3.271 \(\int \frac{(e+f x)^3 \text{sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=463 \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac{3 i f^3 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac{3 i f^3 \text{PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac{3 i f^3 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^4}-\frac{3 i f^3 \text{PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac{3 i f^3 \text{PolyLog}\left (4,i e^{c+d x}\right )}{a d^4}+\frac{3 i f^2 (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{a d^3}-\frac{6 f^2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d^3}-\frac{3 i f (e+f x)^2 \tanh (c+d x)}{2 a d^2}+\frac{3 f (e+f x)^2 \text{sech}(c+d x)}{2 a d^2}+\frac{(e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{i (e+f x)^3 \text{sech}^2(c+d x)}{2 a d}+\frac{(e+f x)^3 \tanh (c+d x) \text{sech}(c+d x)}{2 a d}-\frac{3 i f (e+f x)^2}{2 a d^2} \]

[Out]

(((-3*I)/2)*f*(e + f*x)^2)/(a*d^2) - (6*f^2*(e + f*x)*ArcTan[E^(c + d*x)])/(a*d^3) + ((e + f*x)^3*ArcTan[E^(c
+ d*x)])/(a*d) + ((3*I)*f^2*(e + f*x)*Log[1 + E^(2*(c + d*x))])/(a*d^3) + ((3*I)*f^3*PolyLog[2, (-I)*E^(c + d*
x)])/(a*d^4) - (((3*I)/2)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^2) - ((3*I)*f^3*PolyLog[2, I*E^(c +
d*x)])/(a*d^4) + (((3*I)/2)*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)])/(a*d^2) + (((3*I)/2)*f^3*PolyLog[2, -E^(
2*(c + d*x))])/(a*d^4) + ((3*I)*f^2*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^3) - ((3*I)*f^2*(e + f*x)*Pol
yLog[3, I*E^(c + d*x)])/(a*d^3) - ((3*I)*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(a*d^4) + ((3*I)*f^3*PolyLog[4, I*E
^(c + d*x)])/(a*d^4) + (3*f*(e + f*x)^2*Sech[c + d*x])/(2*a*d^2) + ((I/2)*(e + f*x)^3*Sech[c + d*x]^2)/(a*d) -
(((3*I)/2)*f*(e + f*x)^2*Tanh[c + d*x])/(a*d^2) + ((e + f*x)^3*Sech[c + d*x]*Tanh[c + d*x])/(2*a*d)

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Rubi [A]  time = 0.483055, antiderivative size = 463, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 13, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.448, Rules used = {5571, 4186, 4180, 2279, 2391, 2531, 6609, 2282, 6589, 5451, 4184, 3718, 2190} \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac{3 i f^3 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac{3 i f^3 \text{PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac{3 i f^3 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^4}-\frac{3 i f^3 \text{PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac{3 i f^3 \text{PolyLog}\left (4,i e^{c+d x}\right )}{a d^4}+\frac{3 i f^2 (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{a d^3}-\frac{6 f^2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d^3}-\frac{3 i f (e+f x)^2 \tanh (c+d x)}{2 a d^2}+\frac{3 f (e+f x)^2 \text{sech}(c+d x)}{2 a d^2}+\frac{(e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{i (e+f x)^3 \text{sech}^2(c+d x)}{2 a d}+\frac{(e+f x)^3 \tanh (c+d x) \text{sech}(c+d x)}{2 a d}-\frac{3 i f (e+f x)^2}{2 a d^2} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Sech[c + d*x])/(a + I*a*Sinh[c + d*x]),x]

[Out]

(((-3*I)/2)*f*(e + f*x)^2)/(a*d^2) - (6*f^2*(e + f*x)*ArcTan[E^(c + d*x)])/(a*d^3) + ((e + f*x)^3*ArcTan[E^(c
+ d*x)])/(a*d) + ((3*I)*f^2*(e + f*x)*Log[1 + E^(2*(c + d*x))])/(a*d^3) + ((3*I)*f^3*PolyLog[2, (-I)*E^(c + d*
x)])/(a*d^4) - (((3*I)/2)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^2) - ((3*I)*f^3*PolyLog[2, I*E^(c +
d*x)])/(a*d^4) + (((3*I)/2)*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)])/(a*d^2) + (((3*I)/2)*f^3*PolyLog[2, -E^(
2*(c + d*x))])/(a*d^4) + ((3*I)*f^2*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^3) - ((3*I)*f^2*(e + f*x)*Pol
yLog[3, I*E^(c + d*x)])/(a*d^3) - ((3*I)*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(a*d^4) + ((3*I)*f^3*PolyLog[4, I*E
^(c + d*x)])/(a*d^4) + (3*f*(e + f*x)^2*Sech[c + d*x])/(2*a*d^2) + ((I/2)*(e + f*x)^3*Sech[c + d*x]^2)/(a*d) -
(((3*I)/2)*f*(e + f*x)^2*Tanh[c + d*x])/(a*d^2) + ((e + f*x)^3*Sech[c + d*x]*Tanh[c + d*x])/(2*a*d)

Rule 5571

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Sech[c + d*x]^(n + 2), x], x] + Dist[1/b, Int[(e + f*x)^m*Sech[c + d*x]^(n +
1)*Tanh[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && EqQ[a^2 + b^2, 0]

Rule 4186

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

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 5451

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Sech[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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]

Rubi steps

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

Mathematica [A]  time = 11.3799, size = 767, normalized size = 1.66 \[ -\frac{12 \left (1+i e^c\right ) f^2 \left (4 f^2-d^2 e^2\right ) \text{PolyLog}\left (2,i e^{-c-d x}\right )-12 \left (1+i e^c\right ) f^4 \left (d^2 x^2 \text{PolyLog}\left (2,i e^{-c-d x}\right )+2 \left (d x \text{PolyLog}\left (3,i e^{-c-d x}\right )+\text{PolyLog}\left (4,i e^{-c-d x}\right )\right )\right )-24 \left (1+i e^c\right ) d e f^3 \left (d x \text{PolyLog}\left (2,i e^{-c-d x}\right )+\text{PolyLog}\left (3,i e^{-c-d x}\right )\right )+12 \left (1+i e^c\right ) d f^2 x \left (d^2 e^2-4 f^2\right ) \log \left (1-i e^{-c-d x}\right )-4 \left (1+i e^c\right ) d e f \left (d^2 e^2-12 f^2\right ) \left (d x-\log \left (-e^{c+d x}+i\right )\right )+12 \left (1+i e^c\right ) d^3 e f^3 x^2 \log \left (1-i e^{-c-d x}\right )+4 \left (1+i e^c\right ) d^3 f^4 x^3 \log \left (1-i e^{-c-d x}\right )+\left (d^2 (e+f x)^2-12 f^2\right )^2}{8 a \left (e^c-i\right ) d^4 f}-\frac{\frac{12 i \left (e^c+i\right ) f \left (d^2 (e+f x)^2 \text{PolyLog}\left (2,-i e^{-c-d x}\right )+2 f \left (d (e+f x) \text{PolyLog}\left (3,-i e^{-c-d x}\right )+f \text{PolyLog}\left (4,-i e^{-c-d x}\right )\right )\right )}{d^4}+\frac{4 \left (1-i e^c\right ) (e+f x)^3 \log \left (1+i e^{-c-d x}\right )}{d}+\frac{(e+f x)^4}{f}}{8 a \left (e^c+i\right )}-\frac{3 i \left (e^2 f \sinh \left (\frac{d x}{2}\right )+2 e f^2 x \sinh \left (\frac{d x}{2}\right )+f^3 x^2 \sinh \left (\frac{d x}{2}\right )\right )}{a d^2 \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}\right )+i \sinh \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{i (e+f x)^3}{2 a d \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}\right )+i \sinh \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{x \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right )}{8 a \left (\cosh \left (\frac{c}{2}\right )-i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^3*Sech[c + d*x])/(a + I*a*Sinh[c + d*x]),x]

[Out]

-((e + f*x)^4/f + (4*(1 - I*E^c)*(e + f*x)^3*Log[1 + I*E^(-c - d*x)])/d + ((12*I)*(I + E^c)*f*(d^2*(e + f*x)^2
*PolyLog[2, (-I)*E^(-c - d*x)] + 2*f*(d*(e + f*x)*PolyLog[3, (-I)*E^(-c - d*x)] + f*PolyLog[4, (-I)*E^(-c - d*
x)])))/d^4)/(8*a*(I + E^c)) - ((-12*f^2 + d^2*(e + f*x)^2)^2 + 12*d*(1 + I*E^c)*f^2*(d^2*e^2 - 4*f^2)*x*Log[1
- I*E^(-c - d*x)] + 12*d^3*e*(1 + I*E^c)*f^3*x^2*Log[1 - I*E^(-c - d*x)] + 4*d^3*(1 + I*E^c)*f^4*x^3*Log[1 - I
*E^(-c - d*x)] - 4*d*e*(1 + I*E^c)*f*(d^2*e^2 - 12*f^2)*(d*x - Log[I - E^(c + d*x)]) + 12*(1 + I*E^c)*f^2*(-(d
^2*e^2) + 4*f^2)*PolyLog[2, I*E^(-c - d*x)] - 24*d*e*(1 + I*E^c)*f^3*(d*x*PolyLog[2, I*E^(-c - d*x)] + PolyLog
[3, I*E^(-c - d*x)]) - 12*(1 + I*E^c)*f^4*(d^2*x^2*PolyLog[2, I*E^(-c - d*x)] + 2*(d*x*PolyLog[3, I*E^(-c - d*
x)] + PolyLog[4, I*E^(-c - d*x)])))/(8*a*d^4*(-I + E^c)*f) + (x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3))/(
8*a*(Cosh[c/2] - I*Sinh[c/2])*(Cosh[c/2] + I*Sinh[c/2])) + ((I/2)*(e + f*x)^3)/(a*d*(Cosh[c/2 + (d*x)/2] + I*S
inh[c/2 + (d*x)/2])^2) - ((3*I)*(e^2*f*Sinh[(d*x)/2] + 2*e*f^2*x*Sinh[(d*x)/2] + f^3*x^2*Sinh[(d*x)/2]))/(a*d^
2*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[c/2 + (d*x)/2] + I*Sinh[c/2 + (d*x)/2]))

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Maple [B]  time = 0.244, size = 1152, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

3*I/a/d^2*polylog(2,I*exp(d*x+c))*e*f^2*x-3/2*I/a/d^3*e*f^2*c^2*ln(exp(d*x+c)-I)+3/2*I/a/d^3*e*f^2*c^2*ln(exp(
d*x+c)+I)+3/2*I/a/d^3*ln(1+I*exp(d*x+c))*c^2*e*f^2-3/2*I/a/d^3*ln(1-I*exp(d*x+c))*c^2*e*f^2+3/2*I/a/d^2*e^2*f*
c*ln(exp(d*x+c)-I)-3/2*I/a/d^2*e^2*f*c*ln(exp(d*x+c)+I)+3/2*I/a/d*ln(1-I*exp(d*x+c))*e^2*f*x+3/2*I/a/d^2*ln(1-
I*exp(d*x+c))*c*e^2*f-3/2*I/a/d*ln(1+I*exp(d*x+c))*e^2*f*x-3/2*I/a/d^2*ln(1+I*exp(d*x+c))*c*e^2*f-3/2*I/a/d*ln
(1+I*exp(d*x+c))*e*f^2*x^2-3*I/a/d^2*polylog(2,-I*exp(d*x+c))*e*f^2*x+3/2*I/a/d*ln(1-I*exp(d*x+c))*e*f^2*x^2-3
*I*f^3*polylog(4,-I*exp(d*x+c))/a/d^4+3*I*f^3*polylog(4,I*exp(d*x+c))/a/d^4+1/2*I/a/d*e^3*ln(exp(d*x+c)+I)-3*I
/a/d^2*f^3*x^2-3*I/a/d^4*f^3*c^2+6*I/a/d^4*f^3*polylog(2,-I*exp(d*x+c))-1/2*I/a/d*e^3*ln(exp(d*x+c)-I)-3/2*I/a
/d^2*f^3*polylog(2,-I*exp(d*x+c))*x^2+6*I/a/d^3*e*f^2*ln(exp(d*x+c)-I)-6*I/a/d^3*e*f^2*ln(exp(d*x+c))+3*I/a/d^
3*e*f^2*polylog(3,-I*exp(d*x+c))-3*I/a/d^3*e*f^2*polylog(3,I*exp(d*x+c))-6*I/a/d^3*f^3*c*x-6*I/a/d^4*f^3*c*ln(
exp(d*x+c)-I)-3/2*I/a/d^2*e^2*f*polylog(2,-I*exp(d*x+c))+3/2*I/a/d^2*e^2*f*polylog(2,I*exp(d*x+c))+1/2*I/a/d^4
*f^3*c^3*ln(exp(d*x+c)-I)-1/2*I/a/d^4*f^3*c^3*ln(exp(d*x+c)+I)+6*I/a/d^4*f^3*c*ln(exp(d*x+c))+6*I/a/d^4*f^3*c*
ln(1+I*exp(d*x+c))+1/2*I/a/d^4*f^3*c^3*ln(1-I*exp(d*x+c))-1/2*I/a/d^4*f^3*c^3*ln(1+I*exp(d*x+c))-1/2*I/a/d*f^3
*ln(1+I*exp(d*x+c))*x^3+6*I/a/d^3*f^3*ln(1+I*exp(d*x+c))*x+1/2*I/a/d*f^3*ln(1-I*exp(d*x+c))*x^3+3/2*I/a/d^2*f^
3*polylog(2,I*exp(d*x+c))*x^2-3*I/a/d^3*f^3*polylog(3,I*exp(d*x+c))*x+3*I/a/d^3*f^3*polylog(3,-I*exp(d*x+c))*x
+(d*f^3*x^3*exp(d*x+c)+3*d*e*f^2*x^2*exp(d*x+c)+3*d*e^2*f*x*exp(d*x+c)+d*e^3*exp(d*x+c)+3*f^3*x^2*exp(d*x+c)-3
*I*f^3*x^2+6*e*f^2*x*exp(d*x+c)-6*I*e*f^2*x+3*e^2*f*exp(d*x+c)-3*I*e^2*f)/(exp(d*x+c)-I)^2/d^2/a

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Maxima [A]  time = 2.02883, size = 923, normalized size = 1.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*e^3*(4*e^(-d*x - c)/((4*I*a*e^(-d*x - c) + 2*a*e^(-2*d*x - 2*c) - 2*a)*d) + I*log(e^(-d*x - c) + I)/(a*d)
- I*log(I*e^(-d*x - c) + 1)/(a*d)) + 3/2*I*(d*x*log(-I*e^(d*x + c) + 1) + dilog(I*e^(d*x + c)))*e^2*f/(a*d^2)
- 6*I*e*f^2*x/(a*d^2) + (-3*I*f^3*x^2 - 6*I*e*f^2*x - 3*I*e^2*f + (d*f^3*x^3*e^c + 3*e^2*f*e^c + 3*(d*e*f^2 +
f^3)*x^2*e^c + 3*(d*e^2*f + 2*e*f^2)*x*e^c)*e^(d*x))/(a*d^2*e^(2*d*x + 2*c) - 2*I*a*d^2*e^(d*x + c) - a*d^2)
- 3/2*I*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dilog(-I*e^(d*x + c)) - 2*polylog(3, -I*e^(d*x + c)))*e*f^2/(a
*d^3) + 3/2*I*(d^2*x^2*log(-I*e^(d*x + c) + 1) + 2*d*x*dilog(I*e^(d*x + c)) - 2*polylog(3, I*e^(d*x + c)))*e*f
^2/(a*d^3) + 6*I*e*f^2*log(I*e^(d*x + c) + 1)/(a*d^3) - 1/2*I*(d^3*x^3*log(I*e^(d*x + c) + 1) + 3*d^2*x^2*dilo
g(-I*e^(d*x + c)) - 6*d*x*polylog(3, -I*e^(d*x + c)) + 6*polylog(4, -I*e^(d*x + c)))*f^3/(a*d^4) + 1/2*I*(d^3*
x^3*log(-I*e^(d*x + c) + 1) + 3*d^2*x^2*dilog(I*e^(d*x + c)) - 6*d*x*polylog(3, I*e^(d*x + c)) + 6*polylog(4,
I*e^(d*x + c)))*f^3/(a*d^4) - 3/2*I*(d^2*e^2*f - 4*f^3)*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))/(
a*d^4) - 1/8*(I*d^4*f^3*x^4 + 4*I*d^4*e*f^2*x^3 + 6*I*d^4*e^2*f*x^2)/(a*d^4) + 1/8*(I*d^4*f^3*x^4 + 4*I*d^4*e*
f^2*x^3 + (6*I*d^2*e^2*f - 24*I*f^3)*d^2*x^2)/(a*d^4)

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Fricas [C]  time = 2.45374, size = 3418, normalized size = 7.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(-6*I*d^2*e^2*f + 12*I*c*d*e*f^2 - 6*I*c^2*f^3 + (-3*I*d^2*f^3*x^2 - 6*I*d^2*e*f^2*x - 3*I*d^2*e^2*f + (3*I*d^
2*f^3*x^2 + 6*I*d^2*e*f^2*x + 3*I*d^2*e^2*f)*e^(2*d*x + 2*c) + 6*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*e^(
d*x + c))*dilog(I*e^(d*x + c)) + (3*I*d^2*f^3*x^2 + 6*I*d^2*e*f^2*x + 3*I*d^2*e^2*f - 12*I*f^3 + (-3*I*d^2*f^3
*x^2 - 6*I*d^2*e*f^2*x - 3*I*d^2*e^2*f + 12*I*f^3)*e^(2*d*x + 2*c) - 6*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*
f - 4*f^3)*e^(d*x + c))*dilog(-I*e^(d*x + c)) + (-6*I*d^2*f^3*x^2 - 12*I*d^2*e*f^2*x - 12*I*c*d*e*f^2 + 6*I*c^
2*f^3)*e^(2*d*x + 2*c) + 2*(d^3*f^3*x^3 + d^3*e^3 + 3*d^2*e^2*f - 12*c*d*e*f^2 + 6*c^2*f^3 + 3*(d^3*e*f^2 - d^
2*f^3)*x^2 + 3*(d^3*e^2*f - 2*d^2*e*f^2)*x)*e^(d*x + c) + (-I*d^3*e^3 + 3*I*c*d^2*e^2*f - 3*I*c^2*d*e*f^2 + I*
c^3*f^3 + (I*d^3*e^3 - 3*I*c*d^2*e^2*f + 3*I*c^2*d*e*f^2 - I*c^3*f^3)*e^(2*d*x + 2*c) + 2*(d^3*e^3 - 3*c*d^2*e
^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*e^(d*x + c))*log(e^(d*x + c) + I) + (I*d^3*e^3 - 3*I*c*d^2*e^2*f + (3*I*c^2 -
12*I)*d*e*f^2 + (-I*c^3 + 12*I*c)*f^3 + (-I*d^3*e^3 + 3*I*c*d^2*e^2*f + (-3*I*c^2 + 12*I)*d*e*f^2 + (I*c^3 - 1
2*I*c)*f^3)*e^(2*d*x + 2*c) - 2*(d^3*e^3 - 3*c*d^2*e^2*f + 3*(c^2 - 4)*d*e*f^2 - (c^3 - 12*c)*f^3)*e^(d*x + c)
)*log(e^(d*x + c) - I) + (I*d^3*f^3*x^3 + 3*I*d^3*e*f^2*x^2 + 3*I*c*d^2*e^2*f - 3*I*c^2*d*e*f^2 + (I*c^3 - 12*
I*c)*f^3 + (3*I*d^3*e^2*f - 12*I*d*f^3)*x + (-I*d^3*f^3*x^3 - 3*I*d^3*e*f^2*x^2 - 3*I*c*d^2*e^2*f + 3*I*c^2*d*
e*f^2 + (-I*c^3 + 12*I*c)*f^3 + (-3*I*d^3*e^2*f + 12*I*d*f^3)*x)*e^(2*d*x + 2*c) - 2*(d^3*f^3*x^3 + 3*d^3*e*f^
2*x^2 + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + (c^3 - 12*c)*f^3 + 3*(d^3*e^2*f - 4*d*f^3)*x)*e^(d*x + c))*log(I*e^(d*
x + c) + 1) + (-I*d^3*f^3*x^3 - 3*I*d^3*e*f^2*x^2 - 3*I*d^3*e^2*f*x - 3*I*c*d^2*e^2*f + 3*I*c^2*d*e*f^2 - I*c^
3*f^3 + (I*d^3*f^3*x^3 + 3*I*d^3*e*f^2*x^2 + 3*I*d^3*e^2*f*x + 3*I*c*d^2*e^2*f - 3*I*c^2*d*e*f^2 + I*c^3*f^3)*
e^(2*d*x + 2*c) + 2*(d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*d^3*e^2*f*x + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + c^3*f^3)*
e^(d*x + c))*log(-I*e^(d*x + c) + 1) + (6*I*f^3*e^(2*d*x + 2*c) + 12*f^3*e^(d*x + c) - 6*I*f^3)*polylog(4, I*e
^(d*x + c)) + (-6*I*f^3*e^(2*d*x + 2*c) - 12*f^3*e^(d*x + c) + 6*I*f^3)*polylog(4, -I*e^(d*x + c)) + (6*I*d*f^
3*x + 6*I*d*e*f^2 + (-6*I*d*f^3*x - 6*I*d*e*f^2)*e^(2*d*x + 2*c) - 12*(d*f^3*x + d*e*f^2)*e^(d*x + c))*polylog
(3, I*e^(d*x + c)) + (-6*I*d*f^3*x - 6*I*d*e*f^2 + (6*I*d*f^3*x + 6*I*d*e*f^2)*e^(2*d*x + 2*c) + 12*(d*f^3*x +
d*e*f^2)*e^(d*x + c))*polylog(3, -I*e^(d*x + c)))/(2*a*d^4*e^(2*d*x + 2*c) - 4*I*a*d^4*e^(d*x + c) - 2*a*d^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((f*x+e)**3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*sech(d*x + c)/(I*a*sinh(d*x + c) + a), x)