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

Optimal. Leaf size=224 \[ -\frac{2 f (e+f x) \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{2 f (e+f x) \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac{2 f^2 \text{PolyLog}\left (3,e^{c+d x}\right )}{a d^3}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^2}{a d} \]

[Out]

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

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Rubi [A]  time = 0.345685, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {5575, 4182, 2531, 2282, 6589, 3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac{2 f (e+f x) \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{2 f (e+f x) \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac{2 f^2 \text{PolyLog}\left (3,e^{c+d x}\right )}{a d^3}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^2}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 4.15301, size = 275, normalized size = 1.23 \[ \frac{\frac{2 d (e+f x) \left (2 \left (e^c-i\right ) f \log \left (1-i e^{-c-d x}\right )-i d (e+f x)\right )-4 \left (e^c-i\right ) f^2 \text{PolyLog}\left (2,i e^{-c-d x}\right )}{-1-i e^c}-2 d f (e+f x) \text{PolyLog}\left (2,-e^{c+d x}\right )+2 d f (e+f x) \text{PolyLog}\left (2,e^{c+d x}\right )+2 f^2 \text{PolyLog}\left (3,-e^{c+d x}\right )-2 f^2 \text{PolyLog}\left (3,e^{c+d x}\right )+d^2 (e+f x)^2 \log \left (1-e^{c+d x}\right )-d^2 (e+f x)^2 \log \left (e^{c+d x}+1\right )-\frac{2 i d^2 \sinh \left (\frac{d x}{2}\right ) (e+f x)^2}{\left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}}{a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.131, size = 573, normalized size = 2.6 \begin{align*} 2\,{\frac{{x}^{2}{f}^{2}+2\,efx+{e}^{2}}{da \left ({{\rm e}^{dx+c}}-i \right ) }}+{\frac{4\,i\ln \left ({{\rm e}^{dx+c}}-i \right ) ef}{a{d}^{2}}}-{\frac{4\,i\ln \left ({{\rm e}^{dx+c}} \right ) ef}{a{d}^{2}}}-{\frac{4\,i{f}^{2}cx}{a{d}^{2}}}+{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}+{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}-{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}+{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{dx+c}} \right ) cef}{a{d}^{2}}}-2\,{\frac{\ln \left ({{\rm e}^{dx+c}}+1 \right ) efx}{da}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{dx+c}} \right ) efx}{da}}-2\,{\frac{efc\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{2}}}-2\,{\frac{{f}^{2}{\it polylog} \left ( 3,{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+2\,{\frac{{f}^{2}{\it polylog} \left ( 3,-{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{4\,i{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{e}^{2}\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{da}}-{\frac{{e}^{2}\ln \left ({{\rm e}^{dx+c}}+1 \right ) }{da}}-2\,{\frac{{f}^{2}{\it polylog} \left ( 2,-{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}-{\frac{{c}^{2}{f}^{2}\ln \left ( 1-{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{f}^{2}\ln \left ( 1-{{\rm e}^{dx+c}} \right ){x}^{2}}{da}}+{\frac{{c}^{2}{f}^{2}\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{3}}}-2\,{\frac{ef{\it polylog} \left ( 2,-{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+2\,{\frac{ef{\it polylog} \left ( 2,{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{2\,i{f}^{2}{x}^{2}}{da}}-{\frac{2\,i{f}^{2}{c}^{2}}{a{d}^{3}}}+2\,{\frac{{f}^{2}{\it polylog} \left ( 2,{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}-{\frac{{f}^{2}\ln \left ({{\rm e}^{dx+c}}+1 \right ){x}^{2}}{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.85616, size = 468, normalized size = 2.09 \begin{align*} -e^{2}{\left (\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac{2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} - \frac{2 i \, f^{2} x^{2}}{a d} - \frac{4 i \, e f x}{a d} + \frac{2 \,{\left (f^{2} x^{2} + 2 \, e f x\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac{2 \,{\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac{2 \,{\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac{4 i \, e f \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{2}} - \frac{{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac{{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac{4 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 )} f^{2}}{a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.71007, size = 1368, normalized size = 6.11 \begin{align*} \frac{2 \, d^{2} e^{2} - 4 \, c d e f + 2 \, c^{2} f^{2} +{\left (4 i \, f^{2} e^{\left (d x + c\right )} + 4 \, f^{2}\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) +{\left (2 i \, d f^{2} x + 2 i \, d e f - 2 \,{\left (d f^{2} x + d e f\right )} e^{\left (d x + c\right )}\right )}{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) +{\left (-2 i \, d f^{2} x - 2 i \, d e f + 2 \,{\left (d f^{2} x + d e f\right )} e^{\left (d x + c\right )}\right )}{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) +{\left (-2 i \, d^{2} f^{2} x^{2} - 4 i \, d^{2} e f x - 4 i \, c d e f + 2 i \, c^{2} f^{2}\right )} e^{\left (d x + c\right )} +{\left (i \, d^{2} f^{2} x^{2} + 2 i \, d^{2} e f x + i \, d^{2} e^{2} -{\left (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x + d^{2} e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) +{\left (4 \, d e f - 4 \, c f^{2} +{\left (4 i \, d e f - 4 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (-i \, d^{2} e^{2} + 2 i \, c d e f - i \, c^{2} f^{2} +{\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) +{\left (4 \, d f^{2} x + 4 \, c f^{2} +{\left (4 i \, d f^{2} x + 4 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) +{\left (-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} +{\left (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x + 2 \, c d e f - c^{2} f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \,{\left (f^{2} e^{\left (d x + c\right )} - i \, f^{2}\right )}{\rm polylog}\left (3, -e^{\left (d x + c\right )}\right ) - 2 \,{\left (f^{2} e^{\left (d x + c\right )} - i \, f^{2}\right )}{\rm polylog}\left (3, e^{\left (d x + c\right )}\right )}{a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d^2*e^2 - 4*c*d*e*f + 2*c^2*f^2 + (4*I*f^2*e^(d*x + c) + 4*f^2)*dilog(-I*e^(d*x + c)) + (2*I*d*f^2*x + 2*I*
d*e*f - 2*(d*f^2*x + d*e*f)*e^(d*x + c))*dilog(-e^(d*x + c)) + (-2*I*d*f^2*x - 2*I*d*e*f + 2*(d*f^2*x + d*e*f)
*e^(d*x + c))*dilog(e^(d*x + c)) + (-2*I*d^2*f^2*x^2 - 4*I*d^2*e*f*x - 4*I*c*d*e*f + 2*I*c^2*f^2)*e^(d*x + c)
+ (I*d^2*f^2*x^2 + 2*I*d^2*e*f*x + I*d^2*e^2 - (d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*e^(d*x + c))*log(e^(d*x +
c) + 1) + (4*d*e*f - 4*c*f^2 + (4*I*d*e*f - 4*I*c*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*f^2 + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*e^(d*x + c))*log(e^(d*x + c) - 1) + (4*d*f^2*x + 4*c*f^2
+ (4*I*d*f^2*x + 4*I*c*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 + (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(-e^(d*x + c) + 1) + 2*(f^2
*e^(d*x + c) - I*f^2)*polylog(3, -e^(d*x + c)) - 2*(f^2*e^(d*x + c) - I*f^2)*polylog(3, e^(d*x + 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)/(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{csch}\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*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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