\(\int x^2 \text {arctanh}(c+d \tanh (a+b x)) \, dx\) [284]

Optimal result

Integrand size = 15, antiderivative size = 307 \[ \int x^2 \text {arctanh}(c+d \tanh (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(c+d \tanh (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\operatorname {PolyLog}\left (4,-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\operatorname {PolyLog}\left (4,-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.86 \[ \int x^2 \text {arctanh}(c+d \tanh (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(c+d \tanh (a+b x))+\frac {4 b^3 x^3 \log \left (1+\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-4 b^3 x^3 \log \left (1+\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(1-c+d) e^{-2 (a+b x)}}{-1+c+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(-1-c+d) e^{-2 (a+b x)}}{1+c+d}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {(1-c+d) e^{-2 (a+b x)}}{-1+c+d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {(-1-c+d) e^{-2 (a+b x)}}{1+c+d}\right )-3 \operatorname {PolyLog}\left (4,\frac {(1-c+d) e^{-2 (a+b x)}}{-1+c+d}\right )+3 \operatorname {PolyLog}\left (4,\frac {(-1-c+d) e^{-2 (a+b x)}}{1+c+d}\right )}{24 b^3} \] Input:

Integrate[x^2*ArcTanh[c + d*Tanh[a + b*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6797, 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.

\(\displaystyle \int x^2 \text {arctanh}(d \tanh (a+b x)+c) \, dx\)

\(\Big \downarrow \) 6797

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2*ArcTanh[c + d*Tanh[a + b*x]],x]
 

Output:

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

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 6797

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tanh[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]
 

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 = 13.13 (sec) , antiderivative size = 5316, normalized size of antiderivative = 17.32

method	result	size
risch	\(\text {Expression too large to display}\)	\(5316\)

Input:

int(x^2*arctanh(c+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 900 vs. \(2 (263) = 526\).

Time = 0.12 (sec) , antiderivative size = 900, normalized size of antiderivative = 2.93 \[ \int x^2 \text {arctanh}(c+d \tanh (a+b x)) \, dx=\text {Too large to display} \] Input:

integrate(x^2*arctanh(c+d*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/6*(b^3*x^3*log(-((c + 1)*cosh(b*x + a) + d*sinh(b*x + a))/((c - 1)*cosh( 
b*x + a) + d*sinh(b*x + a))) - 3*b^2*x^2*dilog(sqrt(-(c + d + 1)/(c - d + 
1))*(cosh(b*x + a) + sinh(b*x + a))) - 3*b^2*x^2*dilog(-sqrt(-(c + d + 1)/ 
(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 3*b^2*x^2*dilog(sqrt(-(c + 
 d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) + 3*b^2*x^2*dilog(-s 
qrt(-(c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) + a^3*log(2 
*(c + d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + a) + 2*(c - d + 1)*s 
qrt(-(c + d + 1)/(c - d + 1))) + a^3*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^3*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^3*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))) + 6*b*x*polylog(3, sqrt(-(c + d + 1)/(c - d + 1))*(cosh(b*x + a) 
+ sinh(b*x + a))) + 6*b*x*polylog(3, -sqrt(-(c + d + 1)/(c - d + 1))*(cosh 
(b*x + a) + sinh(b*x + a))) - 6*b*x*polylog(3, sqrt(-(c + d - 1)/(c - d - 
1))*(cosh(b*x + a) + sinh(b*x + a))) - 6*b*x*polylog(3, -sqrt(-(c + d - 1) 
/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) - (b^3*x^3 + a^3)*log(sqrt( 
-(c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b^3*x^3 
+ a^3)*log(-sqrt(-(c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) 
 + 1) + (b^3*x^3 + a^3)*log(sqrt(-(c + d - 1)/(c - d - 1))*(cosh(b*x + ...
 

Sympy [F]

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

integrate(x**2*atanh(c+d*tanh(b*x+a)),x)
 

Output:

Integral(x**2*atanh(c + d*tanh(a + b*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.92 \[ \int x^2 \text {arctanh}(c+d \tanh (a+b x)) \, dx=\frac {1}{3} \, x^{3} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + c\right ) - \frac {1}{18} \, b d {\left (\frac {4 \, b^{3} x^{3} \log \left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - 6 \, b x {\rm Li}_{3}(-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}) + 3 \, {\rm Li}_{4}(-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{4} d} - \frac {4 \, b^{3} x^{3} \log \left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - 6 \, b x {\rm Li}_{3}(-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}) + 3 \, {\rm Li}_{4}(-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{4} d}\right )} \] Input:

integrate(x^2*arctanh(c+d*tanh(b*x+a)),x, algorithm="maxima")
 

Output:

1/3*x^3*arctanh(d*tanh(b*x + a) + c) - 1/18*b*d*((4*b^3*x^3*log((c + d + 1 
)*e^(2*b*x + 2*a)/(c - d + 1) + 1) + 6*b^2*x^2*dilog(-(c + d + 1)*e^(2*b*x 
 + 2*a)/(c - d + 1)) - 6*b*x*polylog(3, -(c + d + 1)*e^(2*b*x + 2*a)/(c - 
d + 1)) + 3*polylog(4, -(c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1)))/(b^4*d) 
- (4*b^3*x^3*log((c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1) + 1) + 6*b^2*x^2* 
dilog(-(c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1)) - 6*b*x*polylog(3, -(c + d 
 - 1)*e^(2*b*x + 2*a)/(c - d - 1)) + 3*polylog(4, -(c + d - 1)*e^(2*b*x + 
2*a)/(c - d - 1)))/(b^4*d))
 

Giac [F]

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

integrate(x^2*arctanh(c+d*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x^2*arctanh(d*tanh(b*x + a) + c), x)
 

Mupad [F(-1)]

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

int(x^2*atanh(c + d*tanh(a + b*x)),x)
 

Output:

int(x^2*atanh(c + d*tanh(a + b*x)), x)
                                                                                    
                                                                                    
 

Reduce [F]

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

int(x^2*atanh(c+d*tanh(b*x+a)),x)
 

Output:

int(atanh(tanh(a + b*x)*d + c)*x**2,x)