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

Optimal. Leaf size=325 \[ -\frac{f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{3 a d^3}+\frac{f^2 \text{PolyLog}\left (2,i e^{c+d x}\right )}{3 a d^3}-\frac{2 f^2 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{3 a d^3}-\frac{4 f (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{3 a d^2}-\frac{2 i f (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{3 a d^2}+\frac{f (e+f x) \text{sech}^2(c+d x)}{3 a d^2}-\frac{i f (e+f x) \tanh (c+d x) \text{sech}(c+d x)}{3 a d^2}-\frac{f^2 \tanh (c+d x)}{3 a d^3}-\frac{i f^2 \text{sech}(c+d x)}{3 a d^3}+\frac{2 (e+f x)^2 \tanh (c+d x)}{3 a d}+\frac{i (e+f x)^2 \text{sech}^3(c+d x)}{3 a d}+\frac{(e+f x)^2 \tanh (c+d x) \text{sech}^2(c+d x)}{3 a d}+\frac{2 (e+f x)^2}{3 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.386601, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.387, Rules used = {5571, 4186, 3767, 8, 4184, 3718, 2190, 2279, 2391, 5451, 4185, 4180} \[ -\frac{f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{3 a d^3}+\frac{f^2 \text{PolyLog}\left (2,i e^{c+d x}\right )}{3 a d^3}-\frac{2 f^2 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{3 a d^3}-\frac{4 f (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{3 a d^2}-\frac{2 i f (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{3 a d^2}+\frac{f (e+f x) \text{sech}^2(c+d x)}{3 a d^2}-\frac{i f (e+f x) \tanh (c+d x) \text{sech}(c+d x)}{3 a d^2}-\frac{f^2 \tanh (c+d x)}{3 a d^3}-\frac{i f^2 \text{sech}(c+d x)}{3 a d^3}+\frac{2 (e+f x)^2 \tanh (c+d x)}{3 a d}+\frac{i (e+f x)^2 \text{sech}^3(c+d x)}{3 a d}+\frac{(e+f x)^2 \tanh (c+d x) \text{sech}^2(c+d x)}{3 a d}+\frac{2 (e+f x)^2}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4184

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

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 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]

Rubi steps

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

Mathematica [A]  time = 8.26164, size = 564, normalized size = 1.74 \[ \frac{12 f^2 \text{PolyLog}\left (2,-i e^{-c-d x}\right )+20 f^2 \text{PolyLog}\left (2,i e^{-c-d x}\right )+\frac{d^2 e^2 \sinh (2 (c+d x))-2 i d^2 e^2 \cosh (c+d x)+4 i d^2 e^2 \cosh (c+2 d x)+2 d^2 e f x \sinh (2 (c+d x))-4 i d^2 e f x \cosh (c+d x)+8 i d^2 e f x \cosh (c+2 d x)+d^2 f^2 x^2 \sinh (2 (c+d x))-2 i d^2 f^2 x^2 \cosh (c+d x)+4 i d^2 f^2 x^2 \cosh (c+2 d x)+2 d e f \cosh (2 c+d x)-2 f^2 \sinh (2 (c+d x))+2 f^2 \sinh (2 c+d x)+2 d f^2 x \cosh (2 c+d x)+4 i f^2 \cosh (c+d x)-2 i f^2 \cosh (c+2 d x)-2 i f^2 \cosh (c)+8 d^2 e^2 \sinh (d x)+16 d^2 e f x \sinh (d x)+8 d^2 f^2 x^2 \sinh (d x)+2 d f \cosh (d x) (e+f x)-2 f^2 \sinh (d x)}{\left (\cosh \left (\frac{c}{2}\right )-i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )-i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{10 i d (e+f x) \left (d (e+f x)+2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right )\right )}{e^c-i}-\frac{6 i d (e+f x) \left (d (e+f x)+2 \left (1-i e^c\right ) f \log \left (1+i e^{-c-d x}\right )\right )}{e^c+i}}{12 a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((10*I)*d*(e + f*x)*(d*(e + f*x) + 2*(1 + I*E^c)*f*Log[1 - I*E^(-c - d*x)]))/(-I + E^c) - ((6*I)*d*(e + f*x)*
(d*(e + f*x) + 2*(1 - I*E^c)*f*Log[1 + I*E^(-c - d*x)]))/(I + E^c) + 12*f^2*PolyLog[2, (-I)*E^(-c - d*x)] + 20
*f^2*PolyLog[2, I*E^(-c - d*x)] + ((-2*I)*f^2*Cosh[c] + 2*d*f*(e + f*x)*Cosh[d*x] - (2*I)*d^2*e^2*Cosh[c + d*x
] + (4*I)*f^2*Cosh[c + d*x] - (4*I)*d^2*e*f*x*Cosh[c + d*x] - (2*I)*d^2*f^2*x^2*Cosh[c + d*x] + 2*d*e*f*Cosh[2
*c + d*x] + 2*d*f^2*x*Cosh[2*c + d*x] + (4*I)*d^2*e^2*Cosh[c + 2*d*x] - (2*I)*f^2*Cosh[c + 2*d*x] + (8*I)*d^2*
e*f*x*Cosh[c + 2*d*x] + (4*I)*d^2*f^2*x^2*Cosh[c + 2*d*x] + 8*d^2*e^2*Sinh[d*x] - 2*f^2*Sinh[d*x] + 16*d^2*e*f
*x*Sinh[d*x] + 8*d^2*f^2*x^2*Sinh[d*x] + d^2*e^2*Sinh[2*(c + d*x)] - 2*f^2*Sinh[2*(c + d*x)] + 2*d^2*e*f*x*Sin
h[2*(c + d*x)] + d^2*f^2*x^2*Sinh[2*(c + d*x)] + 2*f^2*Sinh[2*c + d*x])/((Cosh[c/2] - I*Sinh[c/2])*(Cosh[c/2]
+ I*Sinh[c/2])*(Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^3))/(12*a*d
^3)

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Maple [A]  time = 0.154, size = 509, normalized size = 1.6 \begin{align*}{\frac{{\frac{2\,i}{3}} \left ( -2\,i{d}^{2}{f}^{2}{x}^{2}+4\,{d}^{2}{x}^{2}{f}^{2}{{\rm e}^{dx+c}}-d{f}^{2}x{{\rm e}^{3\,dx+3\,c}}-4\,i{d}^{2}efx+8\,{d}^{2}efx{{\rm e}^{dx+c}}-def{{\rm e}^{3\,dx+3\,c}}-2\,i{d}^{2}{e}^{2}+i{f}^{2}{{\rm e}^{2\,dx+2\,c}}+4\,{d}^{2}{e}^{2}{{\rm e}^{dx+c}}-d{f}^{2}x{{\rm e}^{dx+c}}-{f}^{2}{{\rm e}^{3\,dx+3\,c}}-def{{\rm e}^{dx+c}}+i{f}^{2}-{f}^{2}{{\rm e}^{dx+c}} \right ) }{ \left ({{\rm e}^{dx+c}}+i \right ) \left ({{\rm e}^{dx+c}}-i \right ) ^{3}{d}^{3}a}}-{\frac{5\,f\ln \left ({{\rm e}^{dx+c}}-i \right ) e}{3\,a{d}^{2}}}-{\frac{ef\ln \left ({{\rm e}^{dx+c}}+i \right ) }{a{d}^{2}}}+{\frac{8\,f\ln \left ({{\rm e}^{dx+c}} \right ) e}{3\,a{d}^{2}}}+{\frac{5\,{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{3\,{d}^{3}a}}-{\frac{8\,{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{3\,{d}^{3}a}}+{\frac{{f}^{2}c\ln \left ({{\rm e}^{dx+c}}+i \right ) }{{d}^{3}a}}+{\frac{4\,{x}^{2}{f}^{2}}{3\,da}}+{\frac{8\,{f}^{2}cx}{3\,a{d}^{2}}}+{\frac{4\,{c}^{2}{f}^{2}}{3\,{d}^{3}a}}-{\frac{5\,{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{3\,a{d}^{2}}}-{\frac{5\,{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{3\,{d}^{3}a}}-{\frac{5\,{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{3\,{d}^{3}a}}-{\frac{{f}^{2}\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}-{\frac{{f}^{2}\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) c}{{d}^{3}a}}-{\frac{{f}^{2}{\it polylog} \left ( 2,i{{\rm e}^{dx+c}} \right ) }{{d}^{3}a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 4 \, f^{2}{\left (\frac{2 i \, d^{2} x^{2} +{\left (d x e^{\left (3 \, c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} -{\left (4 \, d^{2} x^{2} e^{c} - d x e^{c} - e^{c}\right )} e^{\left (d x\right )} - i \, e^{\left (2 \, d x + 2 \, c\right )} - i}{6 i \, a d^{3} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a d^{3} e^{\left (3 \, d x + 3 \, c\right )} + 12 \, a d^{3} e^{\left (d x + c\right )} - 6 i \, a d^{3}} + i \, \int \frac{x}{4 \,{\left (a d e^{\left (d x + c\right )} + i \, a d\right )}}\,{d x} - 5 i \, \int \frac{x}{12 \,{\left (a d e^{\left (d x + c\right )} - i \, a d\right )}}\,{d x}\right )} + \frac{1}{3} \, e 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 )} + 4 \, e^{2}{\left (\frac{2 \, e^{\left (-d x - c\right )}}{{\left (6 \, a e^{\left (-d x - c\right )} + 6 \, a e^{\left (-3 \, d x - 3 \, c\right )} - 3 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 i \, a\right )} d} + \frac{i}{{\left (6 \, a e^{\left (-d x - c\right )} + 6 \, a e^{\left (-3 \, d x - 3 \, c\right )} - 3 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 i \, a\right )} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*f^2*((2*I*d^2*x^2 + (d*x*e^(3*c) + e^(3*c))*e^(3*d*x) - (4*d^2*x^2*e^c - d*x*e^c - e^c)*e^(d*x) - I*e^(2*d*x
+ 2*c) - I)/(6*I*a*d^3*e^(4*d*x + 4*c) + 12*a*d^3*e^(3*d*x + 3*c) + 12*a*d^3*e^(d*x + c) - 6*I*a*d^3) + I*int
egrate(1/4*x/(a*d*e^(d*x + c) + I*a*d), x) - 5*I*integrate(1/12*x/(a*d*e^(d*x + c) - I*a*d), x)) + 1/3*e*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) + 2
4*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)) + 4*e^2*(2*e^(-d*x - c)/((6*a*e^(-d*x - c) + 6*a*e^(-3*d*x - 3*c) - 3
*I*a*e^(-4*d*x - 4*c) + 3*I*a)*d) + I/((6*a*e^(-d*x - c) + 6*a*e^(-3*d*x - 3*c) - 3*I*a*e^(-4*d*x - 4*c) + 3*I
*a)*d))

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Fricas [B]  time = 2.2461, size = 1760, normalized size = 5.42 \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)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(4*d^2*e^2 - 8*c*d*e*f + 2*(2*c^2 - 1)*f^2 - 2*f^2*e^(2*d*x + 2*c) - (3*f^2*e^(4*d*x + 4*c) - 6*I*f^2*e^(3*d*x
+ 3*c) - 6*I*f^2*e^(d*x + c) - 3*f^2)*dilog(I*e^(d*x + c)) - (5*f^2*e^(4*d*x + 4*c) - 10*I*f^2*e^(3*d*x + 3*c
) - 10*I*f^2*e^(d*x + c) - 5*f^2)*dilog(-I*e^(d*x + c)) + 4*(d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*
e^(4*d*x + 4*c) + (-8*I*d^2*f^2*x^2 + (-16*I*c - 2*I)*d*e*f + (8*I*c^2 - 2*I)*f^2 + (-16*I*d^2*e*f - 2*I*d*f^2
)*x)*e^(3*d*x + 3*c) + (8*I*d^2*e^2 + (-16*I*c - 2*I)*d*e*f - 2*I*d*f^2*x + (8*I*c^2 - 2*I)*f^2)*e^(d*x + c) +
(3*d*e*f - 3*c*f^2 - 3*(d*e*f - c*f^2)*e^(4*d*x + 4*c) + (6*I*d*e*f - 6*I*c*f^2)*e^(3*d*x + 3*c) + (6*I*d*e*f
- 6*I*c*f^2)*e^(d*x + c))*log(e^(d*x + c) + I) + (5*d*e*f - 5*c*f^2 - 5*(d*e*f - c*f^2)*e^(4*d*x + 4*c) + (10
*I*d*e*f - 10*I*c*f^2)*e^(3*d*x + 3*c) + (10*I*d*e*f - 10*I*c*f^2)*e^(d*x + c))*log(e^(d*x + c) - I) + (5*d*f^
2*x + 5*c*f^2 - 5*(d*f^2*x + c*f^2)*e^(4*d*x + 4*c) + (10*I*d*f^2*x + 10*I*c*f^2)*e^(3*d*x + 3*c) + (10*I*d*f^
2*x + 10*I*c*f^2)*e^(d*x + c))*log(I*e^(d*x + c) + 1) + (3*d*f^2*x + 3*c*f^2 - 3*(d*f^2*x + c*f^2)*e^(4*d*x +
4*c) + (6*I*d*f^2*x + 6*I*c*f^2)*e^(3*d*x + 3*c) + (6*I*d*f^2*x + 6*I*c*f^2)*e^(d*x + c))*log(-I*e^(d*x + c) +
1))/(3*a*d^3*e^(4*d*x + 4*c) - 6*I*a*d^3*e^(3*d*x + 3*c) - 6*I*a*d^3*e^(d*x + c) - 3*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)**2/(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 )^{2}}{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)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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