### 3.377 $$\int x^3 \sinh ^2(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=185 $-\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}-\frac{3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}-\frac{3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}-\frac{x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x^3 \sinh ^2(a+b x)}{2 b}+\frac{3 x}{8 b^3}+\frac{x^3}{4 b}+\frac{x^4}{4}$

[Out]

(3*x)/(8*b^3) + x^3/(4*b) + x^4/4 - (x^3*Log[1 + E^(2*(a + b*x))])/b - (3*x^2*PolyLog[2, -E^(2*(a + b*x))])/(2
*b^2) + (3*x*PolyLog[3, -E^(2*(a + b*x))])/(2*b^3) - (3*PolyLog[4, -E^(2*(a + b*x))])/(4*b^4) - (3*Cosh[a + b*
x]*Sinh[a + b*x])/(8*b^4) - (3*x^2*Cosh[a + b*x]*Sinh[a + b*x])/(4*b^2) + (3*x*Sinh[a + b*x]^2)/(4*b^3) + (x^3
*Sinh[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.246804, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 12, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.667, Rules used = {5449, 5372, 3311, 30, 2635, 8, 3718, 2190, 2531, 6609, 2282, 6589} $-\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}-\frac{3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}-\frac{3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}-\frac{x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x^3 \sinh ^2(a+b x)}{2 b}+\frac{3 x}{8 b^3}+\frac{x^3}{4 b}+\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Sinh[a + b*x]^2*Tanh[a + b*x],x]

[Out]

(3*x)/(8*b^3) + x^3/(4*b) + x^4/4 - (x^3*Log[1 + E^(2*(a + b*x))])/b - (3*x^2*PolyLog[2, -E^(2*(a + b*x))])/(2
*b^2) + (3*x*PolyLog[3, -E^(2*(a + b*x))])/(2*b^3) - (3*PolyLog[4, -E^(2*(a + b*x))])/(4*b^4) - (3*Cosh[a + b*
x]*Sinh[a + b*x])/(8*b^4) - (3*x^2*Cosh[a + b*x]*Sinh[a + b*x])/(4*b^2) + (3*x*Sinh[a + b*x]^2)/(4*b^3) + (x^3
*Sinh[a + b*x]^2)/(2*b)

Rule 5449

Int[((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int
[(c + d*x)^m*Sinh[a + b*x]^n*Tanh[a + b*x]^(p - 2), x] - Int[(c + d*x)^m*Sinh[a + b*x]^(n - 2)*Tanh[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, 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 \sinh ^2(a+b x) \tanh (a+b x) \, dx &=\int x^3 \cosh (a+b x) \sinh (a+b x) \, dx-\int x^3 \tanh (a+b x) \, dx\\ &=\frac{x^4}{4}+\frac{x^3 \sinh ^2(a+b x)}{2 b}-2 \int \frac{e^{2 (a+b x)} x^3}{1+e^{2 (a+b x)}} \, dx-\frac{3 \int x^2 \sinh ^2(a+b x) \, dx}{2 b}\\ &=\frac{x^4}{4}-\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}-\frac{3 \int \sinh ^2(a+b x) \, dx}{4 b^3}+\frac{3 \int x^2 \, dx}{4 b}+\frac{3 \int x^2 \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{x^3}{4 b}+\frac{x^4}{4}-\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}+\frac{3 \int 1 \, dx}{8 b^3}+\frac{3 \int x \text{Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{3 x}{8 b^3}+\frac{x^3}{4 b}+\frac{x^4}{4}-\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}-\frac{3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}-\frac{3 \int \text{Li}_3\left (-e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac{3 x}{8 b^3}+\frac{x^3}{4 b}+\frac{x^4}{4}-\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}-\frac{3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}-\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=\frac{3 x}{8 b^3}+\frac{x^3}{4 b}+\frac{x^4}{4}-\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}-\frac{3 \text{Li}_4\left (-e^{2 (a+b x)}\right )}{4 b^4}-\frac{3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac{3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{3 x \sinh ^2(a+b x)}{4 b^3}+\frac{x^3 \sinh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 2.86173, size = 191, normalized size = 1.03 $\frac{1}{16} \left (\frac{12 \left (2 b^2 x^2 \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+2 b x \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )+\text{PolyLog}\left (4,-e^{-2 (a+b x)}\right )\right )}{b^4}+\frac{\cosh (2 b x) \left (2 b x \cosh (2 a) \left (2 b^2 x^2+3\right )-3 \sinh (2 a) \left (2 b^2 x^2+1\right )\right )}{b^4}+\frac{\sinh (2 b x) \left (2 b x \sinh (2 a) \left (2 b^2 x^2+3\right )-3 \cosh (2 a) \left (2 b^2 x^2+1\right )\right )}{b^4}-\frac{16 x^3 \log \left (e^{-2 (a+b x)}+1\right )}{b}-\frac{8 x^4}{e^{2 a}+1}-4 x^4 \tanh (a)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Sinh[a + b*x]^2*Tanh[a + b*x],x]

[Out]

((-8*x^4)/(1 + E^(2*a)) - (16*x^3*Log[1 + E^(-2*(a + b*x))])/b + (12*(2*b^2*x^2*PolyLog[2, -E^(-2*(a + b*x))]
+ 2*b*x*PolyLog[3, -E^(-2*(a + b*x))] + PolyLog[4, -E^(-2*(a + b*x))]))/b^4 + (Cosh[2*b*x]*(2*b*x*(3 + 2*b^2*x
^2)*Cosh[2*a] - 3*(1 + 2*b^2*x^2)*Sinh[2*a]))/b^4 + ((-3*(1 + 2*b^2*x^2)*Cosh[2*a] + 2*b*x*(3 + 2*b^2*x^2)*Sin
h[2*a])*Sinh[2*b*x])/b^4 - 4*x^4*Tanh[a])/16

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Maple [A]  time = 0.071, size = 189, normalized size = 1. \begin{align*}{\frac{{x}^{4}}{4}}+{\frac{ \left ( 4\,{x}^{3}{b}^{3}-6\,{x}^{2}{b}^{2}+6\,bx-3 \right ){{\rm e}^{2\,bx+2\,a}}}{32\,{b}^{4}}}+{\frac{ \left ( 4\,{x}^{3}{b}^{3}+6\,{x}^{2}{b}^{2}+6\,bx+3 \right ){{\rm e}^{-2\,bx-2\,a}}}{32\,{b}^{4}}}+2\,{\frac{{a}^{3}x}{{b}^{3}}}+{\frac{3\,{a}^{4}}{2\,{b}^{4}}}-{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{3\,{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}+{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}}-{\frac{3\,{\it polylog} \left ( 4,-{{\rm e}^{2\,bx+2\,a}} \right ) }{4\,{b}^{4}}}-2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sech(b*x+a)*sinh(b*x+a)^3,x)

[Out]

1/4*x^4+1/32*(4*b^3*x^3-6*b^2*x^2+6*b*x-3)/b^4*exp(2*b*x+2*a)+1/32*(4*b^3*x^3+6*b^2*x^2+6*b*x+3)/b^4*exp(-2*b*
x-2*a)+2/b^3*a^3*x+3/2/b^4*a^4-x^3*ln(1+exp(2*b*x+2*a))/b-3/2*x^2*polylog(2,-exp(2*b*x+2*a))/b^2+3/2*x*polylog
(3,-exp(2*b*x+2*a))/b^3-3/4*polylog(4,-exp(2*b*x+2*a))/b^4-2/b^4*a^3*ln(exp(b*x+a))

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Maxima [A]  time = 1.3262, size = 244, normalized size = 1.32 \begin{align*} \frac{1}{2} \, x^{4} - \frac{{\left (8 \, b^{4} x^{4} e^{\left (2 \, a\right )} -{\left (4 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 6 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} -{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{32 \, b^{4}} - \frac{4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \,{\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sech(b*x+a)*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/2*x^4 - 1/32*(8*b^4*x^4*e^(2*a) - (4*b^3*x^3*e^(4*a) - 6*b^2*x^2*e^(4*a) + 6*b*x*e^(4*a) - 3*e^(4*a))*e^(2*b
*x) - (4*b^3*x^3 + 6*b^2*x^2 + 6*b*x + 3)*e^(-2*b*x))*e^(-2*a)/b^4 - 1/3*(4*b^3*x^3*log(e^(2*b*x + 2*a) + 1) +
6*b^2*x^2*dilog(-e^(2*b*x + 2*a)) - 6*b*x*polylog(3, -e^(2*b*x + 2*a)) + 3*polylog(4, -e^(2*b*x + 2*a)))/b^4

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Fricas [C]  time = 2.46974, size = 2568, normalized size = 13.88 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sech(b*x+a)*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/32*(4*b^3*x^3 + (4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*cosh(b*x + a)^4 + 4*(4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*
cosh(b*x + a)*sinh(b*x + a)^3 + (4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*sinh(b*x + a)^4 + 6*b^2*x^2 + 8*(b^4*x^4 -
2*a^4)*cosh(b*x + a)^2 + 2*(4*b^4*x^4 - 8*a^4 + 3*(4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*cosh(b*x + a)^2)*sinh(b
*x + a)^2 + 6*b*x - 96*(b^2*x^2*cosh(b*x + a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + b^2*x^2*sinh(b*x + a
)^2)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 96*(b^2*x^2*cosh(b*x + a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x
+ a) + b^2*x^2*sinh(b*x + a)^2)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 32*(a^3*cosh(b*x + a)^2 + 2*a^3*c
osh(b*x + a)*sinh(b*x + a) + a^3*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) + I) + 32*(a^3*cosh(b*x +
a)^2 + 2*a^3*cosh(b*x + a)*sinh(b*x + a) + a^3*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) - 32*((
b^3*x^3 + a^3)*cosh(b*x + a)^2 + 2*(b^3*x^3 + a^3)*cosh(b*x + a)*sinh(b*x + a) + (b^3*x^3 + a^3)*sinh(b*x + a)
^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 32*((b^3*x^3 + a^3)*cosh(b*x + a)^2 + 2*(b^3*x^3 + a^3)*cosh(
b*x + a)*sinh(b*x + a) + (b^3*x^3 + a^3)*sinh(b*x + a)^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - 192*(c
osh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*polylog(4, I*cosh(b*x + a) + I*sinh(b*x + a)
) - 192*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*polylog(4, -I*cosh(b*x + a) - I*si
nh(b*x + a)) + 192*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2)*polylog(3,
I*cosh(b*x + a) + I*sinh(b*x + a)) + 192*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b
*x + a)^2)*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)) + 4*((4*b^3*x^3 - 6*b^2*x^2 + 6*b*x - 3)*cosh(b*x +
a)^3 + 4*(b^4*x^4 - 2*a^4)*cosh(b*x + a))*sinh(b*x + a) + 3)/(b^4*cosh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b
*x + a) + b^4*sinh(b*x + a)^2)

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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*sech(b*x+a)*sinh(b*x+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sech(b*x+a)*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^3*sech(b*x + a)*sinh(b*x + a)^3, x)