\(\int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [279]

Optimal result

Integrand size = 29, antiderivative size = 158 \[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i f \arctan (\sinh (c+d x))}{6 a d^2}-\frac {2 f \log (\cosh (c+d x))}{3 a d^2}+\frac {f \text {sech}^2(c+d x)}{6 a d^2}+\frac {i (e+f x) \text {sech}^3(c+d x)}{3 a d}+\frac {2 (e+f x) \tanh (c+d x)}{3 a d}-\frac {i f \text {sech}(c+d x) \tanh (c+d x)}{6 a d^2}+\frac {(e+f x) \text {sech}^2(c+d x) \tanh (c+d x)}{3 a d} \] Output:

-1/6*I*f*arctan(sinh(d*x+c))/a/d^2-2/3*f*ln(cosh(d*x+c))/a/d^2+1/6*f*sech( 
d*x+c)^2/a/d^2+1/3*I*(f*x+e)*sech(d*x+c)^3/a/d+2/3*(f*x+e)*tanh(d*x+c)/a/d 
-1/6*I*f*sech(d*x+c)*tanh(d*x+c)/a/d^2+1/3*(f*x+e)*sech(d*x+c)^2*tanh(d*x+ 
c)/a/d
 

Mathematica [A] (verified)

Time = 2.87 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.23 \[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {2 d (e+f x) (\cosh (2 (c+d x))-2 i \sinh (c+d x))+\cosh (c+d x) \left (-d e-i f+c f-2 f \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+4 i f \log (\cosh (c+d x))-i \left (d e-c f+2 f \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-4 i f \log (\cosh (c+d x))\right ) \sinh (c+d x)\right )}{6 a d^2 \left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) (-i+\sinh (c+d x))} \] Input:

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

Output:

(2*d*(e + f*x)*(Cosh[2*(c + d*x)] - (2*I)*Sinh[c + d*x]) + Cosh[c + d*x]*( 
-(d*e) - I*f + c*f - 2*f*ArcTan[Tanh[(c + d*x)/2]] + (4*I)*f*Log[Cosh[c + 
d*x]] - I*(d*e - c*f + 2*f*ArcTan[Tanh[(c + d*x)/2]] - (4*I)*f*Log[Cosh[c 
+ d*x]])*Sinh[c + d*x]))/(6*a*d^2*(Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2] 
)*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(-I + Sinh[c + d*x]))
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.483, Rules used = {6105, 3042, 4673, 3042, 4672, 26, 3042, 26, 3956, 5974, 3042, 4255, 3042, 4257}

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.

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

\(\Big \downarrow \) 6105

\(\displaystyle \frac {\int (e+f x) \text {sech}^4(c+d x)dx}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )^4dx}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {2}{3} \int (e+f x) \text {sech}^2(c+d x)dx+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{3} \int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {i f \int -i \tanh (c+d x)dx}{d}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \int \tanh (c+d x)dx}{d}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \int -i \tan (i c+i d x)dx}{d}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}+\frac {i f \int \tan (i c+i d x)dx}{d}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \log (\cosh (c+d x))}{d^2}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \int (e+f x) \text {sech}^3(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 5974

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \log (\cosh (c+d x))}{d^2}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \left (\frac {f \int \text {sech}^3(c+d x)dx}{3 d}-\frac {(e+f x) \text {sech}^3(c+d x)}{3 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \log (\cosh (c+d x))}{d^2}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^3(c+d x)}{3 d}+\frac {f \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^3dx}{3 d}\right )}{a}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \log (\cosh (c+d x))}{d^2}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \left (\frac {f \left (\frac {1}{2} \int \text {sech}(c+d x)dx+\frac {\tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )}{3 d}-\frac {(e+f x) \text {sech}^3(c+d x)}{3 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \log (\cosh (c+d x))}{d^2}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^3(c+d x)}{3 d}+\frac {f \left (\frac {\tanh (c+d x) \text {sech}(c+d x)}{2 d}+\frac {1}{2} \int \csc \left (i c+i d x+\frac {\pi }{2}\right )dx\right )}{3 d}\right )}{a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {2}{3} \left (\frac {(e+f x) \tanh (c+d x)}{d}-\frac {f \log (\cosh (c+d x))}{d^2}\right )+\frac {f \text {sech}^2(c+d x)}{6 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^2(c+d x)}{3 d}}{a}-\frac {i \left (\frac {f \left (\frac {\arctan (\sinh (c+d x))}{2 d}+\frac {\tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )}{3 d}-\frac {(e+f x) \text {sech}^3(c+d x)}{3 d}\right )}{a}\)

Input:

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

Output:

((f*Sech[c + d*x]^2)/(6*d^2) + ((e + f*x)*Sech[c + d*x]^2*Tanh[c + d*x])/( 
3*d) + (2*(-((f*Log[Cosh[c + d*x]])/d^2) + ((e + f*x)*Tanh[c + d*x])/d))/3 
)/a - (I*(-1/3*((e + f*x)*Sech[c + d*x]^3)/d + (f*(ArcTan[Sinh[c + d*x]]/( 
2*d) + (Sech[c + d*x]*Tanh[c + d*x])/(2*d)))/(3*d)))/a
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042

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

rule 3956

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

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

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

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[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 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 = 16.60 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {4 f x}{3 a d}+\frac {4 f c}{3 a \,d^{2}}-\frac {i \left (-8 d f x \,{\mathrm e}^{d x +c}+f \,{\mathrm e}^{3 d x +3 c}-8 d e \,{\mathrm e}^{d x +c}+{\mathrm e}^{d x +c} f +4 i d f x +4 i e d \right )}{3 \left ({\mathrm e}^{d x +c}+i\right ) \left ({\mathrm e}^{d x +c}-i\right )^{3} d^{2} a}-\frac {f \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a \,d^{2}}-\frac {5 f \ln \left ({\mathrm e}^{d x +c}-i\right )}{6 a \,d^{2}}\)	\(143\)

Input:

int((f*x+e)*sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

4/3*f*x/a/d+4/3*f/a/d^2*c-1/3*I*(-8*d*f*x*exp(d*x+c)+f*exp(3*d*x+3*c)-8*d* 
e*exp(d*x+c)+exp(d*x+c)*f+4*I*d*f*x+4*I*e*d)/(exp(d*x+c)+I)/(exp(d*x+c)-I) 
^3/d^2/a-1/2*f/a/d^2*ln(exp(d*x+c)+I)-5/6*f/a/d^2*ln(exp(d*x+c)-I)
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.27 \[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {8 \, d f x e^{\left (4 \, d x + 4 \, c\right )} + 8 \, d e - 2 \, {\left (8 i \, d f x + i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} - 2 \, {\left (-8 i \, d e + i \, f\right )} e^{\left (d x + c\right )} - 3 \, {\left (f e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f e^{\left (d x + c\right )} - f\right )} \log \left (e^{\left (d x + c\right )} + i\right ) - 5 \, {\left (f e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f e^{\left (d x + c\right )} - f\right )} \log \left (e^{\left (d x + c\right )} - i\right )}{6 \, {\left (a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}\right )}} \] Input:

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

Output:

1/6*(8*d*f*x*e^(4*d*x + 4*c) + 8*d*e - 2*(8*I*d*f*x + I*f)*e^(3*d*x + 3*c) 
 - 2*(-8*I*d*e + I*f)*e^(d*x + c) - 3*(f*e^(4*d*x + 4*c) - 2*I*f*e^(3*d*x 
+ 3*c) - 2*I*f*e^(d*x + c) - f)*log(e^(d*x + c) + I) - 5*(f*e^(4*d*x + 4*c 
) - 2*I*f*e^(3*d*x + 3*c) - 2*I*f*e^(d*x + c) - f)*log(e^(d*x + c) - I))/( 
a*d^2*e^(4*d*x + 4*c) - 2*I*a*d^2*e^(3*d*x + 3*c) - 2*I*a*d^2*e^(d*x + c) 
- a*d^2)
 

Sympy [F]

\[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e \operatorname {sech}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f x \operatorname {sech}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \] Input:

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

Output:

-I*(Integral(e*sech(c + d*x)**2/(sinh(c + d*x) - I), x) + Integral(f*x*sec 
h(c + d*x)**2/(sinh(c + d*x) - I), x))/a
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.59 \[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {1}{6} \, f {\left (\frac {24 \, {\left (4 i \, d x e^{\left (4 \, d x + 4 \, c\right )} + {\left (8 \, d x e^{\left (3 \, c\right )} + e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + e^{\left (d x + c\right )}\right )}}{12 i \, a d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} + 24 \, a d^{2} e^{\left (d x + c\right )} - 12 i \, a d^{2}} - \frac {3 \, \log \left ({\left (e^{\left (d x + c\right )} + i\right )} e^{\left (-c\right )}\right )}{a d^{2}} - \frac {5 \, \log \left (-i \, {\left (i \, e^{\left (d x + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {4}{3} \, e {\left (\frac {2 \, e^{\left (-d x - c\right )}}{{\left (2 \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} - i \, a e^{\left (-4 \, d x - 4 \, c\right )} + i \, a\right )} d} + \frac {i}{{\left (2 \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} - i \, a e^{\left (-4 \, d x - 4 \, c\right )} + i \, a\right )} d}\right )} \] Input:

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

Output:

1/6*f*(24*(4*I*d*x*e^(4*d*x + 4*c) + (8*d*x*e^(3*c) + e^(3*c))*e^(3*d*x) + 
 e^(d*x + c))/(12*I*a*d^2*e^(4*d*x + 4*c) + 24*a*d^2*e^(3*d*x + 3*c) + 24* 
a*d^2*e^(d*x + c) - 12*I*a*d^2) - 3*log((e^(d*x + c) + I)*e^(-c))/(a*d^2) 
- 5*log(-I*(I*e^(d*x + c) + 1)*e^(-c))/(a*d^2)) + 4/3*e*(2*e^(-d*x - c)/(( 
2*a*e^(-d*x - c) + 2*a*e^(-3*d*x - 3*c) - I*a*e^(-4*d*x - 4*c) + I*a)*d) + 
 I/((2*a*e^(-d*x - c) + 2*a*e^(-3*d*x - 3*c) - I*a*e^(-4*d*x - 4*c) + I*a) 
*d))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.65 \[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {8 \, d f x e^{\left (4 \, d x + 4 \, c\right )} - 16 i \, d f x e^{\left (3 \, d x + 3 \, c\right )} + 16 i \, d e e^{\left (d x + c\right )} - 3 \, f e^{\left (4 \, d x + 4 \, c\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + 6 i \, f e^{\left (3 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + 6 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} + i\right ) - 5 \, f e^{\left (4 \, d x + 4 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 10 i \, f e^{\left (3 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 10 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 8 \, d e - 2 i \, f e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, f e^{\left (d x + c\right )} + 3 \, f \log \left (e^{\left (d x + c\right )} + i\right ) + 5 \, f \log \left (e^{\left (d x + c\right )} - i\right )}{6 \, {\left (a d^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}\right )}} \] Input:

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

Output:

1/6*(8*d*f*x*e^(4*d*x + 4*c) - 16*I*d*f*x*e^(3*d*x + 3*c) + 16*I*d*e*e^(d* 
x + c) - 3*f*e^(4*d*x + 4*c)*log(e^(d*x + c) + I) + 6*I*f*e^(3*d*x + 3*c)* 
log(e^(d*x + c) + I) + 6*I*f*e^(d*x + c)*log(e^(d*x + c) + I) - 5*f*e^(4*d 
*x + 4*c)*log(e^(d*x + c) - I) + 10*I*f*e^(3*d*x + 3*c)*log(e^(d*x + c) - 
I) + 10*I*f*e^(d*x + c)*log(e^(d*x + c) - I) + 8*d*e - 2*I*f*e^(3*d*x + 3* 
c) - 2*I*f*e^(d*x + c) + 3*f*log(e^(d*x + c) + I) + 5*f*log(e^(d*x + c) - 
I))/(a*d^2*e^(4*d*x + 4*c) - 2*I*a*d^2*e^(3*d*x + 3*c) - 2*I*a*d^2*e^(d*x 
+ c) - a*d^2)
 

Mupad [B] (verification not implemented)

Time = 3.51 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.30 \[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {4\,f\,x}{3\,a\,d}-\frac {f+3\,d\,e+3\,d\,f\,x}{3\,a\,d^2\,\left (1-{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}\right )}-\frac {5\,f\,\ln \left (f+f\,{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}{6\,a\,d^2}-\frac {\left (e+f\,x\right )\,2{}\mathrm {i}}{3\,a\,d\,\left (3\,{\mathrm {e}}^{c+d\,x}+{\mathrm {e}}^{2\,c+2\,d\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,c+3\,d\,x}-\mathrm {i}\right )}-\frac {\left (e+f\,x\right )\,1{}\mathrm {i}}{2\,a\,d\,\left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}-\frac {f\,\ln \left (-1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}{2\,a\,d^2}+\frac {\left (3\,d\,e-2\,f+3\,d\,f\,x\right )\,1{}\mathrm {i}}{6\,a\,d^2\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \] Input:

int((e + f*x)/(cosh(c + d*x)^2*(a + a*sinh(c + d*x)*1i)),x)
 

Output:

(4*f*x)/(3*a*d) - (f + 3*d*e + 3*d*f*x)/(3*a*d^2*(exp(c + d*x)*2i - exp(2* 
c + 2*d*x) + 1)) - (5*f*log(f + f*exp(c + d*x)*1i))/(6*a*d^2) - ((e + f*x) 
*2i)/(3*a*d*(3*exp(c + d*x) + exp(2*c + 2*d*x)*3i - exp(3*c + 3*d*x) - 1i) 
) - ((e + f*x)*1i)/(2*a*d*(exp(c + d*x) + 1i)) - (f*log(exp(c + d*x)*1i - 
1))/(2*a*d^2) + ((3*d*e - 2*f + 3*d*f*x)*1i)/(6*a*d^2*(exp(c + d*x) - 1i))
 

Reduce [F]

\[ \int \frac {(e+f x) \text {sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (\int \frac {\mathrm {sech}\left (d x +c \right )^{2}}{\sinh \left (d x +c \right ) i +1}d x \right ) e +\left (\int \frac {\mathrm {sech}\left (d x +c \right )^{2} x}{\sinh \left (d x +c \right ) i +1}d x \right ) f}{a} \] Input:

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

Output:

(int(sech(c + d*x)**2/(sinh(c + d*x)*i + 1),x)*e + int((sech(c + d*x)**2*x 
)/(sinh(c + d*x)*i + 1),x)*f)/a