\(\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx\) [208]
Optimal result
Integrand size = 16, antiderivative size = 128 \[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3}
\]
[Out]
1/12*b*x^4+1/3*x^3*arccoth(1+d+d*tanh(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)*exp(2*b*x+2*a))/b^2-1/8*polylog(4,-(1+d)*exp(2*b*x+2*a))/b^3
Rubi [A] (verified)
Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6375, 2215, 2221, 2611, 6744,
2320, 6724} \[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=-\frac {\operatorname {PolyLog}\left (4,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {x \operatorname {PolyLog}\left (3,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{6} x^3 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac {b x^4}{12}
\]
[In]
Int[x^2*ArcCoth[1 + d + d*Tanh[a + b*x]],x]
[Out]
(b*x^4)/12 + (x^3*ArcCoth[1 + d + d*Tanh[a + b*x]])/3 - (x^3*Log[1 + (1 + d)*E^(2*a + 2*b*x)])/6 - (x^2*PolyLo
g[2, -((1 + d)*E^(2*a + 2*b*x))])/(4*b) + (x*PolyLog[3, -((1 + d)*E^(2*a + 2*b*x))])/(4*b^2) - PolyLog[4, -((1
+ d)*E^(2*a + 2*b*x))]/(8*b^3)
Rule 2215
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] - Dist[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 2221
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]
- Dist[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 2320
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2611
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] + Dist[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 6375
Int[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcCoth[c + d*Tanh[a + b*x]]/(f*(m + 1))), x] + Dist[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 6724
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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 6744
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] - Dist[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]
Rubi steps \begin{align*}
\text {integral}& = \frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))+\frac {1}{3} b \int \frac {x^3}{1+(1+d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{3} (b (1+d)) \int \frac {e^{2 a+2 b x} x^3}{1+(1+d) e^{2 a+2 b x}} \, dx \\ & = \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )+\frac {1}{2} \int x^2 \log \left (1+(1+d) e^{2 a+2 b x}\right ) \, dx \\ & = \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (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,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {\int x \operatorname {PolyLog}\left (2,(-1-d) e^{2 a+2 b x}\right ) \, dx}{2 b} \\ & = \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (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,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\int \operatorname {PolyLog}\left (3,(-1-d) e^{2 a+2 b x}\right ) \, dx}{4 b^2} \\ & = \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (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,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,(-1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3} \\ & = \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (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,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95
\[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {8 b^3 x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-4 b^3 x^3 \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )+6 b x \operatorname {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )+3 \operatorname {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{24 b^3}
\]
[In]
Integrate[x^2*ArcCoth[1 + d + d*Tanh[a + b*x]],x]
[Out]
(8*b^3*x^3*ArcCoth[1 + d + d*Tanh[a + b*x]] - 4*b^3*x^3*Log[1 + 1/((1 + d)*E^(2*(a + b*x)))] + 6*b^2*x^2*PolyL
og[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)
Maple [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.90 (sec) , antiderivative size = 1625, normalized size of antiderivative =
12.70
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method | result | size |
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risch |
\(\text {Expression too large to display}\) |
\(1625\) |
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[In]
int(x^2*arccoth(1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)
[Out]
1/6*x^3*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)+1/4/b^2*d/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))*x-1/2/b^2*a^2/(
1+d)*x*ln(1+exp(b*x+a)*(-d-1)^(1/2))-1/2/b^2*a^2/(1+d)*x*ln(1-exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*a^3*d/(1+d)*ln(
1+exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*a^3*d/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*a^2*d/(1+d)*dilog(1+exp(b
*x+a)*(-d-1)^(1/2))-1/2/b^3*a^2*d/(1+d)*dilog(1-exp(b*x+a)*(-d-1)^(1/2))+1/6/b^3*d*a^3/(1+d)*ln(d*exp(2*b*x+2*
a)+exp(2*b*x+2*a)+1)+1/2/b^2/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^2*x-1/4/b*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2
*a))*x^2+1/3/b^3*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^3+1/4/b^3*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*a^2+1
/12*b*x^4-1/3*x^3*ln(exp(b*x+a))+1/2/b^2*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^2*x-1/2/b^2*a^2*d/(1+d)*x*ln(1+e
xp(b*x+a)*(-d-1)^(1/2))-1/2/b^2*a^2*d/(1+d)*x*ln(1-exp(b*x+a)*(-d-1)^(1/2))-1/8/b^3/(1+d)*polylog(4,-(1+d)*exp
(2*b*x+2*a))-1/6/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^3-1/6*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^3-1/8/b^3*d/(1+
d)*polylog(4,-(1+d)*exp(2*b*x+2*a))-1/2/b^3*a^3/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*a^3/(1+d)*ln(1-exp
(b*x+a)*(-d-1)^(1/2))-1/2/b^3*a^2/(1+d)*dilog(1+exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*a^2/(1+d)*dilog(1-exp(b*x+a)*
(-d-1)^(1/2))+1/6/b^3*a^3/(1+d)*ln(d*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)-1/4/b/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2
*a))*x^2+1/3/b^3/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^3+1/4/b^3/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*a^2+1/4/b
^2/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))*x-1/12*(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))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+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)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2
*b*x+2*a)+1))+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*P
i*csgn(I*exp(2*b*x+2*a))*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)+1))*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))-I*Pi*csgn(I*
exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*csgn(I*d)*csgn(I/(exp(2*b*x+2*a)+1)*d*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))^3-I*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3
+I*Pi*csgn(I*d)*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))*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)+1)*d*exp(2*b*x+2*a))^3+2*ln(d))*x
^3
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (111) = 222\).
Time = 0.26 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.98
\[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {b^{4} x^{4} + 2 \, b^{3} x^{3} \log \left (\frac {{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d - 4}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d - 4}\right ) + 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{3}}
\]
[In]
integrate(x^2*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="fricas")
[Out]
1/12*(b^4*x^4 + 2*b^3*x^3*log(((d + 2)*cosh(b*x + a) + d*sinh(b*x + a))/(d*cosh(b*x + a) + d*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)*(co
sh(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*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 12*polylog(4, 1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + si
nh(b*x + a))) - 12*polylog(4, -1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))))/b^3
Sympy [F]
\[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\int x^{2} \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx
\]
[In]
integrate(x**2*acoth(1+d+d*tanh(b*x+a)),x)
[Out]
Integral(x**2*acoth(d*tanh(a + b*x) + d + 1), x)
Maxima [A] (verification not implemented)
none
Time = 0.73 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98
\[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\frac {1}{3} \, x^{3} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) + \frac {1}{36} \, {\left (\frac {3 \, x^{4}}{d} - \frac {2 \, {\left (4 \, b^{3} x^{3} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{4} d}\right )} b d
\]
[In]
integrate(x^2*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="maxima")
[Out]
1/3*x^3*arccoth(d*tanh(b*x + a) + d + 1) + 1/36*(3*x^4/d - 2*(4*b^3*x^3*log((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]
\[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\int { x^{2} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \,d x }
\]
[In]
integrate(x^2*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="giac")
[Out]
integrate(x^2*arccoth(d*tanh(b*x + a) + d + 1), x)
Mupad [F(-1)]
Timed out. \[
\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx=\int x^2\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x
\]
[In]
int(x^2*acoth(d + d*tanh(a + b*x) + 1),x)
[Out]
int(x^2*acoth(d + d*tanh(a + b*x) + 1), x)