### 3.474 $$\int x^3 \text{csch}(a+b x) \text{sech}^2(a+b x) \, dx$$

Optimal. Leaf size=226 $-\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{6 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{6 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \text{sech}(a+b x)}{b}$

[Out]

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

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Rubi [A]  time = 0.337148, antiderivative size = 226, 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 = {2622, 321, 207, 5462, 14, 6273, 12, 4182, 2531, 6609, 2282, 6589, 4180} $-\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{6 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{6 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(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 6273

Int[((a_.) + ArcTanh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan
h[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 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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.377735, size = 282, normalized size = 1.25 $\frac{-3 i \left (-2 b x \text{PolyLog}\left (2,-i e^{a+b x}\right )+2 b x \text{PolyLog}\left (2,i e^{a+b x}\right )+2 \text{PolyLog}\left (3,-i e^{a+b x}\right )-2 \text{PolyLog}\left (3,i e^{a+b x}\right )+b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )\right )-3 b^2 x^2 \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))+3 b^2 x^2 \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))+6 b x \text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))-6 b x \text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))-6 \text{PolyLog}(4,-\sinh (a+b x)-\cosh (a+b x))+6 \text{PolyLog}(4,\sinh (a+b x)+\cosh (a+b x))+b^3 x^3 \text{sech}(a+b x)-2 b^3 x^3 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))}{b^4}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.521, size = 0, normalized size = 0. \begin{align*} \int{x}^{3}{\rm csch} \left (bx+a\right ) \left ({\rm sech} \left (bx+a\right ) \right ) ^{2}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \,{\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac{b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x{\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \,{\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} - 24 \, \int \frac{x^{2} e^{\left (b x + a\right )}}{4 \,{\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^3*e^(b*x + a)/(b*e^(2*b*x + 2*a) + b) - (b^3*x^3*log(e^(b*x + a) + 1) + 3*b^2*x^2*dilog(-e^(b*x + a)) - 6*
b*x*polylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))/b^4 + (b^3*x^3*log(-e^(b*x + a) + 1) + 3*b^2*x^2*di
log(e^(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4, e^(b*x + a)))/b^4 - 24*integrate(1/4*x^2*e^(b*
x + a)/(b*e^(2*b*x + 2*a) + b), x)

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Fricas [C]  time = 2.89573, size = 3553, normalized size = 15.72 \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)*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

(2*b^3*x^3*cosh(b*x + a) + 2*b^3*x^3*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)*dilog(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)*dilog(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
)*dilog(-I*cosh(b*x + a) - I*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)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - (b^3*x^3*cosh(b*x + a)^2 + 2*b
^3*x^3*cosh(b*x + a)*sinh(b*x + a) + b^3*x^3*sinh(b*x + a)^2 + b^3*x^3)*log(cosh(b*x + a) + sinh(b*x + a) + 1)
+ (-3*I*a^2*cosh(b*x + a)^2 - 6*I*a^2*cosh(b*x + a)*sinh(b*x + a) - 3*I*a^2*sinh(b*x + a)^2 - 3*I*a^2)*log(co
sh(b*x + a) + sinh(b*x + a) + I) + (3*I*a^2*cosh(b*x + a)^2 + 6*I*a^2*cosh(b*x + a)*sinh(b*x + a) + 3*I*a^2*si
nh(b*x + a)^2 + 3*I*a^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) - (a^3*cosh(b*x + a)^2 + 2*a^3*cosh(b*x + a)*s
inh(b*x + a) + a^3*sinh(b*x + a)^2 + a^3)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + (3*I*b^2*x^2 + (3*I*b^2*x^2
- 3*I*a^2)*cosh(b*x + a)^2 + (6*I*b^2*x^2 - 6*I*a^2)*cosh(b*x + a)*sinh(b*x + a) + (3*I*b^2*x^2 - 3*I*a^2)*si
nh(b*x + a)^2 - 3*I*a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + (-3*I*b^2*x^2 + (-3*I*b^2*x^2 + 3*I*a^2)
*cosh(b*x + a)^2 + (-6*I*b^2*x^2 + 6*I*a^2)*cosh(b*x + a)*sinh(b*x + a) + (-3*I*b^2*x^2 + 3*I*a^2)*sinh(b*x +
a)^2 + 3*I*a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + (b^3*x^3 + a^3 + (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(-cosh(b*x + a) - sinh(
b*x + a) + 1) + 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*polylog(4, cosh(b*x
+ a) + 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(4, -
cosh(b*x + a) - 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)*polylog(3, cosh(b*x + a) + 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(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + (-6*I*cosh(b*x + a)^2 - 12*I*cos
h(b*x + a)*sinh(b*x + a) - 6*I*sinh(b*x + a)^2 - 6*I)*polylog(3, -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)*polylog(3, -cosh(b*x + a) - s
inh(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)

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x