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

Optimal. Leaf size=161 \[ -\frac{i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac{i f \text{PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}-\frac{i f \tanh (c+d x)}{2 a d^2}+\frac{f \text{sech}(c+d x)}{2 a d^2}+\frac{(e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{i (e+f x) \text{sech}^2(c+d x)}{2 a d}+\frac{(e+f x) \tanh (c+d x) \text{sech}(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.142473, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {5571, 4185, 4180, 2279, 2391, 5451, 3767, 8} \[ -\frac{i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac{i f \text{PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}-\frac{i f \tanh (c+d x)}{2 a d^2}+\frac{f \text{sech}(c+d x)}{2 a d^2}+\frac{(e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{i (e+f x) \text{sech}^2(c+d x)}{2 a d}+\frac{(e+f x) \tanh (c+d x) \text{sech}(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

Mathematica [B]  time = 2.69126, size = 707, normalized size = 4.39 \[ \frac{\sqrt{2} f \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (2 (-1)^{3/4} (c+d x)^2+\sqrt{2} \left (-4 i \text{PolyLog}\left (2,-i e^{-c-d x}\right )+(4 i c+4 i d x-2 \pi ) \log \left (1+i e^{-c-d x}\right )+\pi \left (4 \log \left (e^{c+d x}+1\right )+2 \log \left (-\sin \left (\frac{1}{4} (\pi -2 i (c+d x))\right )\right )-4 \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )\right )-3 c-3 d x\right )\right )\right )+\sqrt{2} f \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-2 \sqrt [4]{-1} (c+d x)^2+\sqrt{2} \left (4 i \text{PolyLog}\left (2,i e^{-c-d x}\right )-2 (2 i c+2 i d x+\pi ) \log \left (1-i e^{-c-d x}\right )+\pi \left (-4 \log \left (e^{c+d x}+1\right )+2 \log \left (\sin \left (\frac{1}{4} (\pi +2 i (c+d x))\right )\right )+4 \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )\right )\right )-4 (c+d x) (c f-d (2 e+f x)) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2-4 d e \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-2 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )-i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )-4 d e \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (2 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )+16 f \sinh \left (\frac{1}{2} (c+d x)\right ) \left (\sinh \left (\frac{1}{2} (c+d x)\right )-i \cosh \left (\frac{1}{2} (c+d x)\right )\right )+4 c f \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (-2 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )-i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )+4 c f \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (2 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )+8 i d (e+f x)}{16 d^2 (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.151, size = 268, normalized size = 1.7 \begin{align*}{\frac{dfx{{\rm e}^{dx+c}}+de{{\rm e}^{dx+c}}+f{{\rm e}^{dx+c}}-if}{ \left ({{\rm e}^{dx+c}}-i \right ) ^{2}{d}^{2}a}}-{\frac{{\frac{i}{2}}e\ln \left ({{\rm e}^{dx+c}}-i \right ) }{da}}+{\frac{{\frac{i}{2}}e\ln \left ({{\rm e}^{dx+c}}+i \right ) }{da}}-{\frac{{\frac{i}{2}}f\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{da}}-{\frac{{\frac{i}{2}}f\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{2}}}-{\frac{{\frac{i}{2}}f{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{{\frac{i}{2}}f\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) x}{da}}+{\frac{{\frac{i}{2}}f\ln \left ( 1-i{{\rm e}^{dx+c}} \right ) c}{a{d}^{2}}}+{\frac{{\frac{i}{2}}f{\it polylog} \left ( 2,i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{{\frac{i}{2}}fc\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}}-{\frac{{\frac{i}{2}}fc\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)*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

(d*f*x*exp(d*x+c)+d*e*exp(d*x+c)+f*exp(d*x+c)-I*f)/(exp(d*x+c)-I)^2/d^2/a-1/2*I/a/d*e*ln(exp(d*x+c)-I)+1/2*I/a
/d*e*ln(exp(d*x+c)+I)-1/2*I/a/d*f*ln(1+I*exp(d*x+c))*x-1/2*I/a/d^2*f*ln(1+I*exp(d*x+c))*c-1/2*I*f*polylog(2,-I
*exp(d*x+c))/a/d^2+1/2*I/a/d*f*ln(1-I*exp(d*x+c))*x+1/2*I/a/d^2*f*ln(1-I*exp(d*x+c))*c+1/2*I*f*polylog(2,I*exp
(d*x+c))/a/d^2+1/2*I/a/d^2*f*c*ln(exp(d*x+c)-I)-1/2*I/a/d^2*f*c*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{{\left (d x e^{c} + e^{c}\right )} e^{\left (d x\right )} - i}{2 \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 4 i \, a d^{2} e^{\left (d x + c\right )} - 2 \, a d^{2}} + \int \frac{x}{4 \,{\left (a e^{\left (d x + c\right )} + i \, a\right )}}\,{d x} + \int \frac{x}{4 \,{\left (a e^{\left (d x + c\right )} - i \, a\right )}}\,{d x}\right )} - \frac{1}{2} \, e{\left (\frac{4 \, e^{\left (-d x - c\right )}}{{\left (4 i \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a\right )} d} + \frac{i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{a d} - \frac{i \, \log \left (i \, 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)*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.28914, size = 902, normalized size = 5.6 \begin{align*} \frac{{\left (i \, f e^{\left (2 \, d x + 2 \, c\right )} + 2 \, f e^{\left (d x + c\right )} - i \, f\right )}{\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) +{\left (-i \, f e^{\left (2 \, d x + 2 \, c\right )} - 2 \, f e^{\left (d x + c\right )} + i \, f\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) + 2 \,{\left (d f x + d e + f\right )} e^{\left (d x + c\right )} +{\left (-i \, d e + i \, c f +{\left (i \, d e - i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (d e - c f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + i\right ) +{\left (i \, d e - i \, c f +{\left (-i \, d e + i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \,{\left (d e - c f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (i \, d f x + i \, c f +{\left (-i \, d f x - i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \,{\left (d f x + c f\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) +{\left (-i \, d f x - i \, c f +{\left (i \, d f x + i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (d f x + c f\right )} e^{\left (d x + c\right )}\right )} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) - 2 i \, f}{2 \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 4 i \, a d^{2} e^{\left (d x + c\right )} - 2 \, a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e \operatorname{sech}{\left (c + d x \right )}}{i \sinh{\left (c + d x \right )} + 1}\, dx + \int \frac{f x \operatorname{sech}{\left (c + d x \right )}}{i \sinh{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e*sech(c + d*x)/(I*sinh(c + d*x) + 1), x) + Integral(f*x*sech(c + d*x)/(I*sinh(c + d*x) + 1), x))/a

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \operatorname{sech}\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)*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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