3.3.7 \(\int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx\) [207]

Optimal. Leaf size=155 \[ \frac {b x^5}{20}+\frac {1}{4} x^4 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{8} x^4 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^3 \text {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {3 x^2 \text {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {3 x \text {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 \text {PolyLog}\left (5,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{16 b^4} \]

[Out]

1/20*b*x^5+1/4*x^4*arccoth(1+d+d*tanh(b*x+a))-1/8*x^4*ln(1+(1+d)*exp(2*b*x+2*a))-1/4*x^3*polylog(2,-(1+d)*exp(
2*b*x+2*a))/b+3/8*x^2*polylog(3,-(1+d)*exp(2*b*x+2*a))/b^2-3/8*x*polylog(4,-(1+d)*exp(2*b*x+2*a))/b^3+3/16*pol
ylog(5,-(1+d)*exp(2*b*x+2*a))/b^4

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Rubi [A]
time = 0.21, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 = {6375, 2215, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {3 \text {Li}_5\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{16 b^4}-\frac {3 x \text {Li}_4\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {3 x^2 \text {Li}_3\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {x^3 \text {Li}_2\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{8} x^4 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac {1}{4} x^4 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac {b x^5}{20} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^5)/20 + (x^4*ArcCoth[1 + d + d*Tanh[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*PolyLo
g[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 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*} \int x^3 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx &=\frac {1}{4} x^4 \coth ^{-1}(1+d+d \tanh (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 \tanh (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 \tanh (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 \tanh (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 \tanh (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 \tanh (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 \tanh (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 {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 \tanh (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*}

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Mathematica [A]
time = 0.13, size = 144, normalized size = 0.93 \begin {gather*} \frac {1}{16} \left (4 x^4 \coth ^{-1}(1+d+d \tanh (a+b x))-2 x^4 \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+\frac {4 x^3 \text {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b}+\frac {6 x^2 \text {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b^2}+\frac {6 x \text {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b^3}+\frac {3 \text {PolyLog}\left (5,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^4*ArcCoth[1 + d + d*Tanh[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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.72, size = 1726, normalized size = 11.14

method	result	size
risch	\(\text {Expression too large to display}\)	\(1726\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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+1/20
*b*x^5-1/8/b^4*d*a^4/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)+1/16*I*x^4*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*
x+2*a)+1))*csgn(I*d)*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/16*I*x^4*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))-1/16*I*x^4*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))-1/16*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(2*b*x+2*a))^3+1/1
6*I*x^4*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+1/16*I*x^4*Pi*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1
))^3+1/2/b^4*d*a^4/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))+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)*poly
log(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^2-1/8/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+1/2/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))-1/4/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-3/8
/b^3*d/(1+d)*polylog(4,-(1+d)*exp(2*b*x+2*a))*x-1/4/b^4*d*a^3/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))-3/8/b^4*d
*a^4/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))+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))+1/16*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)+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)/(exp(2*b*x+2*a)+1))*csgn(I*d*exp(2*b*x+2*a)/(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)/(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+1/2/b^4*d*a^4/(1+d)*l
n(1+exp(b*x+a)*(-1-d)^(1/2))-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^3*d*a^3/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))*x-1/16*I*x^4*Pi*csgn(I*exp(2*b*x+2*a))*c
sgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/8*x^4*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)-1/4*x^4*ln(exp(b*x+a)
)-1/8*x^4*ln(d)

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Maxima [A]
time = 0.69, size = 149, normalized size = 0.96 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (d \tanh \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccoth(d*tanh(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (135) = 270\).
time = 0.39, size = 450, normalized size = 2.90 \begin {gather*} \frac {2 \, b^{5} x^{5} + 5 \, b^{4} x^{4} \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 ) - 20 \, b^{3} x^{3} {\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 ) - 20 \, b^{3} x^{3} {\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 ) - 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 ) + \sqrt {-4 \, d - 4}\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 ) - \sqrt {-4 \, d - 4}\right ) + 60 \, b^{2} x^{2} {\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 ) + 60 \, b^{2} x^{2} {\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 ) - 120 \, b x {\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 ) - 120 \, b x {\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 ) - 5 \, {\left (b^{4} x^{4} - a^{4}\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 ) - 5 \, {\left (b^{4} x^{4} - a^{4}\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 ) + 120 \, {\rm polylog}\left (5, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \, {\rm polylog}\left (5, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{40 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acoth(d + d*tanh(a + b*x) + 1),x)

[Out]

int(x^3*acoth(d + d*tanh(a + b*x) + 1), x)

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