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

Optimal. Leaf size=296 \[ \frac{2 i f (e+f x) \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}-\frac{2 i f (e+f x) \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{4 f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{f^2 \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{a d^3}-\frac{2 i f^2 \text{PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}+\frac{2 i f^2 \text{PolyLog}\left (3,e^{c+d x}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac{2 i (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x)^2 \coth (c+d x)}{a d}-\frac{2 (e+f x)^2}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.583369, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 11, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.355, Rules used = {5575, 4184, 3716, 2190, 2279, 2391, 4182, 2531, 2282, 6589, 3318} \[ \frac{2 i f (e+f x) \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}-\frac{2 i f (e+f x) \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{4 f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{f^2 \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{a d^3}-\frac{2 i f^2 \text{PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}+\frac{2 i f^2 \text{PolyLog}\left (3,e^{c+d x}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac{2 i (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x)^2 \coth (c+d x)}{a d}-\frac{2 (e+f x)^2}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5575

Int[(Csch[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Csch[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csch[c + d*x]^(n - 1))/
(a + b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && 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 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

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

Mathematica [B]  time = 13.0869, size = 715, normalized size = 2.42 \[ \frac{-2 \left (e^{2 c}-1\right ) f (f+i d e) \text{PolyLog}\left (2,-e^{-c-d x}\right )+2 i \left (e^{2 c}-1\right ) f (d e+i f) \text{PolyLog}\left (2,e^{-c-d x}\right )-2 i \left (e^{2 c}-1\right ) f^2 \left (d x \text{PolyLog}\left (2,-e^{-c-d x}\right )+\text{PolyLog}\left (3,-e^{-c-d x}\right )\right )+2 i \left (e^{2 c}-1\right ) f^2 \left (d x \text{PolyLog}\left (2,e^{-c-d x}\right )+\text{PolyLog}\left (3,e^{-c-d x}\right )\right )-i \left (e^{2 c}-1\right ) d^2 f^2 x^2 \log \left (1-e^{-c-d x}\right )+i \left (e^{2 c}-1\right ) d^2 f^2 x^2 \log \left (e^{-c-d x}+1\right )+2 \left (e^{2 c}-1\right ) d f x (f-i d e) \log \left (1-e^{-c-d x}\right )+2 \left (e^{2 c}-1\right ) d f x (f+i d e) \log \left (e^{-c-d x}+1\right )+i \left (e^{2 c}-1\right ) d e (d e+2 i f) \left (d x-\log \left (1-e^{c+d x}\right )\right )+\left (1-e^{2 c}\right ) d e (2 f+i d e) \left (d x-\log \left (e^{c+d x}+1\right )\right )-2 d^2 (e+f x)^2}{a \left (e^{2 c}-1\right ) d^3}+\frac{2 \left (\frac{d (e+f x) \left (2 \left (e^c-i\right ) f \log \left (1-i e^{-c-d x}\right )-i d (e+f x)\right )}{e^c-i}-2 f^2 \text{PolyLog}\left (2,i e^{-c-d x}\right )\right )}{a d^3}-\frac{2 \left (e^2 \sinh \left (\frac{d x}{2}\right )+2 e f x \sinh \left (\frac{d x}{2}\right )+f^2 x^2 \sinh \left (\frac{d x}{2}\right )\right )}{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{\text{csch}\left (\frac{c}{2}\right ) \text{csch}\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (e^2 \sinh \left (\frac{d x}{2}\right )+2 e f x \sinh \left (\frac{d x}{2}\right )+f^2 x^2 \sinh \left (\frac{d x}{2}\right )\right )}{2 a d}+\frac{\text{sech}\left (\frac{c}{2}\right ) \text{sech}\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (e^2 \left (-\sinh \left (\frac{d x}{2}\right )\right )-2 e f x \sinh \left (\frac{d x}{2}\right )-f^2 x^2 \sinh \left (\frac{d x}{2}\right )\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((d*(e + f*x)*((-I)*d*(e + f*x) + 2*(-I + E^c)*f*Log[1 - I*E^(-c - d*x)]))/(-I + E^c) - 2*f^2*PolyLog[2, I*
E^(-c - d*x)]))/(a*d^3) + (-2*d^2*(e + f*x)^2 + 2*d*(-1 + E^(2*c))*f*((-I)*d*e + f)*x*Log[1 - E^(-c - d*x)] -
I*d^2*(-1 + E^(2*c))*f^2*x^2*Log[1 - E^(-c - d*x)] + 2*d*(-1 + E^(2*c))*f*(I*d*e + f)*x*Log[1 + E^(-c - d*x)]
+ I*d^2*(-1 + E^(2*c))*f^2*x^2*Log[1 + E^(-c - d*x)] + I*d*e*(-1 + E^(2*c))*(d*e + (2*I)*f)*(d*x - Log[1 - E^(
c + d*x)]) + d*e*(1 - E^(2*c))*(I*d*e + 2*f)*(d*x - Log[1 + E^(c + d*x)]) - 2*(-1 + E^(2*c))*f*(I*d*e + f)*Pol
yLog[2, -E^(-c - d*x)] + (2*I)*(-1 + E^(2*c))*(d*e + I*f)*f*PolyLog[2, E^(-c - d*x)] - (2*I)*(-1 + E^(2*c))*f^
2*(d*x*PolyLog[2, -E^(-c - d*x)] + PolyLog[3, -E^(-c - d*x)]) + (2*I)*(-1 + E^(2*c))*f^2*(d*x*PolyLog[2, E^(-c
- d*x)] + PolyLog[3, E^(-c - d*x)]))/(a*d^3*(-1 + E^(2*c))) + (Sech[c/2]*Sech[c/2 + (d*x)/2]*(-(e^2*Sinh[(d*x
)/2]) - 2*e*f*x*Sinh[(d*x)/2] - f^2*x^2*Sinh[(d*x)/2]))/(2*a*d) + (Csch[c/2]*Csch[c/2 + (d*x)/2]*(e^2*Sinh[(d*
x)/2] + 2*e*f*x*Sinh[(d*x)/2] + f^2*x^2*Sinh[(d*x)/2]))/(2*a*d) - (2*(e^2*Sinh[(d*x)/2] + 2*e*f*x*Sinh[(d*x)/2
] + f^2*x^2*Sinh[(d*x)/2]))/(a*d*(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]  time = 0.151, size = 847, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.0813, size = 814, normalized size = 2.75 \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*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-e^2*(4*(e^(-d*x - c) - I*e^(-2*d*x - 2*c) + 2*I)/((2*a*e^(-d*x - c) - 2*I*a*e^(-2*d*x - 2*c) - 2*a*e^(-3*d*x
- 3*c) + 2*I*a)*d) - I*log(e^(-d*x - c) + 1)/(a*d) + I*log(e^(-d*x - c) - 1)/(a*d)) - 2*f^2*x^2/(a*d) - 8*e*f*
x/(a*d) + (4*I*f^2*x^2 + 8*I*e*f*x - (2*I*f^2*x^2*e^(2*c) + 4*I*e*f*x*e^(2*c))*e^(2*d*x) - 2*(f^2*x^2*e^c + 2*
e*f*x*e^c)*e^(d*x))/(a*d*e^(3*d*x + 3*c) - I*a*d*e^(2*d*x + 2*c) - a*d*e^(d*x + c) + I*a*d) + 2*e*f*log(e^(d*x
+ c) + 1)/(a*d^2) + 4*e*f*log(e^(d*x + c) - I)/(a*d^2) + 2*e*f*log(e^(d*x + c) - 1)/(a*d^2) + I*(d^2*x^2*log(
e^(d*x + c) + 1) + 2*d*x*dilog(-e^(d*x + c)) - 2*polylog(3, -e^(d*x + c)))*f^2/(a*d^3) - I*(d^2*x^2*log(-e^(d*
x + c) + 1) + 2*d*x*dilog(e^(d*x + c)) - 2*polylog(3, e^(d*x + c)))*f^2/(a*d^3) + 4*(d*x*log(I*e^(d*x + c) + 1
) + dilog(-I*e^(d*x + c)))*f^2/(a*d^3) + (2*I*d*e*f + 2*f^2)*(d*x*log(e^(d*x + c) + 1) + dilog(-e^(d*x + c)))/
(a*d^3) - (2*I*d*e*f - 2*f^2)*(d*x*log(-e^(d*x + c) + 1) + dilog(e^(d*x + c)))/(a*d^3) - 1/3*(I*d^3*f^2*x^3 +
(3*I*d*e*f + 3*f^2)*d^2*x^2)/(a*d^3) + 1/3*(I*d^3*f^2*x^3 + (3*I*d*e*f - 3*f^2)*d^2*x^2)/(a*d^3)

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Fricas [C]  time = 2.76264, size = 3237, normalized size = 10.94 \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*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(4*I*d^2*e^2 - 8*I*c*d*e*f + 4*I*c^2*f^2 + (4*f^2*e^(3*d*x + 3*c) - 4*I*f^2*e^(2*d*x + 2*c) - 4*f^2*e^(d*x + c
) + 4*I*f^2)*dilog(-I*e^(d*x + c)) - (2*d*f^2*x + 2*d*e*f - 2*I*f^2 - (2*I*d*f^2*x + 2*I*d*e*f + 2*f^2)*e^(3*d
*x + 3*c) - 2*(d*f^2*x + d*e*f - I*f^2)*e^(2*d*x + 2*c) - (-2*I*d*f^2*x - 2*I*d*e*f - 2*f^2)*e^(d*x + c))*dilo
g(-e^(d*x + c)) + (2*d*f^2*x + 2*d*e*f + 2*I*f^2 + (-2*I*d*f^2*x - 2*I*d*e*f + 2*f^2)*e^(3*d*x + 3*c) - 2*(d*f
^2*x + d*e*f + I*f^2)*e^(2*d*x + 2*c) + (2*I*d*f^2*x + 2*I*d*e*f - 2*f^2)*e^(d*x + c))*dilog(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^(3*d*x + 3*c) + (2*I*d^2*f^2*x^2 + 4*I*d^2*e*f*x - 2*I*d^2
*e^2 + 8*I*c*d*e*f - 4*I*c^2*f^2)*e^(2*d*x + 2*c) + 2*(d^2*f^2*x^2 + 2*d^2*e*f*x - d^2*e^2 + 4*c*d*e*f - 2*c^2
*f^2)*e^(d*x + c) - (d^2*f^2*x^2 + d^2*e^2 - 2*I*d*e*f + (2*d^2*e*f - 2*I*d*f^2)*x - (I*d^2*f^2*x^2 + I*d^2*e^
2 + 2*d*e*f - 2*(-I*d^2*e*f - d*f^2)*x)*e^(3*d*x + 3*c) - (d^2*f^2*x^2 + d^2*e^2 - 2*I*d*e*f + (2*d^2*e*f - 2*
I*d*f^2)*x)*e^(2*d*x + 2*c) - (-I*d^2*f^2*x^2 - I*d^2*e^2 - 2*d*e*f - 2*(I*d^2*e*f + d*f^2)*x)*e^(d*x + c))*lo
g(e^(d*x + c) + 1) + (4*I*d*e*f - 4*I*c*f^2 + 4*(d*e*f - c*f^2)*e^(3*d*x + 3*c) + (-4*I*d*e*f + 4*I*c*f^2)*e^(
2*d*x + 2*c) - 4*(d*e*f - c*f^2)*e^(d*x + c))*log(e^(d*x + c) - I) + (d^2*e^2 - (2*c - 2*I)*d*e*f + (c^2 - 2*I
*c)*f^2 + (-I*d^2*e^2 - 2*(-I*c - 1)*d*e*f + (-I*c^2 - 2*c)*f^2)*e^(3*d*x + 3*c) - (d^2*e^2 - (2*c - 2*I)*d*e*
f + (c^2 - 2*I*c)*f^2)*e^(2*d*x + 2*c) + (I*d^2*e^2 - 2*(I*c + 1)*d*e*f + (I*c^2 + 2*c)*f^2)*e^(d*x + c))*log(
e^(d*x + c) - 1) + (4*I*d*f^2*x + 4*I*c*f^2 + 4*(d*f^2*x + c*f^2)*e^(3*d*x + 3*c) + (-4*I*d*f^2*x - 4*I*c*f^2)
*e^(2*d*x + 2*c) - 4*(d*f^2*x + c*f^2)*e^(d*x + c))*log(I*e^(d*x + c) + 1) + (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 -
2*I*c)*f^2 + (2*d^2*e*f + 2*I*d*f^2)*x + (-I*d^2*f^2*x^2 - 2*I*c*d*e*f + (I*c^2 + 2*c)*f^2 - 2*(I*d^2*e*f - d
*f^2)*x)*e^(3*d*x + 3*c) - (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 - 2*I*c)*f^2 + (2*d^2*e*f + 2*I*d*f^2)*x)*e^(2*d*x
+ 2*c) + (I*d^2*f^2*x^2 + 2*I*c*d*e*f + (-I*c^2 - 2*c)*f^2 - 2*(-I*d^2*e*f + d*f^2)*x)*e^(d*x + c))*log(-e^(d*
x + c) + 1) + (-2*I*f^2*e^(3*d*x + 3*c) - 2*f^2*e^(2*d*x + 2*c) + 2*I*f^2*e^(d*x + c) + 2*f^2)*polylog(3, -e^(
d*x + c)) + (2*I*f^2*e^(3*d*x + 3*c) + 2*f^2*e^(2*d*x + 2*c) - 2*I*f^2*e^(d*x + c) - 2*f^2)*polylog(3, e^(d*x
+ c)))/(a*d^3*e^(3*d*x + 3*c) - I*a*d^3*e^(2*d*x + 2*c) - a*d^3*e^(d*x + c) + I*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*csch(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [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*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out