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

Optimal. Leaf size=126 \[ -\frac{f \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{f \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.147479, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {5575, 4182, 2279, 2391, 3318, 4184, 3475} \[ -\frac{f \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{f \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 1.25237, size = 345, normalized size = 2.74 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (f \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\text{PolyLog}\left (2,-e^{-c-d x}\right )-\text{PolyLog}\left (2,e^{-c-d x}\right )+(c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (e^{-c-d x}+1\right )\right )\right )-2 i d (e+f x) \sinh \left (\frac{1}{2} (c+d x)\right )+d e \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+f (c+d x) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+i f \log (\cosh (c+d x)) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )-c f \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )-2 f \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )\right )}{d^2 (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(f*(c + d*x)*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) - 2*f*ArcTan
[Tanh[(c + d*x)/2]]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) + I*f*Log[Cosh[c + d*x]]*(Cosh[(c + d*x)/2] + I*
Sinh[(c + d*x)/2]) + d*e*Log[Tanh[(c + d*x)/2]]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) - c*f*Log[Tanh[(c +
d*x)/2]]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) + 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]) - (2*I)*
d*(e + f*x)*Sinh[(c + d*x)/2]))/(d^2*(a + I*a*Sinh[c + d*x]))

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Maple [A]  time = 0.132, size = 211, normalized size = 1.7 \begin{align*} 2\,{\frac{fx+e}{da \left ({{\rm e}^{dx+c}}-i \right ) }}+{\frac{f{\it polylog} \left ( 2,{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{f{\it polylog} \left ( 2,-{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{2\,if\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{e\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{da}}-{\frac{e\ln \left ({{\rm e}^{dx+c}}+1 \right ) }{da}}+{\frac{\ln \left ( 1-{{\rm e}^{dx+c}} \right ) fx}{da}}+{\frac{\ln \left ( 1-{{\rm e}^{dx+c}} \right ) cf}{a{d}^{2}}}-{\frac{\ln \left ({{\rm e}^{dx+c}}+1 \right ) fx}{da}}-{\frac{fc\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{2}}}+{\frac{2\,if\ln \left ({{\rm e}^{dx+c}}-i \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)/(a+I*a*sinh(d*x+c)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.62454, size = 548, normalized size = 4.35 \begin{align*} \frac{-2 i \, d f x e^{\left (d x + c\right )} + 2 \, d e -{\left (f e^{\left (d x + c\right )} - i \, f\right )}{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) +{\left (f e^{\left (d x + c\right )} - i \, f\right )}{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) +{\left (i \, d f x + i \, d e -{\left (d f x + d e\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) +{\left (2 i \, f e^{\left (d x + c\right )} + 2 \, f\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (-i \, d e + i \, c f +{\left (d e - c f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) +{\left (-i \, d f x - i \, c f +{\left (d f x + 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 (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)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

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