a))*a^2-1/4/b/(1+d)*polylog(2,(1 
+d)*exp(2*b*x+2*a))*x-1/4/b^2/(1+d)*polylog(2,(1+d)*exp(2*b*x+2*a))*a+1/8/ 
b^2*d/(1+d)*polylog(3,(1+d)*exp(2*b*x+2*a))+1/2/b^2*a^2/(1+d)*ln(1-exp(b*x 
+a)*(1+d)^(1/2))+1/2/b^2*a^2/(1+d)*ln(1+exp(b*x+a)*(1+d)^(1/2))+1/2/b^2*a/ 
(1+d)*dilog(1-exp(b*x+a)*(1+d)^(1/2))-1/4/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a)) 
*x^2-1/8*(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*ex 
p(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))*cs 
gn(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)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (86) = 172\).

Time = 0.10 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.06 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {2 \, b^{3} x^{3} + 3 \, b^{2} x^{2} \log \left (-\frac {d \cosh \left (b x + a\right ) + {\left (d + 2\right )} \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_2\left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm Li}_2\left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d + 1}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d + 1}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 6 \, {\rm polylog}\left (3, \sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (3, -\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{2}} \] Input:

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

Output:

1/12*(2*b^3*x^3 + 3*b^2*x^2*log(-(d*cosh(b*x + a) + (d + 2)*sinh(b*x + a)) 
/(d*cosh(b*x + a) + d*sinh(b*x + a))) - 6*b*x*dilog(sqrt(d + 1)*(cosh(b*x 
+ a) + sinh(b*x + a))) - 6*b*x*dilog(-sqrt(d + 1)*(cosh(b*x + a) + sinh(b* 
x + a))) - 3*a^2*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) + 2 
*sqrt(d + 1)) - 3*a^2*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a 
) - 2*sqrt(d + 1)) - 3*(b^2*x^2 - a^2)*log(sqrt(d + 1)*(cosh(b*x + a) + si 
nh(b*x + a)) + 1) - 3*(b^2*x^2 - a^2)*log(-sqrt(d + 1)*(cosh(b*x + a) + si 
nh(b*x + a)) + 1) + 6*polylog(3, sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a 
))) + 6*polylog(3, -sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))))/b^2
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int x \text {arctanh}(1+d+d \coth (a+b x)) \, dx=\frac {1}{24} \, {\left (\frac {4 \, x^{3}}{d} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{3} d}\right )} b d + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d + 1\right ) \] Input:

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

Output:

1/24*(4*x^3/d - 3*(2*b^2*x^2*log(-(d + 1)*e^(2*b*x + 2*a) + 1) + 2*b*x*dil 
og((d + 1)*e^(2*b*x + 2*a)) - polylog(3, (d + 1)*e^(2*b*x + 2*a)))/(b^3*d) 
)*b*d + 1/2*x^2*arctanh(d*coth(b*x + a) + d + 1)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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