### 3.488 $$\int x^3 \text{csch}^2(a+b x) \text{sech}(a+b x) \, dx$$

Optimal. Leaf size=237 $\frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 x \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^3 \text{csch}(a+b x)}{b}$

[Out]

(-2*x^3*ArcTan[E^(a + b*x)])/b - (6*x^2*ArcTanh[E^(a + b*x)])/b^2 - (x^3*Csch[a + b*x])/b - (6*x*PolyLog[2, -E
^(a + b*x)])/b^3 + ((3*I)*x^2*PolyLog[2, (-I)*E^(a + b*x)])/b^2 - ((3*I)*x^2*PolyLog[2, I*E^(a + b*x)])/b^2 +
(6*x*PolyLog[2, E^(a + b*x)])/b^3 + (6*PolyLog[3, -E^(a + b*x)])/b^4 - ((6*I)*x*PolyLog[3, (-I)*E^(a + b*x)])/
b^3 + ((6*I)*x*PolyLog[3, I*E^(a + b*x)])/b^3 - (6*PolyLog[3, E^(a + b*x)])/b^4 + ((6*I)*PolyLog[4, (-I)*E^(a
+ b*x)])/b^4 - ((6*I)*PolyLog[4, I*E^(a + b*x)])/b^4

________________________________________________________________________________________

Rubi [A]  time = 0.343965, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 13, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.722, Rules used = {2621, 321, 207, 5462, 14, 5205, 12, 4180, 2531, 6609, 2282, 6589, 4182} $\frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 x \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^3 \text{csch}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Csch[a + b*x]^2*Sech[a + b*x],x]

[Out]

(-2*x^3*ArcTan[E^(a + b*x)])/b - (6*x^2*ArcTanh[E^(a + b*x)])/b^2 - (x^3*Csch[a + b*x])/b - (6*x*PolyLog[2, -E
^(a + b*x)])/b^3 + ((3*I)*x^2*PolyLog[2, (-I)*E^(a + b*x)])/b^2 - ((3*I)*x^2*PolyLog[2, I*E^(a + b*x)])/b^2 +
(6*x*PolyLog[2, E^(a + b*x)])/b^3 + (6*PolyLog[3, -E^(a + b*x)])/b^4 - ((6*I)*x*PolyLog[3, (-I)*E^(a + b*x)])/
b^3 + ((6*I)*x*PolyLog[3, I*E^(a + b*x)])/b^3 - (6*PolyLog[3, E^(a + b*x)])/b^4 + ((6*I)*PolyLog[4, (-I)*E^(a
+ b*x)])/b^4 - ((6*I)*PolyLog[4, I*E^(a + b*x)])/b^4

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 5462

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Wit
h[{u = IntHide[Csch[a + b*x]^n*Sech[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)
*u, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5205

Int[((a_.) + ArcTan[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 + u^2), x], x]
, x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m +
1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rubi steps

\begin{align*} \int x^3 \text{csch}^2(a+b x) \text{sech}(a+b x) \, dx &=-\frac{x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^3 \text{csch}(a+b x)}{b}-3 \int x^2 \left (-\frac{\tan ^{-1}(\sinh (a+b x))}{b}-\frac{\text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^3 \text{csch}(a+b x)}{b}-3 \int \left (-\frac{x^2 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^2 \text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^3 \text{csch}(a+b x)}{b}+\frac{3 \int x^2 \tan ^{-1}(\sinh (a+b x)) \, dx}{b}+\frac{3 \int x^2 \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac{6 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b^2}-\frac{\int b x^3 \text{sech}(a+b x) \, dx}{b}\\ &=-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \int \text{Li}_2\left (-e^{a+b x}\right ) \, dx}{b^3}-\frac{6 \int \text{Li}_2\left (e^{a+b x}\right ) \, dx}{b^3}-\int x^3 \text{sech}(a+b x) \, dx\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac{(3 i) \int x^2 \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{(3 i) \int x^2 \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac{6 \text{Li}_3\left (e^{a+b x}\right )}{b^4}-\frac{(6 i) \int x \text{Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}+\frac{(6 i) \int x \text{Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac{6 \text{Li}_3\left (e^{a+b x}\right )}{b^4}+\frac{(6 i) \int \text{Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}-\frac{(6 i) \int \text{Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac{6 \text{Li}_3\left (e^{a+b x}\right )}{b^4}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}-\frac{6 \text{Li}_3\left (e^{a+b x}\right )}{b^4}+\frac{6 i \text{Li}_4\left (-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{Li}_4\left (i e^{a+b x}\right )}{b^4}\\ \end{align*}

Mathematica [A]  time = 1.76773, size = 333, normalized size = 1.41 $\frac{-2 i \left (-3 b^2 x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )+3 b^2 x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )+6 b x \text{PolyLog}\left (3,-i e^{a+b x}\right )-6 b x \text{PolyLog}\left (3,i e^{a+b x}\right )-6 \text{PolyLog}\left (4,-i e^{a+b x}\right )+6 \text{PolyLog}\left (4,i e^{a+b x}\right )+b^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 x^3 \log \left (1+i e^{a+b x}\right )\right )-12 \left (b x \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))-b x \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))-\text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))+\text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))+b^2 x^2 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))\right )-2 b^3 x^3 \text{csch}(a)+b^3 x^3 \text{csch}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{csch}\left (\frac{1}{2} (a+b x)\right )+b^3 x^3 \text{sech}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right )}{2 b^4}$

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Csch[a + b*x]^2*Sech[a + b*x],x]

[Out]

(-2*b^3*x^3*Csch[a] - 12*(b^2*x^2*ArcTanh[Cosh[a + b*x] + Sinh[a + b*x]] + b*x*PolyLog[2, -Cosh[a + b*x] - Sin
h[a + b*x]] - b*x*PolyLog[2, Cosh[a + b*x] + Sinh[a + b*x]] - PolyLog[3, -Cosh[a + b*x] - Sinh[a + b*x]] + Pol
yLog[3, Cosh[a + b*x] + Sinh[a + b*x]]) - (2*I)*(b^3*x^3*Log[1 - I*E^(a + b*x)] - b^3*x^3*Log[1 + I*E^(a + b*x
)] - 3*b^2*x^2*PolyLog[2, (-I)*E^(a + b*x)] + 3*b^2*x^2*PolyLog[2, I*E^(a + b*x)] + 6*b*x*PolyLog[3, (-I)*E^(a
+ b*x)] - 6*b*x*PolyLog[3, I*E^(a + b*x)] - 6*PolyLog[4, (-I)*E^(a + b*x)] + 6*PolyLog[4, I*E^(a + b*x)]) + b
^3*x^3*Csch[a/2]*Csch[(a + b*x)/2]*Sinh[(b*x)/2] + b^3*x^3*Sech[a/2]*Sech[(a + b*x)/2]*Sinh[(b*x)/2])/(2*b^4)

________________________________________________________________________________________

Maple [F]  time = 0.454, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ({\rm csch} \left (bx+a\right ) \right ) ^{2}{\rm sech} \left (bx+a\right )\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*csch(b*x+a)^2*sech(b*x+a),x)

[Out]

int(x^3*csch(b*x+a)^2*sech(b*x+a),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac{3 \,{\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac{3 \,{\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} - 8 \, \int \frac{x^{3} e^{\left (b x + a\right )}}{4 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(b*x+a)^2*sech(b*x+a),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [C]  time = 2.80188, size = 3559, normalized size = 15.02 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(b*x+a)^2*sech(b*x+a),x, algorithm="fricas")

[Out]

-(2*b^3*x^3*cosh(b*x + a) + 2*b^3*x^3*sinh(b*x + a) - 6*(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 - b*x)*dilog(cosh(b*x + a) + sinh(b*x + a)) - (-3*I*b^2*x^2*cosh(b*x + a)^2 - 6*I*b^2
*x^2*cosh(b*x + a)*sinh(b*x + a) - 3*I*b^2*x^2*sinh(b*x + a)^2 + 3*I*b^2*x^2)*dilog(I*cosh(b*x + a) + I*sinh(b
*x + a)) - (3*I*b^2*x^2*cosh(b*x + a)^2 + 6*I*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + 3*I*b^2*x^2*sinh(b*x + a)^
2 - 3*I*b^2*x^2)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 6*(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 - b*x)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + 3*(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 - b^2*x^2)*log(cosh(b*x + a) + sinh(b*x + a) + 1
) - (I*a^3*cosh(b*x + a)^2 + 2*I*a^3*cosh(b*x + a)*sinh(b*x + a) + I*a^3*sinh(b*x + a)^2 - I*a^3)*log(cosh(b*x
+ a) + sinh(b*x + a) + I) - (-I*a^3*cosh(b*x + a)^2 - 2*I*a^3*cosh(b*x + a)*sinh(b*x + a) - I*a^3*sinh(b*x +
a)^2 + I*a^3)*log(cosh(b*x + a) + sinh(b*x + a) - I) - 3*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x +
a) + a^2*sinh(b*x + a)^2 - a^2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - (-I*b^3*x^3 - I*a^3 + (I*b^3*x^3 + I
*a^3)*cosh(b*x + a)^2 + (2*I*b^3*x^3 + 2*I*a^3)*cosh(b*x + a)*sinh(b*x + a) + (I*b^3*x^3 + I*a^3)*sinh(b*x + a
)^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - (I*b^3*x^3 + I*a^3 + (-I*b^3*x^3 - I*a^3)*cosh(b*x + a)^2 +
(-2*I*b^3*x^3 - 2*I*a^3)*cosh(b*x + a)*sinh(b*x + a) + (-I*b^3*x^3 - I*a^3)*sinh(b*x + a)^2)*log(-I*cosh(b*x +
a) - I*sinh(b*x + a) + 1) + 3*(b^2*x^2 - (b^2*x^2 - a^2)*cosh(b*x + a)^2 - 2*(b^2*x^2 - a^2)*cosh(b*x + a)*si
nh(b*x + a) - (b^2*x^2 - a^2)*sinh(b*x + a)^2 - a^2)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) - (-6*I*cosh(b*x
+ a)^2 - 12*I*cosh(b*x + a)*sinh(b*x + a) - 6*I*sinh(b*x + a)^2 + 6*I)*polylog(4, I*cosh(b*x + a) + I*sinh(b*x
+ a)) - (6*I*cosh(b*x + a)^2 + 12*I*cosh(b*x + a)*sinh(b*x + a) + 6*I*sinh(b*x + a)^2 - 6*I)*polylog(4, -I*co
sh(b*x + a) - I*sinh(b*x + a)) + 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*pol
ylog(3, cosh(b*x + a) + sinh(b*x + a)) - (6*I*b*x*cosh(b*x + a)^2 + 12*I*b*x*cosh(b*x + a)*sinh(b*x + a) + 6*I
*b*x*sinh(b*x + a)^2 - 6*I*b*x)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) - (-6*I*b*x*cosh(b*x + a)^2 - 12
*I*b*x*cosh(b*x + a)*sinh(b*x + a) - 6*I*b*x*sinh(b*x + a)^2 + 6*I*b*x)*polylog(3, -I*cosh(b*x + a) - I*sinh(b
*x + a)) - 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*polylog(3, -cosh(b*x + a)
- sinh(b*x + a)))/(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 - b^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{csch}^{2}{\left (a + b x \right )} \operatorname{sech}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*csch(b*x+a)**2*sech(b*x+a),x)

[Out]

Integral(x**3*csch(a + b*x)**2*sech(a + b*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{csch}\left (b x + a\right )^{2} \operatorname{sech}\left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csch(b*x+a)^2*sech(b*x+a),x, algorithm="giac")

[Out]

integrate(x^3*csch(b*x + a)^2*sech(b*x + a), x)