\(\int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [273]

Optimal result

Integrand size = 27, antiderivative size = 161 \[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {(e+f x) \arctan \left (e^{c+d x}\right )}{a d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {f \text {sech}(c+d x)}{2 a d^2}+\frac {i (e+f x) \text {sech}^2(c+d x)}{2 a d}-\frac {i f \tanh (c+d x)}{2 a d^2}+\frac {(e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 a d} \] Output:

(f*x+e)*arctan(exp(d*x+c))/a/d-1/2*I*f*polylog(2,-I*exp(d*x+c))/a/d^2+1/2* 
I*f*polylog(2,I*exp(d*x+c))/a/d^2+1/2*f*sech(d*x+c)/a/d^2+1/2*I*(f*x+e)*se 
ch(d*x+c)^2/a/d-1/2*I*f*tanh(d*x+c)/a/d^2+1/2*(f*x+e)*sech(d*x+c)*tanh(d*x 
+c)/a/d
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(400\) vs. \(2(161)=322\).

Time = 2.33 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.48 \[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {-2 i d (e+f x)+(c+d x) (c f-d (2 e+f x)) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2+(1-i) \left (\frac {1}{2} d^2 f x^2+d e (c+d x)-(1-i) (d e-c f) (c+d x)+(1-i) f (c+d x) \log \left (1+i e^{-c-d x}\right )+(1-i) (d e-c f) \log \left (i+e^{c+d x}\right )-(1-i) f \operatorname {PolyLog}\left (2,-i e^{-c-d x}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2+(1+i) \left (\frac {1}{2} d^2 f x^2+d e (c+d x)-(1+i) (d e-c f) (c+d x)+(1+i) f (c+d x) \log \left (1-i e^{-c-d x}\right )+(1+i) (d e-c f) \log \left (i-e^{c+d x}\right )-(1+i) f \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-4 f \sinh \left (\frac {1}{2} (c+d x)\right ) \left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d^2 (a+i a \sinh (c+d x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {6105, 3042, 4673, 3042, 4668, 2715, 2838, 5974, 3042, 4254, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6105

\(\displaystyle \frac {\int (e+f x) \text {sech}^3(c+d x)dx}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )^3dx}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {1}{2} \int (e+f x) \text {sech}(c+d x)dx+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )dx+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {\frac {1}{2} \left (-\frac {i f \int \log \left (1-i e^{c+d x}\right )dx}{d}+\frac {i f \int \log \left (1+i e^{c+d x}\right )dx}{d}+\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {1}{2} \left (-\frac {i f \int e^{-c-d x} \log \left (1-i e^{c+d x}\right )de^{c+d x}}{d^2}+\frac {i f \int e^{-c-d x} \log \left (1+i e^{c+d x}\right )de^{c+d x}}{d^2}+\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 5974

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (\frac {f \int \text {sech}^2(c+d x)dx}{2 d}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^2(c+d x)}{2 d}+\frac {f \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{2 d}\right )}{a}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^2(c+d x)}{2 d}+\frac {i f \int 1d(-i \tanh (c+d x))}{2 d^2}\right )}{a}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (\frac {f \tanh (c+d x)}{2 d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 d}\right )}{a}\)

Input:

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

Output:

((-I)*(-1/2*((e + f*x)*Sech[c + d*x]^2)/d + (f*Tanh[c + d*x])/(2*d^2)))/a 
+ (((2*(e + f*x)*ArcTan[E^(c + d*x)])/d - (I*f*PolyLog[2, (-I)*E^(c + d*x) 
])/d^2 + (I*f*PolyLog[2, I*E^(c + d*x)])/d^2)/2 + (f*Sech[c + d*x])/(2*d^2 
) + ((e + f*x)*Sech[c + d*x]*Tanh[c + d*x])/(2*d))/a
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4668

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] + (-Simp[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] + Simp[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 4673

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] + (-Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x] + S 
imp[b^2*((n - 2)/(n - 1))   Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2), x], x]) 
 /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2]
 

rule 5974

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + 
(b_.)*(x_)]^(p_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Sech[a + b*x]^n/(b*n)) 
, x] + Simp[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 6105

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_ 
.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[(e + f*x)^m*Sech[ 
c + d*x]^(n + 2), x], x] + Simp[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]
 

Maple [A] (verified)

Time = 5.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.35

method	result	size
risch	\(\frac {d f x \,{\mathrm e}^{d x +c}+d e \,{\mathrm e}^{d x +c}+{\mathrm e}^{d x +c} f -i f}{\left ({\mathrm e}^{d x +c}-i\right )^{2} d^{2} a}+\frac {e \arctan \left ({\mathrm e}^{d x +c}\right )}{d a}+\frac {i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) x}{2 d a}+\frac {i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) c}{2 d^{2} a}+\frac {i f \operatorname {polylog}\left (2, i {\mathrm e}^{d x +c}\right )}{2 a \,d^{2}}-\frac {i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{2 d a}-\frac {i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{2 d^{2} a}-\frac {i f \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{2 a \,d^{2}}-\frac {f c \arctan \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}\)	\(218\)

Input:

int((f*x+e)*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (135) = 270\).

Time = 0.11 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.19 \[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {{\left (i \, f e^{\left (2 \, d x + 2 \, c\right )} + 2 \, f e^{\left (d x + c\right )} - i \, f\right )} {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) + {\left (-i \, f e^{\left (2 \, d x + 2 \, c\right )} - 2 \, f e^{\left (d x + c\right )} + i \, f\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) + 2 \, {\left (d f x + d e + f\right )} e^{\left (d x + c\right )} + {\left (-i \, d e + i \, c f + {\left (i \, d e - i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d e - c f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + {\left (i \, d e - i \, c f + {\left (-i \, d e + i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, {\left (d e - c f\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (i \, d f x + i \, c f + {\left (-i \, d f x - i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, {\left (d f x + c f\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\left (-i \, d f x - i \, c f + {\left (i \, d f x + i \, c f\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d f x + c f\right )} e^{\left (d x + c\right )}\right )} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) - 2 i \, f}{2 \, {\left (a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}\right )}} \] Input:

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

Output:

1/2*((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)/(a*d^2*e^(2*d*x + 2*c) - 2*I*a*d^2*e^(d*x + c) - a*d^2)
 

Sympy [F]

\[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f x \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \operatorname {sech}\left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \operatorname {sech}\left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {e+f\,x}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:

int((e + f*x)/(cosh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)
 

Output:

int((e + f*x)/(cosh(c + d*x)*(a + a*sinh(c + d*x)*1i)), x)
 

Reduce [F]

\[ \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (\int \frac {\mathrm {sech}\left (d x +c \right )}{\sinh \left (d x +c \right ) i +1}d x \right ) e +\left (\int \frac {\mathrm {sech}\left (d x +c \right ) x}{\sinh \left (d x +c \right ) i +1}d x \right ) f}{a} \] Input:

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

Output:

(int(sech(c + d*x)/(sinh(c + d*x)*i + 1),x)*e + int((sech(c + d*x)*x)/(sin 
h(c + d*x)*i + 1),x)*f)/a