3.272 \(\int \frac{(e+f x)^2 \text{sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=268 \[ -\frac{i f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac{i f (e+f x) \text{PolyLog}\left (2,i e^{c+d x}\right )}{a d^2}+\frac{i f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac{i f^2 \text{PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac{i f (e+f x) \tanh (c+d x)}{a d^2}+\frac{f (e+f x) \text{sech}(c+d x)}{a d^2}-\frac{f^2 \tan ^{-1}(\sinh (c+d x))}{a d^3}+\frac{i f^2 \log (\cosh (c+d x))}{a d^3}+\frac{(e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{i (e+f x)^2 \text{sech}^2(c+d x)}{2 a d}+\frac{(e+f x)^2 \tanh (c+d x) \text{sech}(c+d x)}{2 a d} \]

[Out]

((e + f*x)^2*ArcTan[E^(c + d*x)])/(a*d) - (f^2*ArcTan[Sinh[c + d*x]])/(a*d^3) + (I*f^2*Log[Cosh[c + d*x]])/(a*
d^3) - (I*f*(e + f*x)*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^2) + (I*f*(e + f*x)*PolyLog[2, I*E^(c + d*x)])/(a*d^2
) + (I*f^2*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^3) - (I*f^2*PolyLog[3, I*E^(c + d*x)])/(a*d^3) + (f*(e + f*x)*Se
ch[c + d*x])/(a*d^2) + ((I/2)*(e + f*x)^2*Sech[c + d*x]^2)/(a*d) - (I*f*(e + f*x)*Tanh[c + d*x])/(a*d^2) + ((e
+ f*x)^2*Sech[c + d*x]*Tanh[c + d*x])/(2*a*d)

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Rubi [A]  time = 0.263905, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {5571, 4186, 3770, 4180, 2531, 2282, 6589, 5451, 4184, 3475} \[ -\frac{i f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac{i f (e+f x) \text{PolyLog}\left (2,i e^{c+d x}\right )}{a d^2}+\frac{i f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac{i f^2 \text{PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac{i f (e+f x) \tanh (c+d x)}{a d^2}+\frac{f (e+f x) \text{sech}(c+d x)}{a d^2}-\frac{f^2 \tan ^{-1}(\sinh (c+d x))}{a d^3}+\frac{i f^2 \log (\cosh (c+d x))}{a d^3}+\frac{(e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{i (e+f x)^2 \text{sech}^2(c+d x)}{2 a d}+\frac{(e+f x)^2 \tanh (c+d x) \text{sech}(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)^2*ArcTan[E^(c + d*x)])/(a*d) - (f^2*ArcTan[Sinh[c + d*x]])/(a*d^3) + (I*f^2*Log[Cosh[c + d*x]])/(a*
d^3) - (I*f*(e + f*x)*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^2) + (I*f*(e + f*x)*PolyLog[2, I*E^(c + d*x)])/(a*d^2
) + (I*f^2*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^3) - (I*f^2*PolyLog[3, I*E^(c + d*x)])/(a*d^3) + (f*(e + f*x)*Se
ch[c + d*x])/(a*d^2) + ((I/2)*(e + f*x)^2*Sech[c + d*x]^2)/(a*d) - (I*f*(e + f*x)*Tanh[c + d*x])/(a*d^2) + ((e
+ f*x)^2*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 3770

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

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 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 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 3475

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

Rubi steps

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

Mathematica [A]  time = 11.3621, size = 501, normalized size = 1.87 \[ -\frac{\frac{-6 \left (1+i e^c\right ) e f \text{PolyLog}\left (2,i e^{-c-d x}\right )-6 \left (1+i e^c\right ) f^2 \left (x \text{PolyLog}\left (2,i e^{-c-d x}\right )+\frac{\text{PolyLog}\left (3,i e^{-c-d x}\right )}{d}\right )-\frac{3 \left (1+i e^c\right ) \left (d^2 e^2-4 f^2\right ) \left (d x-\log \left (-e^{c+d x}+i\right )\right )}{d}+6 \left (1+i e^c\right ) d e f x \log \left (1-i e^{-c-d x}\right )+3 \left (1+i e^c\right ) d f^2 x^2 \log \left (1-i e^{-c-d x}\right )+3 x \left (d^2 e^2-4 f^2\right )+3 d^2 e f x^2+d^2 f^2 x^3}{\left (e^c-i\right ) d^2}+\frac{\frac{6 i \left (e^c+i\right ) f \left (d (e+f x) \text{PolyLog}\left (2,-i e^{-c-d x}\right )+f \text{PolyLog}\left (3,-i e^{-c-d x}\right )\right )}{d^3}+\frac{3 \left (1-i e^c\right ) (e+f x)^2 \log \left (1+i e^{-c-d x}\right )}{d}+\frac{(e+f x)^3}{f}}{e^c+i}+\frac{12 i f \sinh \left (\frac{d x}{2}\right ) (e+f x)}{d^2 \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{3 i (e+f x)^2}{d \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2}-x \text{sech}(c) \left (3 e^2+3 e f x+f^2 x^2\right )}{6 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.132, size = 613, normalized size = 2.3 \begin{align*}{\frac{d{x}^{2}{f}^{2}{{\rm e}^{dx+c}}+2\,defx{{\rm e}^{dx+c}}+d{e}^{2}{{\rm e}^{dx+c}}-2\,i{f}^{2}x+2\,{f}^{2}x{{\rm e}^{dx+c}}-2\,ief+2\,ef{{\rm e}^{dx+c}}}{ \left ({{\rm e}^{dx+c}}-i \right ) ^{2}{d}^{2}a}}-{\frac{ief{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{i\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) efx}{da}}+{\frac{{\frac{i}{2}}{e}^{2}\ln \left ({{\rm e}^{dx+c}}+i \right ) }{da}}-{\frac{{\frac{i}{2}}{e}^{2}\ln \left ({{\rm e}^{dx+c}}-i \right ) }{da}}-{\frac{{\frac{i}{2}}{c}^{2}{f}^{2}\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-{\frac{iefc\ln \left ({{\rm e}^{dx+c}}+i \right ) }{a{d}^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( 1-i{{\rm e}^{dx+c}} \right ){c}^{2}{f}^{2}}{a{d}^{3}}}+{\frac{i{f}^{2}{\it polylog} \left ( 3,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}-{\frac{2\,i{f}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{\frac{i}{2}}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ){c}^{2}{f}^{2}}{a{d}^{3}}}+{\frac{i\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) cef}{a{d}^{2}}}+{\frac{i{\it polylog} \left ( 2,i{{\rm e}^{dx+c}} \right ){f}^{2}x}{a{d}^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ){f}^{2}{x}^{2}}{da}}-{\frac{i{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ){f}^{2}x}{a{d}^{2}}}+{\frac{i\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) efx}{da}}+{\frac{{\frac{i}{2}}\ln \left ( 1-i{{\rm e}^{dx+c}} \right ){f}^{2}{x}^{2}}{da}}+{\frac{{\frac{i}{2}}{c}^{2}{f}^{2}\ln \left ({{\rm e}^{dx+c}}+i \right ) }{a{d}^{3}}}+{\frac{2\,i{f}^{2}\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-{\frac{i\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) cef}{a{d}^{2}}}+{\frac{iefc\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}}-{\frac{i{f}^{2}{\it polylog} \left ( 3,i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{ief{\it polylog} \left ( 2,i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.99471, size = 522, normalized size = 1.95 \begin{align*} -\frac{1}{2} \, e^{2}{\left (\frac{4 \, e^{\left (-d x - c\right )}}{{\left (4 i \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, 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{-2 i \, f^{2} x - 2 i \, e f +{\left (d f^{2} x^{2} e^{c} + 2 \, e f e^{c} + 2 \,{\left (d e f + 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{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 f}{a d^{2}} + \frac{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 f}{a d^{2}} - \frac{2 i \, f^{2} x}{a d^{2}} - \frac{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 )} f^{2}}{2 \, a d^{3}} + \frac{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 )} f^{2}}{2 \, a d^{3}} + \frac{2 i \, f^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^2*(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)) + (-2*I*f^2*x - 2*I*e*f + (d*f^2*x^2*e^c + 2*e*f*e^c + 2*(d*e*f + 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) - I*(d*x*log(I*e^(d*x + c) + 1) + dilog(-
I*e^(d*x + c)))*e*f/(a*d^2) + I*(d*x*log(-I*e^(d*x + c) + 1) + dilog(I*e^(d*x + c)))*e*f/(a*d^2) - 2*I*f^2*x/(
a*d^2) - 1/2*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)))*f
^2/(a*d^3) + 1/2*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))
)*f^2/(a*d^3) + 2*I*f^2*log(I*e^(d*x + c) + 1)/(a*d^3)

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Fricas [C]  time = 2.38792, size = 1958, normalized size = 7.31 \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)^2*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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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)**2*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 )}^{2} \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)^2*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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