3.3.77 \(\int \frac {(e+f x)^3 \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [277]

Optimal. Leaf size=450 \[ \frac {2 (e+f x)^3}{3 a d}-\frac {i f (e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right )}{a d^2}+\frac {i f^3 \text {ArcTan}(\sinh (c+d x))}{a d^4}-\frac {2 f (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{a d^2}+\frac {f^3 \log (\cosh (c+d x))}{a d^4}-\frac {f^2 (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {f^2 (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )}{a d^3}-\frac {2 f^2 (e+f x) \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{a d^3}+\frac {f^3 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac {f^3 \text {PolyLog}\left (3,i e^{c+d x}\right )}{a d^4}+\frac {f^3 \text {PolyLog}\left (3,-e^{2 (c+d x)}\right )}{a d^4}-\frac {i f^2 (e+f x) \text {sech}(c+d x)}{a d^3}+\frac {f (e+f x)^2 \text {sech}^2(c+d x)}{2 a d^2}+\frac {i (e+f x)^3 \text {sech}^3(c+d x)}{3 a d}-\frac {f^2 (e+f x) \tanh (c+d x)}{a d^3}+\frac {2 (e+f x)^3 \tanh (c+d x)}{3 a d}-\frac {i f (e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)}{3 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.42, antiderivative size = 450, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {5690, 4271, 4269, 3556, 3799, 2221, 2611, 2320, 6724, 5559, 3855, 4265} \begin {gather*} \frac {i f^3 \text {ArcTan}(\sinh (c+d x))}{a d^4}-\frac {i f (e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right )}{a d^2}+\frac {f^3 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^4}-\frac {f^3 \text {Li}_3\left (i e^{c+d x}\right )}{a d^4}+\frac {f^3 \text {Li}_3\left (-e^{2 (c+d x)}\right )}{a d^4}+\frac {f^3 \log (\cosh (c+d x))}{a d^4}-\frac {f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {f^2 (e+f x) \text {Li}_2\left (i e^{c+d x}\right )}{a d^3}-\frac {2 f^2 (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right )}{a d^3}-\frac {f^2 (e+f x) \tanh (c+d x)}{a d^3}-\frac {i f^2 (e+f x) \text {sech}(c+d x)}{a d^3}-\frac {2 f (e+f x)^2 \log \left (e^{2 (c+d x)}+1\right )}{a d^2}+\frac {f (e+f x)^2 \text {sech}^2(c+d x)}{2 a d^2}-\frac {i f (e+f x)^2 \tanh (c+d x) \text {sech}(c+d x)}{2 a d^2}+\frac {2 (e+f x)^3 \tanh (c+d x)}{3 a d}+\frac {i (e+f x)^3 \text {sech}^3(c+d x)}{3 a d}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}^2(c+d x)}{3 a d}+\frac {2 (e+f x)^3}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Sech[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

(2*(e + f*x)^3)/(3*a*d) - (I*f*(e + f*x)^2*ArcTan[E^(c + d*x)])/(a*d^2) + (I*f^3*ArcTan[Sinh[c + d*x]])/(a*d^4
) - (2*f*(e + f*x)^2*Log[1 + E^(2*(c + d*x))])/(a*d^2) + (f^3*Log[Cosh[c + d*x]])/(a*d^4) - (f^2*(e + f*x)*Pol
yLog[2, (-I)*E^(c + d*x)])/(a*d^3) + (f^2*(e + f*x)*PolyLog[2, I*E^(c + d*x)])/(a*d^3) - (2*f^2*(e + f*x)*Poly
Log[2, -E^(2*(c + d*x))])/(a*d^3) + (f^3*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^4) - (f^3*PolyLog[3, I*E^(c + d*x)
])/(a*d^4) + (f^3*PolyLog[3, -E^(2*(c + d*x))])/(a*d^4) - (I*f^2*(e + f*x)*Sech[c + d*x])/(a*d^3) + (f*(e + f*
x)^2*Sech[c + d*x]^2)/(2*a*d^2) + ((I/3)*(e + f*x)^3*Sech[c + d*x]^3)/(a*d) - (f^2*(e + f*x)*Tanh[c + d*x])/(a
*d^3) + (2*(e + f*x)^3*Tanh[c + d*x])/(3*a*d) - ((I/2)*f*(e + f*x)^2*Sech[c + d*x]*Tanh[c + d*x])/(a*d^2) + ((
e + f*x)^3*Sech[c + d*x]^2*Tanh[c + d*x])/(3*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 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 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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1127\) vs. \(2(450)=900\).
time = 10.08, size = 1127, normalized size = 2.50 \begin {gather*} \frac {f \left (15 d^3 e^2 e^c x-12 d e^c f^2 x+15 d^3 e e^c f x^2+5 d^3 e^c f^2 x^3+15 i d^2 e^2 \log \left (i-e^{c+d x}\right )-15 d^2 e^2 e^c \log \left (i-e^{c+d x}\right )-12 i f^2 \log \left (i-e^{c+d x}\right )+12 e^c f^2 \log \left (i-e^{c+d x}\right )+30 i d^2 e f x \log \left (1+i e^{c+d x}\right )-30 d^2 e e^c f x \log \left (1+i e^{c+d x}\right )+15 i d^2 f^2 x^2 \log \left (1+i e^{c+d x}\right )-15 d^2 e^c f^2 x^2 \log \left (1+i e^{c+d x}\right )-30 d \left (-i+e^c\right ) f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )+30 \left (-i+e^c\right ) f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right )\right )}{6 a d^4 \left (-i+e^c\right )}-\frac {i f \left (d^2 \left (i d e^c x \left (3 e^2+3 e f x+f^2 x^2\right )+3 \left (1-i e^c\right ) (e+f x)^2 \log \left (1-i e^{c+d x}\right )\right )+6 d \left (1-i e^c\right ) f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )+6 i \left (i+e^c\right ) f^2 \text {PolyLog}\left (3,i e^{c+d x}\right )\right )}{2 a d^4 \left (i+e^c\right )}+\frac {e^3 \sinh \left (\frac {d x}{2}\right )+3 e^2 f x \sinh \left (\frac {d x}{2}\right )+3 e f^2 x^2 \sinh \left (\frac {d x}{2}\right )+f^3 x^3 \sinh \left (\frac {d x}{2}\right )}{2 a d \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 {e^3 \sinh \left (\frac {d x}{2}\right )+3 e^2 f x \sinh \left (\frac {d x}{2}\right )+3 e f^2 x^2 \sinh \left (\frac {d x}{2}\right )+f^3 x^3 \sinh \left (\frac {d x}{2}\right )}{3 a d \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 )^3}+\frac {i d e^3 \cosh \left (\frac {c}{2}\right )+3 e^2 f \cosh \left (\frac {c}{2}\right )+3 i d e^2 f x \cosh \left (\frac {c}{2}\right )+6 e f^2 x \cosh \left (\frac {c}{2}\right )+3 i d e f^2 x^2 \cosh \left (\frac {c}{2}\right )+3 f^3 x^2 \cosh \left (\frac {c}{2}\right )+i d f^3 x^3 \cosh \left (\frac {c}{2}\right )+d e^3 \sinh \left (\frac {c}{2}\right )+3 i e^2 f \sinh \left (\frac {c}{2}\right )+3 d e^2 f x \sinh \left (\frac {c}{2}\right )+6 i e f^2 x \sinh \left (\frac {c}{2}\right )+3 d e f^2 x^2 \sinh \left (\frac {c}{2}\right )+3 i f^3 x^2 \sinh \left (\frac {c}{2}\right )+d f^3 x^3 \sinh \left (\frac {c}{2}\right )}{6 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 )^2}+\frac {5 d^2 e^3 \sinh \left (\frac {d x}{2}\right )-12 e f^2 \sinh \left (\frac {d x}{2}\right )+15 d^2 e^2 f x \sinh \left (\frac {d x}{2}\right )-12 f^3 x \sinh \left (\frac {d x}{2}\right )+15 d^2 e f^2 x^2 \sinh \left (\frac {d x}{2}\right )+5 d^2 f^3 x^3 \sinh \left (\frac {d x}{2}\right )}{6 a d^3 \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 )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^3*Sech[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

(f*(15*d^3*e^2*E^c*x - 12*d*E^c*f^2*x + 15*d^3*e*E^c*f*x^2 + 5*d^3*E^c*f^2*x^3 + (15*I)*d^2*e^2*Log[I - E^(c +
d*x)] - 15*d^2*e^2*E^c*Log[I - E^(c + d*x)] - (12*I)*f^2*Log[I - E^(c + d*x)] + 12*E^c*f^2*Log[I - E^(c + d*x
)] + (30*I)*d^2*e*f*x*Log[1 + I*E^(c + d*x)] - 30*d^2*e*E^c*f*x*Log[1 + I*E^(c + d*x)] + (15*I)*d^2*f^2*x^2*Lo
g[1 + I*E^(c + d*x)] - 15*d^2*E^c*f^2*x^2*Log[1 + I*E^(c + d*x)] - 30*d*(-I + E^c)*f*(e + f*x)*PolyLog[2, (-I)
*E^(c + d*x)] + 30*(-I + E^c)*f^2*PolyLog[3, (-I)*E^(c + d*x)]))/(6*a*d^4*(-I + E^c)) - ((I/2)*f*(d^2*(I*d*E^c
*x*(3*e^2 + 3*e*f*x + f^2*x^2) + 3*(1 - I*E^c)*(e + f*x)^2*Log[1 - I*E^(c + d*x)]) + 6*d*(1 - I*E^c)*f*(e + f*
x)*PolyLog[2, I*E^(c + d*x)] + (6*I)*(I + E^c)*f^2*PolyLog[3, I*E^(c + d*x)]))/(a*d^4*(I + E^c)) + (e^3*Sinh[(
d*x)/2] + 3*e^2*f*x*Sinh[(d*x)/2] + 3*e*f^2*x^2*Sinh[(d*x)/2] + f^3*x^3*Sinh[(d*x)/2])/(2*a*d*(Cosh[c/2] - I*S
inh[c/2])*(Cosh[c/2 + (d*x)/2] - I*Sinh[c/2 + (d*x)/2])) + (e^3*Sinh[(d*x)/2] + 3*e^2*f*x*Sinh[(d*x)/2] + 3*e*
f^2*x^2*Sinh[(d*x)/2] + f^3*x^3*Sinh[(d*x)/2])/(3*a*d*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[c/2 + (d*x)/2] + I*Sinh[
c/2 + (d*x)/2])^3) + (I*d*e^3*Cosh[c/2] + 3*e^2*f*Cosh[c/2] + (3*I)*d*e^2*f*x*Cosh[c/2] + 6*e*f^2*x*Cosh[c/2]
+ (3*I)*d*e*f^2*x^2*Cosh[c/2] + 3*f^3*x^2*Cosh[c/2] + I*d*f^3*x^3*Cosh[c/2] + d*e^3*Sinh[c/2] + (3*I)*e^2*f*Si
nh[c/2] + 3*d*e^2*f*x*Sinh[c/2] + (6*I)*e*f^2*x*Sinh[c/2] + 3*d*e*f^2*x^2*Sinh[c/2] + (3*I)*f^3*x^2*Sinh[c/2]
+ d*f^3*x^3*Sinh[c/2])/(6*a*d^2*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[c/2 + (d*x)/2] + I*Sinh[c/2 + (d*x)/2])^2) + (
5*d^2*e^3*Sinh[(d*x)/2] - 12*e*f^2*Sinh[(d*x)/2] + 15*d^2*e^2*f*x*Sinh[(d*x)/2] - 12*f^3*x*Sinh[(d*x)/2] + 15*
d^2*e*f^2*x^2*Sinh[(d*x)/2] + 5*d^2*f^3*x^3*Sinh[(d*x)/2])/(6*a*d^3*(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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1000 vs. \(2 (424 ) = 848\).
time = 5.18, size = 1001, normalized size = 2.22

method	result	size
risch	\(-\frac {8 f^{2} e c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {5 f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c e}{a \,d^{3}}+\frac {8 f^{2} c e x}{a \,d^{2}}-\frac {5 f^{3} \polylog \left (2, -i {\mathrm e}^{d x +c}\right ) x}{a \,d^{3}}-\frac {5 f^{3} c^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 a \,d^{4}}-\frac {3 f^{3} c^{2} \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a \,d^{4}}+\frac {3 f^{3} \ln \left (1-i {\mathrm e}^{d x +c}\right ) c^{2}}{2 a \,d^{4}}-\frac {3 f^{3} \ln \left (1-i {\mathrm e}^{d x +c}\right ) x^{2}}{2 a \,d^{2}}-\frac {5 f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{2 a \,d^{2}}+\frac {5 f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2}}{2 a \,d^{4}}-\frac {3 f \,e^{2} \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a \,d^{2}}-\frac {3 f^{2} e \polylog \left (2, i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {5 f^{2} e \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {3 f^{3} \polylog \left (3, i {\mathrm e}^{d x +c}\right )}{a \,d^{4}}-\frac {3 f^{2} \ln \left (1-i {\mathrm e}^{d x +c}\right ) e x}{a \,d^{2}}+\frac {3 f^{2} e c \ln \left ({\mathrm e}^{d x +c}+i\right )}{a \,d^{3}}+\frac {5 f^{2} e c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}-\frac {3 f^{2} \ln \left (1-i {\mathrm e}^{d x +c}\right ) c e}{a \,d^{3}}+\frac {i \left (6 i e \,f^{2} {\mathrm e}^{2 d x +2 c}+8 d^{2} f^{3} x^{3} {\mathrm e}^{d x +c}-3 d \,f^{3} x^{2} {\mathrm e}^{3 d x +3 c}+6 i f^{3} x \,{\mathrm e}^{2 d x +2 c}+24 d^{2} e \,f^{2} x^{2} {\mathrm e}^{d x +c}-6 d e \,f^{2} x \,{\mathrm e}^{3 d x +3 c}-4 i d^{2} f^{3} x^{3}+6 i f^{2} e +24 d^{2} e^{2} f x \,{\mathrm e}^{d x +c}-3 d \,e^{2} f \,{\mathrm e}^{3 d x +3 c}-3 d \,f^{3} x^{2} {\mathrm e}^{d x +c}-6 f^{3} x \,{\mathrm e}^{3 d x +3 c}-4 i d^{2} e^{3}-12 i d^{2} e^{2} f x +8 d^{2} e^{3} {\mathrm e}^{d x +c}-6 d e \,f^{2} x \,{\mathrm e}^{d x +c}-6 e \,f^{2} {\mathrm e}^{3 d x +3 c}+6 i f^{3} x -3 d \,e^{2} f \,{\mathrm e}^{d x +c}-6 f^{3} x \,{\mathrm e}^{d x +c}-12 i d^{2} e \,f^{2} x^{2}-6 e \,f^{2} {\mathrm e}^{d x +c}\right )}{3 \left ({\mathrm e}^{d x +c}+i\right ) \left ({\mathrm e}^{d x +c}-i\right )^{3} d^{3} a}-\frac {5 f \ln \left ({\mathrm e}^{d x +c}-i\right ) e^{2}}{2 a \,d^{2}}-\frac {5 e \,f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{d^{2} a}+\frac {4 f \,e^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {4 f^{3} c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {4 f^{2} e \,x^{2}}{a d}+\frac {4 f^{2} c^{2} e}{a \,d^{3}}-\frac {3 f^{3} \polylog \left (2, i {\mathrm e}^{d x +c}\right ) x}{a \,d^{3}}-\frac {8 f^{3} c^{3}}{3 a \,d^{4}}+\frac {4 f^{3} x^{3}}{3 a d}-\frac {2 f^{3} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {2 f^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{4}}+\frac {5 f^{3} \polylog \left (3, -i {\mathrm e}^{d x +c}\right )}{a \,d^{4}}-\frac {4 f^{3} c^{2} x}{d^{3} a}\)	\(1001\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-5/a/d^3*f^2*ln(1+I*exp(d*x+c))*c*e-3/a/d^2*f^2*ln(1-I*exp(d*x+c))*e*x+3/a/d^3*f^2*e*c*ln(exp(d*x+c)+I)+5/a/d^
3*f^2*e*c*ln(exp(d*x+c)-I)-8/a/d^3*f^2*e*c*ln(exp(d*x+c))-3/a/d^3*f^2*ln(1-I*exp(d*x+c))*c*e+8/a/d^2*f^2*c*e*x
-5/a/d^2*f^2*ln(1+I*exp(d*x+c))*e*x+1/3*I*(6*I*e*f^2*exp(2*d*x+2*c)+8*d^2*f^3*x^3*exp(d*x+c)-3*d*f^3*x^2*exp(3
*d*x+3*c)+6*I*f^3*x*exp(2*d*x+2*c)+24*d^2*e*f^2*x^2*exp(d*x+c)-6*d*e*f^2*x*exp(3*d*x+3*c)-4*I*d^2*f^3*x^3+6*I*
f^2*e+24*d^2*e^2*f*x*exp(d*x+c)-3*d*e^2*f*exp(3*d*x+3*c)-3*d*f^3*x^2*exp(d*x+c)-6*f^3*x*exp(3*d*x+3*c)-4*I*d^2
*e^3-12*I*d^2*e^2*f*x+8*d^2*e^3*exp(d*x+c)-6*d*e*f^2*x*exp(d*x+c)-6*e*f^2*exp(3*d*x+3*c)+6*I*f^3*x-3*d*e^2*f*e
xp(d*x+c)-6*f^3*x*exp(d*x+c)-12*I*d^2*e*f^2*x^2-6*e*f^2*exp(d*x+c))/(exp(d*x+c)+I)/(exp(d*x+c)-I)^3/d^3/a+5*f^
3*polylog(3,-I*exp(d*x+c))/a/d^4+3*f^3*polylog(3,I*exp(d*x+c))/a/d^4-5/2/a/d^4*f^3*c^2*ln(exp(d*x+c)-I)-3/2/a/
d^4*f^3*c^2*ln(exp(d*x+c)+I)+4/a/d^2*f*e^2*ln(exp(d*x+c))+4/a/d^4*f^3*c^2*ln(exp(d*x+c))+3/2/a/d^4*f^3*ln(1-I*
exp(d*x+c))*c^2+4/a/d*f^2*e*x^2+4/a/d^3*f^2*c^2*e-3/2/a/d^2*f^3*ln(1-I*exp(d*x+c))*x^2-3/a/d^3*f^3*polylog(2,I
*exp(d*x+c))*x-5/2/a/d^2*f^3*ln(1+I*exp(d*x+c))*x^2-5/a/d^3*f^3*polylog(2,-I*exp(d*x+c))*x+5/2/a/d^4*f^3*ln(1+
I*exp(d*x+c))*c^2-3/a/d^3*f^2*e*polylog(2,I*exp(d*x+c))-5/a/d^3*f^2*e*polylog(2,-I*exp(d*x+c))-5/2/a/d^2*f*e^2
*ln(exp(d*x+c)-I)-3/2/a/d^2*f*e^2*ln(exp(d*x+c)+I)-8/3/a/d^4*f^3*c^3+4/3/a/d*f^3*x^3-2/a/d^4*f^3*ln(exp(d*x+c)
)+2/a/d^4*f^3*ln(exp(d*x+c)-I)-4/d^3/a*f^3*c^2*x

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Maxima [A]
time = 0.51, size = 749, normalized size = 1.66 \begin {gather*} \frac {1}{2} \, f {\left (\frac {24 \, {\left (4 i \, d x e^{\left (4 \, d x + 4 \, c\right )} + {\left (8 \, d x e^{\left (3 \, c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + e^{\left (d x + c\right )}\right )}}{12 i \, a d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} + 24 \, a d^{2} e^{\left (d x + c\right )} - 12 i \, a d^{2}} - \frac {3 \, \log \left ({\left (e^{\left (d x + c\right )} + i\right )} e^{\left (-c\right )}\right )}{a d^{2}} - \frac {5 \, \log \left (-i \, {\left (i \, e^{\left (d x + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} e^{2} + \frac {4}{3} \, {\left (\frac {2 \, e^{\left (-d x - c\right )}}{{\left (2 \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} - i \, a e^{\left (-4 \, d x - 4 \, c\right )} + i \, a\right )} d} + \frac {i}{{\left (2 \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} - i \, a e^{\left (-4 \, d x - 4 \, c\right )} + i \, a\right )} d}\right )} e^{3} + \frac {4 i \, d^{2} f^{3} x^{3} + 12 i \, d^{2} f^{2} x^{2} e - 6 i \, f^{3} x - 6 i \, f^{2} e + 3 \, {\left (d f^{3} x^{2} e^{\left (3 \, c\right )} + 2 \, f^{2} e^{\left (3 \, c + 1\right )} + 2 \, {\left (f^{3} e^{\left (3 \, c\right )} + d f^{2} e^{\left (3 \, c + 1\right )}\right )} x\right )} e^{\left (3 \, d x\right )} - 6 \, {\left (i \, f^{3} x e^{\left (2 \, c\right )} + i \, f^{2} e^{\left (2 \, c + 1\right )}\right )} e^{\left (2 \, d x\right )} - {\left (8 \, d^{2} f^{3} x^{3} e^{c} + 3 \, {\left (8 \, d^{2} f^{2} e^{\left (c + 1\right )} - d f^{3} e^{c}\right )} x^{2} - 6 \, f^{2} e^{\left (c + 1\right )} - 6 \, {\left (d f^{2} e^{\left (c + 1\right )} + f^{3} e^{c}\right )} x\right )} e^{\left (d x\right )}}{3 i \, a d^{3} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a d^{3} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a d^{3} e^{\left (d x + c\right )} - 3 i \, a d^{3}} - \frac {2 \, f^{3} x}{a d^{3}} - \frac {5 \, {\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 )} f^{2} e}{a d^{3}} - \frac {3 \, {\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 )} f^{2} e}{a d^{3}} - \frac {5 \, {\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 )} f^{3}}{2 \, a d^{4}} - \frac {3 \, {\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 )} f^{3}}{2 \, a d^{4}} + \frac {2 \, f^{3} \log \left (e^{\left (d x + c\right )} - i\right )}{a d^{4}} + \frac {4 \, {\left (d^{3} f^{3} x^{3} + 3 \, d^{3} f^{2} x^{2} e\right )}}{3 \, a d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

1/2*f*(24*(4*I*d*x*e^(4*d*x + 4*c) + (8*d*x*e^(3*c) + e^(3*c))*e^(3*d*x) + e^(d*x + c))/(12*I*a*d^2*e^(4*d*x +
4*c) + 24*a*d^2*e^(3*d*x + 3*c) + 24*a*d^2*e^(d*x + c) - 12*I*a*d^2) - 3*log((e^(d*x + c) + I)*e^(-c))/(a*d^2
) - 5*log(-I*(I*e^(d*x + c) + 1)*e^(-c))/(a*d^2))*e^2 + 4/3*(2*e^(-d*x - c)/((2*a*e^(-d*x - c) + 2*a*e^(-3*d*x
- 3*c) - I*a*e^(-4*d*x - 4*c) + I*a)*d) + I/((2*a*e^(-d*x - c) + 2*a*e^(-3*d*x - 3*c) - I*a*e^(-4*d*x - 4*c)
+ I*a)*d))*e^3 + (4*I*d^2*f^3*x^3 + 12*I*d^2*f^2*x^2*e - 6*I*f^3*x - 6*I*f^2*e + 3*(d*f^3*x^2*e^(3*c) + 2*f^2*
e^(3*c + 1) + 2*(f^3*e^(3*c) + d*f^2*e^(3*c + 1))*x)*e^(3*d*x) - 6*(I*f^3*x*e^(2*c) + I*f^2*e^(2*c + 1))*e^(2*
d*x) - (8*d^2*f^3*x^3*e^c + 3*(8*d^2*f^2*e^(c + 1) - d*f^3*e^c)*x^2 - 6*f^2*e^(c + 1) - 6*(d*f^2*e^(c + 1) + f
^3*e^c)*x)*e^(d*x))/(3*I*a*d^3*e^(4*d*x + 4*c) + 6*a*d^3*e^(3*d*x + 3*c) + 6*a*d^3*e^(d*x + c) - 3*I*a*d^3) -
2*f^3*x/(a*d^3) - 5*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*f^2*e/(a*d^3) - 3*(d*x*log(-I*e^(d*x
+ c) + 1) + dilog(I*e^(d*x + c)))*f^2*e/(a*d^3) - 5/2*(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)))*f^3/(a*d^4) - 3/2*(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)))*f^3/(a*d^4) + 2*f^3*log(e^(d*x + c) - I)/(a*d^4) + 4/3*(d^3*f^3*x^3 + 3
*d^3*f^2*x^2*e)/(a*d^4)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1424 vs. \(2 (425) = 850\).
time = 0.37, size = 1424, normalized size = 3.16 \begin {gather*} -\frac {24 \, c d^{2} f e^{2} - 12 \, {\left (2 \, c^{2} - 1\right )} d f^{2} e + 4 \, {\left (2 \, c^{3} - 3 \, c\right )} f^{3} - 8 \, d^{3} e^{3} - 18 \, {\left (d f^{3} x + d f^{2} e - {\left (d f^{3} x + d f^{2} e\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (-i \, d f^{3} x - i \, d f^{2} e\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-i \, d f^{3} x - i \, d f^{2} e\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) - 30 \, {\left (d f^{3} x + d f^{2} e - {\left (d f^{3} x + d f^{2} e\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (-i \, d f^{3} x - i \, d f^{2} e\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-i \, d f^{3} x - i \, d f^{2} e\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 4 \, {\left (2 \, d^{3} f^{3} x^{3} - 3 \, d f^{3} x + {\left (2 \, c^{3} - 3 \, c\right )} f^{3} + 6 \, {\left (d^{3} f x + c d^{2} f\right )} e^{2} + 6 \, {\left (d^{3} f^{2} x^{2} - c^{2} d f^{2}\right )} e\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (8 i \, d^{3} f^{3} x^{3} + 3 i \, d^{2} f^{3} x^{2} - 6 i \, d f^{3} x + 4 \, {\left (2 i \, c^{3} - 3 i \, c\right )} f^{3} + 3 \, {\left (8 i \, d^{3} f x + {\left (8 i \, c + i\right )} d^{2} f\right )} e^{2} + 6 \, {\left (4 i \, d^{3} f^{2} x^{2} + i \, d^{2} f^{2} x + {\left (-4 i \, c^{2} + i\right )} d f^{2}\right )} e\right )} e^{\left (3 \, d x + 3 \, c\right )} + 12 \, {\left (d f^{3} x + d f^{2} e\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (3 i \, d^{2} f^{3} x^{2} - 6 i \, d f^{3} x + 3 \, {\left (8 i \, c + i\right )} d^{2} f e^{2} + 4 \, {\left (2 i \, c^{3} - 3 i \, c\right )} f^{3} - 8 i \, d^{3} e^{3} + 6 \, {\left (i \, d^{2} f^{2} x + {\left (-4 i \, c^{2} + i\right )} d f^{2}\right )} e\right )} e^{\left (d x + c\right )} - 9 \, {\left (c^{2} f^{3} - 2 \, c d f^{2} e + d^{2} f e^{2} - {\left (c^{2} f^{3} - 2 \, c d f^{2} e + d^{2} f e^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (-i \, c^{2} f^{3} + 2 i \, c d f^{2} e - i \, d^{2} f e^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-i \, c^{2} f^{3} + 2 i \, c d f^{2} e - i \, d^{2} f e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + 3 \, {\left (10 \, c d f^{2} e - {\left (5 \, c^{2} - 4\right )} f^{3} - 5 \, d^{2} f e^{2} - {\left (10 \, c d f^{2} e - {\left (5 \, c^{2} - 4\right )} f^{3} - 5 \, d^{2} f e^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (10 i \, c d f^{2} e + {\left (-5 i \, c^{2} + 4 i\right )} f^{3} - 5 i \, d^{2} f e^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )} + 2 \, {\left (10 i \, c d f^{2} e + {\left (-5 i \, c^{2} + 4 i\right )} f^{3} - 5 i \, d^{2} f e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 15 \, {\left (d^{2} f^{3} x^{2} - c^{2} f^{3} + 2 \, {\left (d^{2} f^{2} x + c d f^{2}\right )} e - {\left (d^{2} f^{3} x^{2} - c^{2} f^{3} + 2 \, {\left (d^{2} f^{2} x + c d f^{2}\right )} e\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (-i \, d^{2} f^{3} x^{2} + i \, c^{2} f^{3} + 2 \, {\left (-i \, d^{2} f^{2} x - i \, c d f^{2}\right )} e\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-i \, d^{2} f^{3} x^{2} + i \, c^{2} f^{3} + 2 \, {\left (-i \, d^{2} f^{2} x - i \, c d f^{2}\right )} e\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) - 9 \, {\left (d^{2} f^{3} x^{2} - c^{2} f^{3} + 2 \, {\left (d^{2} f^{2} x + c d f^{2}\right )} e - {\left (d^{2} f^{3} x^{2} - c^{2} f^{3} + 2 \, {\left (d^{2} f^{2} x + c d f^{2}\right )} e\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (-i \, d^{2} f^{3} x^{2} + i \, c^{2} f^{3} + 2 \, {\left (-i \, d^{2} f^{2} x - i \, c d f^{2}\right )} e\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-i \, d^{2} f^{3} x^{2} + i \, c^{2} f^{3} + 2 \, {\left (-i \, d^{2} f^{2} x - i \, c d f^{2}\right )} e\right )} e^{\left (d x + c\right )}\right )} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) - 18 \, {\left (f^{3} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, f^{3} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f^{3} e^{\left (d x + c\right )} - f^{3}\right )} {\rm polylog}\left (3, i \, e^{\left (d x + c\right )}\right ) - 30 \, {\left (f^{3} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, f^{3} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f^{3} e^{\left (d x + c\right )} - f^{3}\right )} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{6 \, {\left (a d^{4} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, a d^{4} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, a d^{4} e^{\left (d x + c\right )} - a d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(24*c*d^2*f*e^2 - 12*(2*c^2 - 1)*d*f^2*e + 4*(2*c^3 - 3*c)*f^3 - 8*d^3*e^3 - 18*(d*f^3*x + d*f^2*e - (d*f
^3*x + d*f^2*e)*e^(4*d*x + 4*c) - 2*(-I*d*f^3*x - I*d*f^2*e)*e^(3*d*x + 3*c) - 2*(-I*d*f^3*x - I*d*f^2*e)*e^(d
*x + c))*dilog(I*e^(d*x + c)) - 30*(d*f^3*x + d*f^2*e - (d*f^3*x + d*f^2*e)*e^(4*d*x + 4*c) - 2*(-I*d*f^3*x -
I*d*f^2*e)*e^(3*d*x + 3*c) - 2*(-I*d*f^3*x - I*d*f^2*e)*e^(d*x + c))*dilog(-I*e^(d*x + c)) - 4*(2*d^3*f^3*x^3
- 3*d*f^3*x + (2*c^3 - 3*c)*f^3 + 6*(d^3*f*x + c*d^2*f)*e^2 + 6*(d^3*f^2*x^2 - c^2*d*f^2)*e)*e^(4*d*x + 4*c) +
2*(8*I*d^3*f^3*x^3 + 3*I*d^2*f^3*x^2 - 6*I*d*f^3*x + 4*(2*I*c^3 - 3*I*c)*f^3 + 3*(8*I*d^3*f*x + (8*I*c + I)*d
^2*f)*e^2 + 6*(4*I*d^3*f^2*x^2 + I*d^2*f^2*x + (-4*I*c^2 + I)*d*f^2)*e)*e^(3*d*x + 3*c) + 12*(d*f^3*x + d*f^2*
e)*e^(2*d*x + 2*c) + 2*(3*I*d^2*f^3*x^2 - 6*I*d*f^3*x + 3*(8*I*c + I)*d^2*f*e^2 + 4*(2*I*c^3 - 3*I*c)*f^3 - 8*
I*d^3*e^3 + 6*(I*d^2*f^2*x + (-4*I*c^2 + I)*d*f^2)*e)*e^(d*x + c) - 9*(c^2*f^3 - 2*c*d*f^2*e + d^2*f*e^2 - (c^
2*f^3 - 2*c*d*f^2*e + d^2*f*e^2)*e^(4*d*x + 4*c) - 2*(-I*c^2*f^3 + 2*I*c*d*f^2*e - I*d^2*f*e^2)*e^(3*d*x + 3*c
) - 2*(-I*c^2*f^3 + 2*I*c*d*f^2*e - I*d^2*f*e^2)*e^(d*x + c))*log(e^(d*x + c) + I) + 3*(10*c*d*f^2*e - (5*c^2
- 4)*f^3 - 5*d^2*f*e^2 - (10*c*d*f^2*e - (5*c^2 - 4)*f^3 - 5*d^2*f*e^2)*e^(4*d*x + 4*c) + 2*(10*I*c*d*f^2*e +
(-5*I*c^2 + 4*I)*f^3 - 5*I*d^2*f*e^2)*e^(3*d*x + 3*c) + 2*(10*I*c*d*f^2*e + (-5*I*c^2 + 4*I)*f^3 - 5*I*d^2*f*e
^2)*e^(d*x + c))*log(e^(d*x + c) - I) - 15*(d^2*f^3*x^2 - c^2*f^3 + 2*(d^2*f^2*x + c*d*f^2)*e - (d^2*f^3*x^2 -
c^2*f^3 + 2*(d^2*f^2*x + c*d*f^2)*e)*e^(4*d*x + 4*c) - 2*(-I*d^2*f^3*x^2 + I*c^2*f^3 + 2*(-I*d^2*f^2*x - I*c*
d*f^2)*e)*e^(3*d*x + 3*c) - 2*(-I*d^2*f^3*x^2 + I*c^2*f^3 + 2*(-I*d^2*f^2*x - I*c*d*f^2)*e)*e^(d*x + c))*log(I
*e^(d*x + c) + 1) - 9*(d^2*f^3*x^2 - c^2*f^3 + 2*(d^2*f^2*x + c*d*f^2)*e - (d^2*f^3*x^2 - c^2*f^3 + 2*(d^2*f^2
*x + c*d*f^2)*e)*e^(4*d*x + 4*c) - 2*(-I*d^2*f^3*x^2 + I*c^2*f^3 + 2*(-I*d^2*f^2*x - I*c*d*f^2)*e)*e^(3*d*x +
3*c) - 2*(-I*d^2*f^3*x^2 + I*c^2*f^3 + 2*(-I*d^2*f^2*x - I*c*d*f^2)*e)*e^(d*x + c))*log(-I*e^(d*x + c) + 1) -
18*(f^3*e^(4*d*x + 4*c) - 2*I*f^3*e^(3*d*x + 3*c) - 2*I*f^3*e^(d*x + c) - f^3)*polylog(3, I*e^(d*x + c)) - 30*
(f^3*e^(4*d*x + 4*c) - 2*I*f^3*e^(3*d*x + 3*c) - 2*I*f^3*e^(d*x + c) - f^3)*polylog(3, -I*e^(d*x + c)))/(a*d^4
*e^(4*d*x + 4*c) - 2*I*a*d^4*e^(3*d*x + 3*c) - 2*I*a*d^4*e^(d*x + c) - a*d^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \left (\int \frac {e^{3} \operatorname {sech}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{3} x^{3} \operatorname {sech}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e f^{2} x^{2} \operatorname {sech}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e^{2} f x \operatorname {sech}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^3}{{\mathrm {cosh}\left (c+d\,x\right )}^2\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^3/(cosh(c + d*x)^2*(a + a*sinh(c + d*x)*1i)),x)

[Out]

int((e + f*x)^3/(cosh(c + d*x)^2*(a + a*sinh(c + d*x)*1i)), x)

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