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

Optimal. Leaf size=163 \[ \frac{i f \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}-\frac{i f \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{f \log (\sinh (c+d x))}{a d^2}+\frac{2 f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}+\frac{2 i (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{(e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x) \coth (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.228043, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {5575, 4184, 3475, 4182, 2279, 2391, 3318} \[ \frac{i f \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}-\frac{i f \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{f \log (\sinh (c+d x))}{a d^2}+\frac{2 f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}+\frac{2 i (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{(e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x) \coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*(e + f*x)*ArcTanh[E^(c + d*x)])/(a*d) - ((e + f*x)*Coth[c + d*x])/(a*d) + (2*f*Log[Cosh[c/2 + (I/4)*Pi
+ (d*x)/2]])/(a*d^2) + (f*Log[Sinh[c + d*x]])/(a*d^2) + (I*f*PolyLog[2, -E^(c + d*x)])/(a*d^2) - (I*f*PolyLog[
2, E^(c + d*x)])/(a*d^2) - ((e + f*x)*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 3475

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

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

Mathematica [B]  time = 5.43604, size = 454, normalized size = 2.79 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (-2 i f \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\text{PolyLog}\left (2,-e^{-c-d x}\right )-\text{PolyLog}\left (2,e^{-c-d x}\right )+(c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (e^{-c-d x}+1\right )\right )\right )-4 d (e+f x) \sinh \left (\frac{1}{2} (c+d x)\right )-i d (e+f x) \sinh \left (\frac{1}{2} (c+d x)\right ) \left (\tanh \left (\frac{1}{2} (c+d x)\right )-i\right )-d (e+f x) \cosh \left (\frac{1}{2} (c+d x)\right ) \left (\coth \left (\frac{1}{2} (c+d x)\right )+i\right )+2 d e \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sinh \left (\frac{1}{2} (c+d x)\right )-i \cosh \left (\frac{1}{2} (c+d x)\right )\right )+2 f (c+d x) \left (\sinh \left (\frac{1}{2} (c+d x)\right )-i \cosh \left (\frac{1}{2} (c+d x)\right )\right )+2 f \log (\cosh (c+d x)) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+2 f \log (\sinh (c+d x)) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+2 i c f \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+4 i f \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 d^2 (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(-(d*(e + f*x)*Cosh[(c + d*x)/2]*(I + Coth[(c + d*x)/2])) + (4*I)*f
*ArcTan[Tanh[(c + d*x)/2]]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) + 2*f*Log[Cosh[c + d*x]]*(Cosh[(c + d*x)/
2] + I*Sinh[(c + d*x)/2]) + 2*f*Log[Sinh[c + d*x]]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) + (2*I)*c*f*Log[T
anh[(c + d*x)/2]]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) - (2*I)*f*((c + d*x)*(Log[1 - E^(-c - d*x)] - Log[
1 + E^(-c - d*x)]) + PolyLog[2, -E^(-c - d*x)] - PolyLog[2, E^(-c - d*x)])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*
x)/2]) - 4*d*(e + f*x)*Sinh[(c + d*x)/2] + 2*f*(c + d*x)*((-I)*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2]) + 2*d*e*
Log[Tanh[(c + d*x)/2]]*((-I)*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2]) - I*d*(e + f*x)*Sinh[(c + d*x)/2]*(-I + Ta
nh[(c + d*x)/2])))/(2*d^2*(a + I*a*Sinh[c + d*x]))

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Maple [B]  time = 0.151, size = 316, normalized size = 1.9 \begin{align*}{\frac{-2\,i \left ( fx{{\rm e}^{2\,dx+2\,c}}+e{{\rm e}^{2\,dx+2\,c}}-2\,fx-i{{\rm e}^{dx+c}}fx-2\,e-i{{\rm e}^{dx+c}}e \right ) }{ \left ({{\rm e}^{2\,dx+2\,c}}-1 \right ) \left ({{\rm e}^{dx+c}}-i \right ) da}}-{\frac{if{\it polylog} \left ( 2,{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{if{\it polylog} \left ( 2,-{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-4\,{\frac{f\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{f\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{2}}}+{\frac{f\ln \left ({{\rm e}^{dx+c}}+1 \right ) }{a{d}^{2}}}-{\frac{ie\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{da}}+{\frac{ie\ln \left ({{\rm e}^{dx+c}}+1 \right ) }{da}}-{\frac{i\ln \left ( 1-{{\rm e}^{dx+c}} \right ) fx}{da}}-{\frac{i\ln \left ( 1-{{\rm e}^{dx+c}} \right ) cf}{a{d}^{2}}}+{\frac{if\ln \left ({{\rm e}^{dx+c}}+1 \right ) x}{da}}+2\,{\frac{f\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}}+{\frac{ifc\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.59178, size = 1272, normalized size = 7.8 \begin{align*} \frac{4 i \, d e - 2 i \, c f +{\left (i \, f e^{\left (3 \, d x + 3 \, c\right )} + f e^{\left (2 \, d x + 2 \, c\right )} - i \, f e^{\left (d x + c\right )} - f\right )}{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) +{\left (-i \, f e^{\left (3 \, d x + 3 \, c\right )} - f e^{\left (2 \, d x + 2 \, c\right )} + i \, f e^{\left (d x + c\right )} + f\right )}{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \,{\left (2 \, d f x + c f\right )} e^{\left (3 \, d x + 3 \, c\right )} +{\left (2 i \, d f x - 2 i \, d e + 2 i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (d f x - d e + c f\right )} e^{\left (d x + c\right )} -{\left (d f x + d e -{\left (i \, d f x + i \, d e + f\right )} e^{\left (3 \, d x + 3 \, c\right )} -{\left (d f x + d e - i \, f\right )} e^{\left (2 \, d x + 2 \, c\right )} -{\left (-i \, d f x - i \, d e - f\right )} e^{\left (d x + c\right )} - i \, f\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) +{\left (2 \, f e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f e^{\left (2 \, d x + 2 \, c\right )} - 2 \, f e^{\left (d x + c\right )} + 2 i \, f\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (d e -{\left (c - i\right )} f +{\left (-i \, d e +{\left (i \, c + 1\right )} f\right )} e^{\left (3 \, d x + 3 \, c\right )} -{\left (d e -{\left (c - i\right )} f\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (i \, d e +{\left (-i \, c - 1\right )} f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) +{\left (d f x + c f +{\left (-i \, d f x - i \, c f\right )} e^{\left (3 \, d x + 3 \, c\right )} -{\left (d f x + c f\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (i \, d f x + i \, c f\right )} e^{\left (d x + c\right )}\right )} \log \left (-e^{\left (d x + c\right )} + 1\right )}{a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - i \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - a d^{2} e^{\left (d x + c\right )} + i \, a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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