3.221 \(\int x^3 \coth ^{-1}(1+d+d \coth (a+b x)) \, dx\)

Optimal. Leaf size=152 \[ \frac{3 x^2 \text{PolyLog}\left (3,(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{PolyLog}\left (4,(d+1) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \text{PolyLog}\left (5,(d+1) e^{2 a+2 b x}\right )}{16 b^4}-\frac{x^3 \text{PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{8} x^4 \log \left (1-(d+1) e^{2 a+2 b x}\right )+\frac{1}{4} x^4 \coth ^{-1}(d \coth (a+b x)+d+1)+\frac{b x^5}{20} \]

[Out]

(b*x^5)/20 + (x^4*ArcCoth[1 + d + d*Coth[a + b*x]])/4 - (x^4*Log[1 - (1 + d)*E^(2*a + 2*b*x)])/8 - (x^3*PolyLo
g[2, (1 + d)*E^(2*a + 2*b*x)])/(4*b) + (3*x^2*PolyLog[3, (1 + d)*E^(2*a + 2*b*x)])/(8*b^2) - (3*x*PolyLog[4, (
1 + d)*E^(2*a + 2*b*x)])/(8*b^3) + (3*PolyLog[5, (1 + d)*E^(2*a + 2*b*x)])/(16*b^4)

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Rubi [A]  time = 0.30751, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6242, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 x^2 \text{PolyLog}\left (3,(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{PolyLog}\left (4,(d+1) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \text{PolyLog}\left (5,(d+1) e^{2 a+2 b x}\right )}{16 b^4}-\frac{x^3 \text{PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{8} x^4 \log \left (1-(d+1) e^{2 a+2 b x}\right )+\frac{1}{4} x^4 \coth ^{-1}(d \coth (a+b x)+d+1)+\frac{b x^5}{20} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCoth[1 + d + d*Coth[a + b*x]],x]

[Out]

(b*x^5)/20 + (x^4*ArcCoth[1 + d + d*Coth[a + b*x]])/4 - (x^4*Log[1 - (1 + d)*E^(2*a + 2*b*x)])/8 - (x^3*PolyLo
g[2, (1 + d)*E^(2*a + 2*b*x)])/(4*b) + (3*x^2*PolyLog[3, (1 + d)*E^(2*a + 2*b*x)])/(8*b^2) - (3*x*PolyLog[4, (
1 + d)*E^(2*a + 2*b*x)])/(8*b^3) + (3*PolyLog[5, (1 + d)*E^(2*a + 2*b*x)])/(16*b^4)

Rule 6242

Int[ArcCoth[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*ArcCoth[c + d*Coth[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^3 \coth ^{-1}(1+d+d \coth (a+b x)) \, dx &=\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))+\frac{1}{4} b \int \frac{x^4}{1+(-1-d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))+\frac{1}{4} (b (1+d)) \int \frac{e^{2 a+2 b x} x^4}{1+(-1-d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )+\frac{1}{2} \int x^3 \log \left (1+(-1-d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 \int x^2 \text{Li}_2\left (-(-1-d) e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 \int x \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \int \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right ) \, dx}{8 b^3}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_4((1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \coth ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \text{Li}_5\left ((1+d) e^{2 a+2 b x}\right )}{16 b^4}\\ \end{align*}

Mathematica [A]  time = 0.174493, size = 141, normalized size = 0.93 \[ \frac{1}{16} \left (\frac{6 x^2 \text{PolyLog}\left (3,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^2}+\frac{6 x \text{PolyLog}\left (4,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^3}+\frac{3 \text{PolyLog}\left (5,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^4}+\frac{4 x^3 \text{PolyLog}\left (2,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b}-2 x^4 \log \left (1-\frac{e^{-2 (a+b x)}}{d+1}\right )+4 x^4 \coth ^{-1}(d \coth (a+b x)+d+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCoth[1 + d + d*Coth[a + b*x]],x]

[Out]

(4*x^4*ArcCoth[1 + d + d*Coth[a + b*x]] - 2*x^4*Log[1 - 1/((1 + d)*E^(2*(a + b*x)))] + (4*x^3*PolyLog[2, 1/((1
+ d)*E^(2*(a + b*x)))])/b + (6*x^2*PolyLog[3, 1/((1 + d)*E^(2*(a + b*x)))])/b^2 + (6*x*PolyLog[4, 1/((1 + d)*
E^(2*(a + b*x)))])/b^3 + (3*PolyLog[5, 1/((1 + d)*E^(2*(a + b*x)))])/b^4)/16

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Maple [C]  time = 32.161, size = 1698, normalized size = 11.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(1+d+d*coth(b*x+a)),x)

[Out]

-1/8/b^4*d*a^4/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)-1)+1/2/b^3*a^3/(1+d)*ln(1-exp(b*x+a)*(1+d)^(1/2))*x+1/
2/b^3*a^3/(1+d)*ln(1+exp(b*x+a)*(1+d)^(1/2))*x-1/2/b^3*a^3/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a))*x-1/4/b*d/(1+d)*po
lylog(2,(1+d)*exp(2*b*x+2*a))*x^3+3/8/b^2*d/(1+d)*polylog(3,(1+d)*exp(2*b*x+2*a))*x^2-3/8/b^3*d/(1+d)*polylog(
4,(1+d)*exp(2*b*x+2*a))*x+1/2/b^4*d*a^3/(1+d)*dilog(1-exp(b*x+a)*(1+d)^(1/2))+1/2/b^4*d*a^3/(1+d)*dilog(1+exp(
b*x+a)*(1+d)^(1/2))+1/2/b^4*d*a^4/(1+d)*ln(1+exp(b*x+a)*(1+d)^(1/2))-3/8/b^4*d*a^4/(1+d)*ln(1-(1+d)*exp(2*b*x+
2*a))-1/4/b^4*d*a^3/(1+d)*polylog(2,(1+d)*exp(2*b*x+2*a))-1/16*I*x^4*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+1/16*I*x^4*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+1/16*I*x^4*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))^2-1/16*I*x^4*Pi*csgn(I*d)*csgn(I*d/(exp(2*b*x+2*a)-1)*exp(2*b*x+2*a))^
2-1/16*I*x^4*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)*exp(2*b*x+2*a))^2+1/2/b^
3*d*a^3/(1+d)*ln(1+exp(b*x+a)*(1+d)^(1/2))*x-1/2/b^3*d*a^3/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a))*x+1/2/b^3*d*a^3/(1
+d)*ln(1-exp(b*x+a)*(1+d)^(1/2))*x+1/2/b^4*d*a^4/(1+d)*ln(1-exp(b*x+a)*(1+d)^(1/2))-1/16*I*x^4*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+1/16*I*x^4*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))
-1/8*I*x^4*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+1/16*I*x^4*Pi*csgn(I*exp(2*b*x+2*a))^3+1/20*b*x^5+1/
16*I*x^4*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)-1))^3+1/8*x^4*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)-1)-1/8/b^4*
a^4/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)-1)+3/16/b^4/(1+d)*polylog(5,(1+d)*exp(2*b*x+2*a))-1/8/(1+d)*ln(1-
(1+d)*exp(2*b*x+2*a))*x^4-1/8*d/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a))*x^4-1/4/b^4*a^3/(1+d)*polylog(2,(1+d)*exp(2*b
*x+2*a))-1/4/b/(1+d)*polylog(2,(1+d)*exp(2*b*x+2*a))*x^3+3/8/b^2/(1+d)*polylog(3,(1+d)*exp(2*b*x+2*a))*x^2-3/8
/b^3/(1+d)*polylog(4,(1+d)*exp(2*b*x+2*a))*x+1/2/b^4*a^3/(1+d)*dilog(1-exp(b*x+a)*(1+d)^(1/2))+1/2/b^4*a^3/(1+
d)*dilog(1+exp(b*x+a)*(1+d)^(1/2))+3/16/b^4*d/(1+d)*polylog(5,(1+d)*exp(2*b*x+2*a))+1/2/b^4*a^4/(1+d)*ln(1-exp
(b*x+a)*(1+d)^(1/2))+1/2/b^4*a^4/(1+d)*ln(1+exp(b*x+a)*(1+d)^(1/2))-3/8/b^4*a^4/(1+d)*ln(1-(1+d)*exp(2*b*x+2*a
))+1/16*I*x^4*Pi*csgn(I*d/(exp(2*b*x+2*a)-1)*exp(2*b*x+2*a))^3-1/4*x^4*ln(exp(b*x+a))-1/8*x^4*ln(d)-1/16*I*x^4
*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+1/16*I*x^4*Pi*csgn(I*exp(2*b*x+2*a))*cs
gn(I/(exp(2*b*x+2*a)-1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)-1))+1/16*I*x^4*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))-1/16*I*x^4*Pi*csgn(I/(exp(2*b*x+2*a)-1))*cs
gn(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))

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Maxima [A]  time = 3.42852, size = 197, normalized size = 1.3 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arcoth}\left (d \coth \left (b x + a\right ) + d + 1\right ) + \frac{1}{40} \,{\left (\frac{2 \, x^{5}}{d} - \frac{5 \,{\left (2 \, b^{4} x^{4} \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3}{\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2}{\rm Li}_{3}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x{\rm Li}_{4}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \,{\rm Li}_{5}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(1+d+d*coth(b*x+a)),x, algorithm="maxima")

[Out]

1/4*x^4*arccoth(d*coth(b*x + a) + d + 1) + 1/40*(2*x^5/d - 5*(2*b^4*x^4*log(-(d + 1)*e^(2*b*x + 2*a) + 1) + 4*
b^3*x^3*dilog((d + 1)*e^(2*b*x + 2*a)) - 6*b^2*x^2*polylog(3, (d + 1)*e^(2*b*x + 2*a)) + 6*b*x*polylog(4, (d +
1)*e^(2*b*x + 2*a)) - 3*polylog(5, (d + 1)*e^(2*b*x + 2*a)))/(b^5*d))*b*d

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Fricas [C]  time = 2.26028, size = 1280, normalized size = 8.42 \begin{align*} \frac{2 \, b^{5} x^{5} + 5 \, b^{4} x^{4} \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 ) - 20 \, b^{3} x^{3}{\rm Li}_2\left (\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 20 \, b^{3} x^{3}{\rm Li}_2\left (-\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \, a^{4} \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 ) - 5 \, a^{4} \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 ) + 60 \, b^{2} x^{2}{\rm polylog}\left (3, \sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 60 \, b^{2} x^{2}{\rm polylog}\left (3, -\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x{\rm polylog}\left (4, \sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x{\rm polylog}\left (4, -\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \,{\left (b^{4} x^{4} - a^{4}\right )} \log \left (\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 5 \,{\left (b^{4} x^{4} - a^{4}\right )} \log \left (-\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 120 \,{\rm polylog}\left (5, \sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \,{\rm polylog}\left (5, -\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{40 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(1+d+d*coth(b*x+a)),x, algorithm="fricas")

[Out]

1/40*(2*b^5*x^5 + 5*b^4*x^4*log((d*cosh(b*x + a) + (d + 2)*sinh(b*x + a))/(d*cosh(b*x + a) + d*sinh(b*x + a)))
- 20*b^3*x^3*dilog(sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))) - 20*b^3*x^3*dilog(-sqrt(d + 1)*(cosh(b*x + a
) + sinh(b*x + a))) - 5*a^4*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) + 2*sqrt(d + 1)) - 5*a^4*log
(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) - 2*sqrt(d + 1)) + 60*b^2*x^2*polylog(3, sqrt(d + 1)*(cosh(
b*x + a) + sinh(b*x + a))) + 60*b^2*x^2*polylog(3, -sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))) - 120*b*x*pol
ylog(4, sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))) - 120*b*x*polylog(4, -sqrt(d + 1)*(cosh(b*x + a) + sinh(b
*x + a))) - 5*(b^4*x^4 - a^4)*log(sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 5*(b^4*x^4 - a^4)*log(-sq
rt(d + 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + 120*polylog(5, sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))) +
120*polylog(5, -sqrt(d + 1)*(cosh(b*x + a) + sinh(b*x + a))))/b^4

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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**3*acoth(1+d+d*coth(b*x+a)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arcoth}\left (d \coth \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^3*arccoth(1+d+d*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^3*arccoth(d*coth(b*x + a) + d + 1), x)