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

Optimal. Leaf size=233 \[ -\frac{3 i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{8 a d^2}+\frac{3 i f \text{PolyLog}\left (2,i e^{c+d x}\right )}{8 a d^2}+\frac{i f \tanh ^3(c+d x)}{12 a d^2}-\frac{i f \tanh (c+d x)}{4 a d^2}+\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{3 f \text{sech}(c+d x)}{8 a d^2}+\frac{3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{4 a d}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}+\frac{(e+f x) \tanh (c+d x) \text{sech}^3(c+d x)}{4 a d}+\frac{3 (e+f x) \tanh (c+d x) \text{sech}(c+d x)}{8 a d} \]

[Out]

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

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Rubi [A]  time = 0.189825, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {5571, 4185, 4180, 2279, 2391, 5451, 3767} \[ -\frac{3 i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{8 a d^2}+\frac{3 i f \text{PolyLog}\left (2,i e^{c+d x}\right )}{8 a d^2}+\frac{i f \tanh ^3(c+d x)}{12 a d^2}-\frac{i f \tanh (c+d x)}{4 a d^2}+\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{3 f \text{sech}(c+d x)}{8 a d^2}+\frac{3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{4 a d}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}+\frac{(e+f x) \tanh (c+d x) \text{sech}^3(c+d x)}{4 a d}+\frac{3 (e+f x) \tanh (c+d x) \text{sech}(c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4185

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

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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{(e+f x) \text{sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac{i \int (e+f x) \text{sech}^4(c+d x) \tanh (c+d x) \, dx}{a}+\frac{\int (e+f x) \text{sech}^5(c+d x) \, dx}{a}\\ &=\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}+\frac{(e+f x) \text{sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac{3 \int (e+f x) \text{sech}^3(c+d x) \, dx}{4 a}-\frac{(i f) \int \text{sech}^4(c+d x) \, dx}{4 a d}\\ &=\frac{3 f \text{sech}(c+d x)}{8 a d^2}+\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}+\frac{3 (e+f x) \text{sech}(c+d x) \tanh (c+d x)}{8 a d}+\frac{(e+f x) \text{sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac{3 \int (e+f x) \text{sech}(c+d x) \, dx}{8 a}+\frac{f \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-i \tanh (c+d x)\right )}{4 a d^2}\\ &=\frac{3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{4 a d}+\frac{3 f \text{sech}(c+d x)}{8 a d^2}+\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}-\frac{i f \tanh (c+d x)}{4 a d^2}+\frac{3 (e+f x) \text{sech}(c+d x) \tanh (c+d x)}{8 a d}+\frac{(e+f x) \text{sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac{i f \tanh ^3(c+d x)}{12 a d^2}-\frac{(3 i f) \int \log \left (1-i e^{c+d x}\right ) \, dx}{8 a d}+\frac{(3 i f) \int \log \left (1+i e^{c+d x}\right ) \, dx}{8 a d}\\ &=\frac{3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{4 a d}+\frac{3 f \text{sech}(c+d x)}{8 a d^2}+\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}-\frac{i f \tanh (c+d x)}{4 a d^2}+\frac{3 (e+f x) \text{sech}(c+d x) \tanh (c+d x)}{8 a d}+\frac{(e+f x) \text{sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac{i f \tanh ^3(c+d x)}{12 a d^2}-\frac{(3 i f) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{8 a d^2}+\frac{(3 i f) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{8 a d^2}\\ &=\frac{3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{4 a d}-\frac{3 i f \text{Li}_2\left (-i e^{c+d x}\right )}{8 a d^2}+\frac{3 i f \text{Li}_2\left (i e^{c+d x}\right )}{8 a d^2}+\frac{3 f \text{sech}(c+d x)}{8 a d^2}+\frac{f \text{sech}^3(c+d x)}{12 a d^2}+\frac{i (e+f x) \text{sech}^4(c+d x)}{4 a d}-\frac{i f \tanh (c+d x)}{4 a d^2}+\frac{3 (e+f x) \text{sech}(c+d x) \tanh (c+d x)}{8 a d}+\frac{(e+f x) \text{sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac{i f \tanh ^3(c+d x)}{12 a d^2}\\ \end{align*}

Mathematica [B]  time = 6.66753, size = 1290, normalized size = 5.54 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/24)*(6*d*e - I*f - 6*c*f + 6*f*(c + d*x)))/(d^2*(a + I*a*Sinh[c + d*x])) + ((I/8)*(d*e - c*f + f*(c + d*x)
))/(d^2*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2*(a + I*a*Sinh[c + d*x])) + (3*(c + d*x)*(2*d*e - 2*c*f + f
*(c + d*x))*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(16*d^2*(a + I*a*Sinh[c + d*x])) + (((3*I)/8)*e*((I/2
)*(c + d*x) + Log[Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(d*(a
+ I*a*Sinh[c + d*x])) - (((3*I)/8)*c*f*((I/2)*(c + d*x) + Log[Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2]])*(Cosh
[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(d^2*(a + I*a*Sinh[c + d*x])) - (((3*I)/8)*e*((-I/2)*(c + d*x) + Log[C
osh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(d*(a + I*a*Sinh[c + d*x
])) + (((3*I)/8)*c*f*((-I/2)*(c + d*x) + Log[Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*
Sinh[(c + d*x)/2])^2)/(d^2*(a + I*a*Sinh[c + d*x])) + (3*f*(-(c + d*x)^2/(4*E^((I/4)*Pi)) - ((3*Pi*(c + d*x))/
4 - Pi*Log[1 + E^(c + d*x)] - 2*(-Pi/4 + (I/2)*(c + d*x))*Log[1 - E^((2*I)*(-Pi/4 + (I/2)*(c + d*x)))] + Pi*Lo
g[Cosh[(c + d*x)/2]] - (Pi*Log[-Sin[Pi/4 - (I/2)*(c + d*x)]])/2 + I*PolyLog[2, E^((2*I)*(-Pi/4 + (I/2)*(c + d*
x)))])/Sqrt[2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(4*Sqrt[2]*d^2*(a + I*a*Sinh[c + d*x])) + (3*f*(-
(E^((I/4)*Pi)*(c + d*x)^2)/4 + ((Pi*(c + d*x))/4 - Pi*Log[1 + E^(c + d*x)] - 2*(Pi/4 + (I/2)*(c + d*x))*Log[1
- E^((2*I)*(Pi/4 + (I/2)*(c + d*x)))] + Pi*Log[Cosh[(c + d*x)/2]] + (Pi*Log[Sin[Pi/4 + (I/2)*(c + d*x)]])/2 +
I*PolyLog[2, E^((2*I)*(Pi/4 + (I/2)*(c + d*x)))])/Sqrt[2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(4*Sqr
t[2]*d^2*(a + I*a*Sinh[c + d*x])) - ((I/8)*(d*e - c*f + f*(c + d*x))*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])
^2)/(d^2*(Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2])^2*(a + I*a*Sinh[c + d*x])) - ((I/12)*f*Sinh[(c + d*x)/2])/(
d^2*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(a + I*a*Sinh[c + d*x])) - (((7*I)/12)*f*(Cosh[(c + d*x)/2] + I*
Sinh[(c + d*x)/2])*Sinh[(c + d*x)/2])/(d^2*(a + I*a*Sinh[c + d*x])) + ((I/4)*f*(Cosh[(c + d*x)/2] + I*Sinh[(c
+ d*x)/2])^2*Sinh[(c + d*x)/2])/(d^2*(Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2])*(a + I*a*Sinh[c + d*x]))

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Maple [B]  time = 0.213, size = 445, normalized size = 1.9 \begin{align*}{\frac{9\,f{{\rm e}^{5\,dx+5\,c}}+8\,f{{\rm e}^{3\,dx+3\,c}}-f{{\rm e}^{dx+c}}-18\,if{{\rm e}^{4\,dx+4\,c}}-18\,idfx{{\rm e}^{4\,dx+4\,c}}-18\,ide{{\rm e}^{4\,dx+4\,c}}-22\,if{{\rm e}^{2\,dx+2\,c}}+6\,de{{\rm e}^{3\,dx+3\,c}}+9\,dfx{{\rm e}^{5\,dx+5\,c}}+6\,dfx{{\rm e}^{3\,dx+3\,c}}+18\,idfx{{\rm e}^{2\,dx+2\,c}}-4\,if+9\,dfx{{\rm e}^{dx+c}}+18\,ide{{\rm e}^{2\,dx+2\,c}}+9\,de{{\rm e}^{5\,dx+5\,c}}+9\,de{{\rm e}^{dx+c}}}{12\, \left ({{\rm e}^{dx+c}}+i \right ) ^{2} \left ({{\rm e}^{dx+c}}-i \right ) ^{4}{d}^{2}a}}-{\frac{{\frac{3\,i}{8}}e\ln \left ({{\rm e}^{dx+c}}-i \right ) }{da}}+{\frac{{\frac{3\,i}{8}}e\ln \left ({{\rm e}^{dx+c}}+i \right ) }{da}}-{\frac{{\frac{3\,i}{8}}f\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{da}}-{\frac{{\frac{3\,i}{8}}f\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{2}}}-{\frac{{\frac{3\,i}{8}}f{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{{\frac{3\,i}{8}}f\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) x}{da}}+{\frac{{\frac{3\,i}{8}}f\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) c}{a{d}^{2}}}+{\frac{{\frac{3\,i}{8}}f{\it polylog} \left ( 2,i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{{\frac{3\,i}{8}}fc\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}}-{\frac{{\frac{3\,i}{8}}fc\ln \left ({{\rm e}^{dx+c}}+i \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(9*f*exp(5*d*x+5*c)+8*f*exp(3*d*x+3*c)-f*exp(d*x+c)-18*I*f*exp(4*d*x+4*c)-18*I*d*f*x*exp(4*d*x+4*c)-18*I*
d*e*exp(4*d*x+4*c)-22*I*f*exp(2*d*x+2*c)+6*d*e*exp(3*d*x+3*c)+9*d*f*x*exp(5*d*x+5*c)+6*d*f*x*exp(3*d*x+3*c)+18
*I*d*f*x*exp(2*d*x+2*c)-4*I*f+9*d*f*x*exp(d*x+c)+18*I*d*e*exp(2*d*x+2*c)+9*d*e*exp(5*d*x+5*c)+9*d*e*exp(d*x+c)
)/(exp(d*x+c)+I)^2/(exp(d*x+c)-I)^4/d^2/a-3/8*I/a/d*e*ln(exp(d*x+c)-I)+3/8*I/a/d*e*ln(exp(d*x+c)+I)-3/8*I/a/d*
f*ln(1+I*exp(d*x+c))*x-3/8*I/a/d^2*f*ln(1+I*exp(d*x+c))*c-3/8*I*f*polylog(2,-I*exp(d*x+c))/a/d^2+3/8*I/a/d*f*l
n(1-I*exp(d*x+c))*x+3/8*I/a/d^2*f*ln(1-I*exp(d*x+c))*c+3/8*I*f*polylog(2,I*exp(d*x+c))/a/d^2+3/8*I/a/d^2*f*c*l
n(exp(d*x+c)-I)-3/8*I/a/d^2*f*c*ln(exp(d*x+c)+I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 8 \, f{\left (\frac{9 \,{\left (d x e^{\left (5 \, c\right )} + e^{\left (5 \, c\right )}\right )} e^{\left (5 \, d x\right )} +{\left (-18 i \, d x e^{\left (4 \, c\right )} - 18 i \, e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \,{\left (3 \, d x e^{\left (3 \, c\right )} + 4 \, e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} +{\left (18 i \, d x e^{\left (2 \, c\right )} - 22 i \, e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} +{\left (9 \, d x e^{c} - e^{c}\right )} e^{\left (d x\right )} - 4 i}{96 \, a d^{2} e^{\left (6 \, d x + 6 \, c\right )} - 192 i \, a d^{2} e^{\left (5 \, d x + 5 \, c\right )} + 96 \, a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 384 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 96 \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 192 i \, a d^{2} e^{\left (d x + c\right )} - 96 \, a d^{2}} + 3 \, \int \frac{x}{64 \,{\left (a e^{\left (d x + c\right )} + i \, a\right )}}\,{d x} + 3 \, \int \frac{x}{64 \,{\left (a e^{\left (d x + c\right )} - i \, a\right )}}\,{d x}\right )} - \frac{1}{8} \, e{\left (\frac{64 \,{\left (3 \, e^{\left (-d x - c\right )} - 6 i \, e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + 6 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}\right )}}{{\left (64 i \, a e^{\left (-d x - c\right )} - 32 \, a e^{\left (-2 \, d x - 2 \, c\right )} + 128 i \, a e^{\left (-3 \, d x - 3 \, c\right )} + 32 \, a e^{\left (-4 \, d x - 4 \, c\right )} + 64 i \, a e^{\left (-5 \, d x - 5 \, c\right )} + 32 \, a e^{\left (-6 \, d x - 6 \, c\right )} - 32 \, a\right )} d} + \frac{3 i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{a d} - \frac{3 i \, \log \left (e^{\left (-d x - c\right )} - i\right )}{a d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*f*((9*(d*x*e^(5*c) + e^(5*c))*e^(5*d*x) + (-18*I*d*x*e^(4*c) - 18*I*e^(4*c))*e^(4*d*x) + 2*(3*d*x*e^(3*c) +
4*e^(3*c))*e^(3*d*x) + (18*I*d*x*e^(2*c) - 22*I*e^(2*c))*e^(2*d*x) + (9*d*x*e^c - e^c)*e^(d*x) - 4*I)/(96*a*d^
2*e^(6*d*x + 6*c) - 192*I*a*d^2*e^(5*d*x + 5*c) + 96*a*d^2*e^(4*d*x + 4*c) - 384*I*a*d^2*e^(3*d*x + 3*c) - 96*
a*d^2*e^(2*d*x + 2*c) - 192*I*a*d^2*e^(d*x + c) - 96*a*d^2) + 3*integrate(1/64*x/(a*e^(d*x + c) + I*a), x) + 3
*integrate(1/64*x/(a*e^(d*x + c) - I*a), x)) - 1/8*e*(64*(3*e^(-d*x - c) - 6*I*e^(-2*d*x - 2*c) + 2*e^(-3*d*x
- 3*c) + 6*I*e^(-4*d*x - 4*c) + 3*e^(-5*d*x - 5*c))/((64*I*a*e^(-d*x - c) - 32*a*e^(-2*d*x - 2*c) + 128*I*a*e^
(-3*d*x - 3*c) + 32*a*e^(-4*d*x - 4*c) + 64*I*a*e^(-5*d*x - 5*c) + 32*a*e^(-6*d*x - 6*c) - 32*a)*d) + 3*I*log(
e^(-d*x - c) + I)/(a*d) - 3*I*log(e^(-d*x - c) - I)/(a*d))

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Fricas [B]  time = 2.44546, size = 2423, normalized size = 10.4 \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)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

((9*I*f*e^(6*d*x + 6*c) + 18*f*e^(5*d*x + 5*c) + 9*I*f*e^(4*d*x + 4*c) + 36*f*e^(3*d*x + 3*c) - 9*I*f*e^(2*d*x
+ 2*c) + 18*f*e^(d*x + c) - 9*I*f)*dilog(I*e^(d*x + c)) + (-9*I*f*e^(6*d*x + 6*c) - 18*f*e^(5*d*x + 5*c) - 9*
I*f*e^(4*d*x + 4*c) - 36*f*e^(3*d*x + 3*c) + 9*I*f*e^(2*d*x + 2*c) - 18*f*e^(d*x + c) + 9*I*f)*dilog(-I*e^(d*x
+ c)) + 18*(d*f*x + d*e + f)*e^(5*d*x + 5*c) + (-36*I*d*f*x - 36*I*d*e - 36*I*f)*e^(4*d*x + 4*c) + 4*(3*d*f*x
+ 3*d*e + 4*f)*e^(3*d*x + 3*c) + (36*I*d*f*x + 36*I*d*e - 44*I*f)*e^(2*d*x + 2*c) + 2*(9*d*f*x + 9*d*e - f)*e
^(d*x + c) + (-9*I*d*e + 9*I*c*f + (9*I*d*e - 9*I*c*f)*e^(6*d*x + 6*c) + 18*(d*e - c*f)*e^(5*d*x + 5*c) + (9*I
*d*e - 9*I*c*f)*e^(4*d*x + 4*c) + 36*(d*e - c*f)*e^(3*d*x + 3*c) + (-9*I*d*e + 9*I*c*f)*e^(2*d*x + 2*c) + 18*(
d*e - c*f)*e^(d*x + c))*log(e^(d*x + c) + I) + (9*I*d*e - 9*I*c*f + (-9*I*d*e + 9*I*c*f)*e^(6*d*x + 6*c) - 18*
(d*e - c*f)*e^(5*d*x + 5*c) + (-9*I*d*e + 9*I*c*f)*e^(4*d*x + 4*c) - 36*(d*e - c*f)*e^(3*d*x + 3*c) + (9*I*d*e
- 9*I*c*f)*e^(2*d*x + 2*c) - 18*(d*e - c*f)*e^(d*x + c))*log(e^(d*x + c) - I) + (9*I*d*f*x + 9*I*c*f + (-9*I*
d*f*x - 9*I*c*f)*e^(6*d*x + 6*c) - 18*(d*f*x + c*f)*e^(5*d*x + 5*c) + (-9*I*d*f*x - 9*I*c*f)*e^(4*d*x + 4*c) -
36*(d*f*x + c*f)*e^(3*d*x + 3*c) + (9*I*d*f*x + 9*I*c*f)*e^(2*d*x + 2*c) - 18*(d*f*x + c*f)*e^(d*x + c))*log(
I*e^(d*x + c) + 1) + (-9*I*d*f*x - 9*I*c*f + (9*I*d*f*x + 9*I*c*f)*e^(6*d*x + 6*c) + 18*(d*f*x + c*f)*e^(5*d*x
+ 5*c) + (9*I*d*f*x + 9*I*c*f)*e^(4*d*x + 4*c) + 36*(d*f*x + c*f)*e^(3*d*x + 3*c) + (-9*I*d*f*x - 9*I*c*f)*e^
(2*d*x + 2*c) + 18*(d*f*x + c*f)*e^(d*x + c))*log(-I*e^(d*x + c) + 1) - 8*I*f)/(24*a*d^2*e^(6*d*x + 6*c) - 48*
I*a*d^2*e^(5*d*x + 5*c) + 24*a*d^2*e^(4*d*x + 4*c) - 96*I*a*d^2*e^(3*d*x + 3*c) - 24*a*d^2*e^(2*d*x + 2*c) - 4
8*I*a*d^2*e^(d*x + c) - 24*a*d^2)

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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)*sech(d*x+c)**3/(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 )} \operatorname{sech}\left (d x + c\right )^{3}}{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)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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