\(\int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [271]

Optimal result

Integrand size = 29, antiderivative size = 463 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 f^2 (e+f x) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \left (e^{c+d x}\right )}{a d}+\frac {3 i f^2 (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^4}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac {3 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (4,i e^{c+d x}\right )}{a d^4}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 a d^2}+\frac {i (e+f x)^3 \text {sech}^2(c+d x)}{2 a d}-\frac {3 i f (e+f x)^2 \tanh (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \text {sech}(c+d x) \tanh (c+d x)}{2 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {5690, 4271, 4265, 2317, 2438, 2611, 6744, 2320, 6724, 5559, 4269, 3799, 2221} \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {6 f^2 (e+f x) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \left (e^{c+d x}\right )}{a d}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^4}-\frac {3 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (4,i e^{c+d x}\right )}{a d^4}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}+\frac {3 i f^2 (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{a d^3}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}-\frac {3 i f (e+f x)^2 \tanh (c+d x)}{2 a d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 a d^2}+\frac {i (e+f x)^3 \text {sech}^2(c+d x)}{2 a d}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 a d}-\frac {3 i f (e+f x)^2}{2 a d^2} \]

[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 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 2317

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 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 2438

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 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 3799

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 4265

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 4269

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 4271

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]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 5559

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Sim
p[(-(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 5690

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 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*} \text {integral}& = -\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) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \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) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \left (e^{c+d x}\right )}{a d}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,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) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right ) \, dx}{a d^2}-\frac {\left (3 i f^2\right ) \int (e+f x) \operatorname {PolyLog}\left (2,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 ) \text {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 ) \text {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) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \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 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,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 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right ) \, dx}{a d^3}+\frac {\left (3 i f^3\right ) \int \operatorname {PolyLog}\left (3,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) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \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 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,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 ) \text {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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i x)}{x} \, dx,x,e^{c+d x}\right )}{a d^4}+\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \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 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^4}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac {3 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (4,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] (verified)

Time = 8.07 (sec) , antiderivative size = 828, normalized size of antiderivative = 1.79 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\frac {(e+f x)^4}{f}+\frac {4 \left (1-i e^c\right ) (e+f x)^3 \log \left (1+i e^{-c-d x}\right )}{d}+\frac {12 i \left (i+e^c\right ) f \left (d^2 (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{-c-d x}\right )+2 f \left (d (e+f x) \operatorname {PolyLog}\left (3,-i e^{-c-d x}\right )+f \operatorname {PolyLog}\left (4,-i e^{-c-d x}\right )\right )\right )}{d^4}}{8 a \left (i+e^c\right )}-\frac {-4 d^2 e \left (1+i e^c\right ) f \left (d^2 e^2-12 f^2\right ) x+\left (-12 f^2+d^2 (e+f x)^2\right )^2+12 d \left (1+i e^c\right ) f^2 \left (d^2 e^2-4 f^2\right ) x \log \left (1-i e^{-c-d x}\right )+12 d^3 e \left (1+i e^c\right ) f^3 x^2 \log \left (1-i e^{-c-d x}\right )+4 d^3 \left (1+i e^c\right ) f^4 x^3 \log \left (1-i e^{-c-d x}\right )+4 d e \left (1+i e^c\right ) f \left (d^2 e^2-12 f^2\right ) \log \left (i-e^{c+d x}\right )+12 \left (1+i e^c\right ) f^2 \left (-d^2 e^2+4 f^2\right ) \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )-24 d^2 e \left (1+i e^c\right ) f^3 x \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )-12 d^2 \left (1+i e^c\right ) f^4 x^2 \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )-24 d e \left (1+i e^c\right ) f^3 \operatorname {PolyLog}\left (3,i e^{-c-d x}\right )-24 d \left (1+i e^c\right ) f^4 x \operatorname {PolyLog}\left (3,i e^{-c-d x}\right )-24 \left (1+i e^c\right ) f^4 \operatorname {PolyLog}\left (4,i e^{-c-d x}\right )}{8 a d^4 \left (-i+e^c\right ) f}+\frac {x \left (4 e^3+6 e^2 f x+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 )}+\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 {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 )} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1079 vs. \(2 (416 ) = 832\).

Time = 22.22 (sec) , antiderivative size = 1080, normalized size of antiderivative = 2.33

method	result	size
risch	\(\text {Expression too large to display}\)	\(1080\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1461 vs. \(2 (390) = 780\).

Time = 0.26 (sec) , antiderivative size = 1461, normalized size of antiderivative = 3.16 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e^{3} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{3} x^{3} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e f^{2} x^{2} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e^{2} f x \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \]

[In]

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

[Out]

-I*(Integral(e**3*sech(c + d*x)/(sinh(c + d*x) - I), x) + Integral(f**3*x**3*sech(c + d*x)/(sinh(c + d*x) - I)
, x) + Integral(3*e*f**2*x**2*sech(c + d*x)/(sinh(c + d*x) - I), x) + Integral(3*e**2*f*x*sech(c + d*x)/(sinh(
c + d*x) - I), x))/a

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 685, normalized size of antiderivative = 1.48 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {1}{2} \, e^{3} {\left (\frac {4 \, e^{\left (-d x - c\right )}}{-2 \, {\left (-2 i \, a e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} + a\right )} d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{a d} - \frac {i \, \log \left (i \, e^{\left (-d x - c\right )} + 1\right )}{a d}\right )} + \frac {3 i \, {\left (d x \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right )\right )} e^{2} f}{2 \, a d^{2}} - \frac {6 i \, e f^{2} x}{a d^{2}} + \frac {-3 i \, f^{3} x^{2} - 6 i \, e f^{2} x - 3 i \, e^{2} f + {\left (d f^{3} x^{3} e^{c} + 3 \, e^{2} f e^{c} + 3 \, {\left (d e f^{2} + f^{3}\right )} x^{2} e^{c} + 3 \, {\left (d e^{2} f + 2 \, e f^{2}\right )} x e^{c}\right )} e^{\left (d x\right )}}{a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}} - \frac {3 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} e f^{2}}{2 \, a d^{3}} + \frac {3 i \, {\left (d^{2} x^{2} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(i \, e^{\left (d x + c\right )})\right )} e f^{2}}{2 \, a d^{3}} + \frac {6 i \, e f^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{3}} - \frac {i \, {\left (d^{3} x^{3} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 6 \, d x {\rm Li}_{3}(-i \, e^{\left (d x + c\right )}) + 6 \, {\rm Li}_{4}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{2 \, a d^{4}} + \frac {i \, {\left (d^{3} x^{3} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) - 6 \, d x {\rm Li}_{3}(i \, e^{\left (d x + c\right )}) + 6 \, {\rm Li}_{4}(i \, e^{\left (d x + c\right )})\right )} f^{3}}{2 \, a d^{4}} - \frac {3 i \, {\left (d^{2} e^{2} f - 4 \, f^{3}\right )} {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )}}{2 \, a d^{4}} - \frac {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}}{8 \, a d^{4}} + \frac {i \, d^{4} f^{3} x^{4} + 4 i \, d^{4} e f^{2} x^{3} - 6 \, {\left (-i \, d^{2} e^{2} f + 4 i \, f^{3}\right )} d^{2} x^{2}}{8 \, a d^{4}} \]

[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 + 4*I*f^3)*d^2*x^2)/(a*d^4)

Giac [F]

\[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \operatorname {sech}\left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {{\left (e+f\,x\right )}^3}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

[In]

int((e + f*x)^3/(cosh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)

[Out]

int((e + f*x)^3/(cosh(c + d*x)*(a + a*sinh(c + d*x)*1i)), x)