3.208 \(\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx\)
Optimal. Leaf size=128 \[ \frac{x \text{PolyLog}\left (3,-(d+1) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,-(d+1) e^{2 a+2 b x}\right )}{8 b^3}-\frac{x^2 \text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{6} x^3 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac{1}{3} x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^4}{12} \]
[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)
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Rubi [A] time = 0.269003, antiderivative size = 128, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used =
{6240, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac{x \text{PolyLog}\left (3,-(d+1) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,-(d+1) e^{2 a+2 b x}\right )}{8 b^3}-\frac{x^2 \text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{6} x^3 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac{1}{3} x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^4}{12} \]
Antiderivative was successfully verified.
[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 6240
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 2184
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 2190
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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2531
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 6609
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]
Rule 2282
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 6589
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]
Rubi steps
\begin{align*} \int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx &=\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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{\int x \text{Li}_2\left (-(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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{x \text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\int \text{Li}_3\left ((-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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{x \text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_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 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{x \text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\text{Li}_4\left (-(1+d) e^{2 a+2 b x}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 0.113944, size = 118, normalized size = 0.92 \[ \frac{1}{24} \left (\frac{6 x \text{PolyLog}\left (3,-\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^2}+\frac{3 \text{PolyLog}\left (4,-\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^3}+\frac{6 x^2 \text{PolyLog}\left (2,-\frac{e^{-2 (a+b x)}}{d+1}\right )}{b}-4 x^3 \log \left (\frac{e^{-2 (a+b x)}}{d+1}+1\right )+8 x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^2*ArcCoth[1 + d + d*Tanh[a + b*x]],x]
[Out]
(8*x^3*ArcCoth[1 + d + d*Tanh[a + b*x]] - 4*x^3*Log[1 + 1/((1 + d)*E^(2*(a + b*x)))] + (6*x^2*PolyLog[2, -(1/(
(1 + d)*E^(2*(a + b*x))))])/b + (6*x*PolyLog[3, -(1/((1 + d)*E^(2*(a + b*x))))])/b^2 + (3*PolyLog[4, -(1/((1 +
d)*E^(2*(a + b*x))))])/b^3)/24
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Maple [C] time = 37.152, size = 1667, normalized size = 13. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*arccoth(1+d+d*tanh(b*x+a)),x)
[Out]
-1/3*x^3*ln(exp(b*x+a))-1/6*x^3*ln(d)-1/4/b*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*x^2-1/6*I*x^3*Pi*csgn(I*e
xp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/12*I*x^3*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))^2-1/2/b^3*a^3/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))-1/8/b^3*d/(1+d)*polylog(4,-(1+d)*exp(2*b*x+2*a))-1/
6*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^3-1/2/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))+1/3/b^3/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^3-1/4/b/(1+d)*polylog(2,-(1+d)*exp
(2*b*x+2*a))*x^2+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-1/2/b^3*a^3/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))+1/4/b^3*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*a^2+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)*ln(1+exp(b*x+a)*(-d-1)^(1/2))*x-1/2/b^2*a^2
/(1+d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))*x-1/2/b^3*d*a^3/(1+d)*ln(1+exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*d*a^3/(1+d)*l
n(1-exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*d*a^2/(1+d)*dilog(1+exp(b*x+a)*(-d-1)^(1/2))-1/2/b^3*d*a^2/(1+d)*dilog(1-
exp(b*x+a)*(-d-1)^(1/2))+1/2/b^2/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x*a^2+1/3/b^3*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2
*a))*a^3+1/12*b*x^4+1/6/b^3*a^3/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)+1/6/b^3*d*a^3/(1+d)*ln(exp(2*b*x+2
*a)*d+exp(2*b*x+2*a)+1)-1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*csgn(I*d/(exp(2*b*x+2*a)+1)*ex
p(2*b*x+2*a))^2-1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/12*I*x^3*Pi
*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))^2-1/2/b^2*d*a^2/(1+
d)*ln(1-exp(b*x+a)*(-d-1)^(1/2))*x+1/2/b^2*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x*a^2-1/2/b^2*d*a^2/(1+d)*ln(1+e
xp(b*x+a)*(-d-1)^(1/2))*x-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+1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a))^3+1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/12*I*x^
3*Pi*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))^3+1/12*I*x^3*Pi*csgn(I*exp(b*x+a))^2*csgn(
I*exp(2*b*x+2*a))+1/12*I*x^3*Pi*csgn(I*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b
*x+2*a)*d+exp(2*b*x+2*a)+1))^2-1/12*I*x^3*Pi*csgn(I*d)*csgn(I*d/(exp(2*b*x+2*a)+1)*exp(2*b*x+2*a))^2+1/6*x^3*l
n(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)+1/12*I*x^3*Pi*csgn(I*d/(exp(2*b*x+2*a)+1)*exp(2*b*x+2*a))^3-1/12*I*x^3*Pi
*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2
*a)*d+exp(2*b*x+2*a)+1))+1/12*I*x^3*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))+1/12*I*x^3*Pi*csgn(I*d)*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*csgn(I*d/(exp(2*b*x+2*a
)+1)*exp(2*b*x+2*a))
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Maxima [A] time = 3.46662, size = 169, normalized size = 1.32 \begin{align*} \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 \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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Fricas [C] time = 1.76501, size = 1160, normalized size = 9.06 \begin{align*} \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*acoth(1+d+d*tanh(b*x+a)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)