\(\int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [285]

Optimal result

Integrand size = 29, antiderivative size = 233 \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {3 (e+f x) \arctan \left (e^{c+d x}\right )}{4 a d}-\frac {3 i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{8 a d^2}+\frac {3 i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{8 a d^2}+\frac {3 f \text {sech}(c+d x)}{8 a d^2}+\frac {f \text {sech}^3(c+d x)}{12 a d^2}+\frac {i (e+f x) \text {sech}^4(c+d x)}{4 a d}-\frac {i f \tanh (c+d x)}{4 a d^2}+\frac {3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{8 a d}+\frac {(e+f x) \text {sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac {i f \tanh ^3(c+d x)}{12 a d^2} \] Output:

3/4*(f*x+e)*arctan(exp(d*x+c))/a/d-3/8*I*f*polylog(2,-I*exp(d*x+c))/a/d^2+ 
3/8*I*f*polylog(2,I*exp(d*x+c))/a/d^2+3/8*f*sech(d*x+c)/a/d^2+1/12*f*sech( 
d*x+c)^3/a/d^2+1/4*I*(f*x+e)*sech(d*x+c)^4/a/d-1/4*I*f*tanh(d*x+c)/a/d^2+3 
/8*(f*x+e)*sech(d*x+c)*tanh(d*x+c)/a/d+1/4*(f*x+e)*sech(d*x+c)^3*tanh(d*x+ 
c)/a/d+1/12*I*f*tanh(d*x+c)^3/a/d^2
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(617\) vs. \(2(233)=466\).

Time = 4.20 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.65 \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {2 (f+6 i d (e+f x))+\frac {6 i d (e+f x)}{\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}-9 (c+d x) (c f-d (2 e+f x)) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-(9-9 i) \left (\frac {1}{2} d^2 f x^2+d e (c+d x)-(1-i) (d e-c f) (c+d x)+(1-i) f (c+d x) \log \left (1+i e^{-c-d x}\right )+(1-i) (d e-c f) \log \left (i+e^{c+d x}\right )-(1-i) f \operatorname {PolyLog}\left (2,-i e^{-c-d x}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-(9+9 i) \left (\frac {1}{2} d^2 f x^2+d e (c+d x)-(1+i) (d e-c f) (c+d x)+(1+i) f (c+d x) \log \left (1-i e^{-c-d x}\right )+(1+i) (d e-c f) \log \left (i-e^{c+d x}\right )-(1+i) f \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-\frac {6 i d (e+f x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}{\left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 i f \sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )}+\frac {12 i f \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )}+28 f \sinh \left (\frac {1}{2} (c+d x)\right ) \left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{48 d^2 (a+i a \sinh (c+d x))} \] Input:

Integrate[((e + f*x)*Sech[c + d*x]^3)/(a + I*a*Sinh[c + d*x]),x]
 

Output:

(2*(f + (6*I)*d*(e + f*x)) + ((6*I)*d*(e + f*x))/(Cosh[(c + d*x)/2] + I*Si 
nh[(c + d*x)/2])^2 - 9*(c + d*x)*(c*f - d*(2*e + f*x))*(Cosh[(c + d*x)/2] 
+ I*Sinh[(c + d*x)/2])^2 - (9 - 9*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 - (9 + 9*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 - ((6*I)*d*(e + f*x)*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(C 
osh[(c + d*x)/2] - I*Sinh[(c + d*x)/2])^2 - ((4*I)*f*Sinh[(c + d*x)/2])/(C 
osh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) + ((12*I)*f*(Cosh[(c + d*x)/2] + I 
*Sinh[(c + d*x)/2])^2*Sinh[(c + d*x)/2])/(Cosh[(c + d*x)/2] - I*Sinh[(c + 
d*x)/2]) + 28*f*Sinh[(c + d*x)/2]*((-I)*Cosh[(c + d*x)/2] + Sinh[(c + d*x) 
/2]))/(48*d^2*(a + I*a*Sinh[c + d*x]))
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.94, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {6105, 3042, 4673, 3042, 4673, 3042, 4668, 2715, 2838, 5974, 3042, 4254, 2009}

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.

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

\(\Big \downarrow \) 6105

\(\displaystyle \frac {\int (e+f x) \text {sech}^5(c+d x)dx}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )^5dx}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {3}{4} \int (e+f x) \text {sech}^3(c+d x)dx+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )^3dx+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int (e+f x) \text {sech}(c+d x)dx+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )dx+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {i f \int \log \left (1-i e^{c+d x}\right )dx}{d}+\frac {i f \int \log \left (1+i e^{c+d x}\right )dx}{d}+\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {i f \int e^{-c-d x} \log \left (1-i e^{c+d x}\right )de^{c+d x}}{d^2}+\frac {i f \int e^{-c-d x} \log \left (1+i e^{c+d x}\right )de^{c+d x}}{d^2}+\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \int (e+f x) \text {sech}^4(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 5974

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \left (\frac {f \int \text {sech}^4(c+d x)dx}{4 d}-\frac {(e+f x) \text {sech}^4(c+d x)}{4 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^4(c+d x)}{4 d}+\frac {f \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^4dx}{4 d}\right )}{a}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^4(c+d x)}{4 d}+\frac {i f \int \left (1-\tanh ^2(c+d x)\right )d(-i \tanh (c+d x))}{4 d^2}\right )}{a}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^4(c+d x)}{4 d}+\frac {i f \left (\frac {1}{3} i \tanh ^3(c+d x)-i \tanh (c+d x)\right )}{4 d^2}\right )}{a}\)

Input:

Int[((e + f*x)*Sech[c + d*x]^3)/(a + I*a*Sinh[c + d*x]),x]
 

Output:

((f*Sech[c + d*x]^3)/(12*d^2) + ((e + f*x)*Sech[c + d*x]^3*Tanh[c + d*x])/ 
(4*d) + (3*(((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)))/4)/a - (I*(-1 
/4*((e + f*x)*Sech[c + d*x]^4)/d + ((I/4)*f*((-I)*Tanh[c + d*x] + (I/3)*Ta 
nh[c + d*x]^3))/d^2))/a
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 = 40.15 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.70

method	result	size
risch	\(\frac {9 d e \,{\mathrm e}^{d x +c}+9 f \,{\mathrm e}^{5 d x +5 c}+8 f \,{\mathrm e}^{3 d x +3 c}-{\mathrm e}^{d x +c} f -18 i d e \,{\mathrm e}^{4 d x +4 c}-18 i d f x \,{\mathrm e}^{4 d x +4 c}+9 d e \,{\mathrm e}^{5 d x +5 c}+6 d f x \,{\mathrm e}^{3 d x +3 c}-22 i f \,{\mathrm e}^{2 d x +2 c}+18 i d e \,{\mathrm e}^{2 d x +2 c}+9 d f x \,{\mathrm e}^{d x +c}+9 d f x \,{\mathrm e}^{5 d x +5 c}-18 i f \,{\mathrm e}^{4 d x +4 c}+18 i d f x \,{\mathrm e}^{2 d x +2 c}+6 d e \,{\mathrm e}^{3 d x +3 c}-4 i f}{12 \left ({\mathrm e}^{d x +c}+i\right )^{2} \left ({\mathrm e}^{d x +c}-i\right )^{4} d^{2} a}+\frac {3 e \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d a}+\frac {3 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) x}{8 d a}+\frac {3 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) c}{8 d^{2} a}+\frac {3 i f \operatorname {polylog}\left (2, i {\mathrm e}^{d x +c}\right )}{8 a \,d^{2}}-\frac {3 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{8 d a}-\frac {3 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{8 d^{2} a}-\frac {3 i f \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{8 a \,d^{2}}-\frac {3 f c \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d^{2} a}\)	\(396\)

Input:

int((f*x+e)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/12*(9*d*e*exp(d*x+c)+9*f*exp(5*d*x+5*c)+8*f*exp(3*d*x+3*c)-exp(d*x+c)*f- 
18*I*d*e*exp(4*d*x+4*c)-18*I*d*f*x*exp(4*d*x+4*c)+9*d*e*exp(5*d*x+5*c)+6*d 
*f*x*exp(3*d*x+3*c)-22*I*f*exp(2*d*x+2*c)+18*I*d*e*exp(2*d*x+2*c)+9*d*f*x* 
exp(d*x+c)+9*d*f*x*exp(5*d*x+5*c)-18*I*f*exp(4*d*x+4*c)+18*I*d*f*x*exp(2*d 
*x+2*c)+6*d*e*exp(3*d*x+3*c)-4*I*f)/(exp(d*x+c)+I)^2/(exp(d*x+c)-I)^4/d^2/ 
a+3/4/d/a*e*arctan(exp(d*x+c))+3/8*I/d/a*f*ln(1-I*exp(d*x+c))*x+3/8*I/d^2/ 
a*f*ln(1-I*exp(d*x+c))*c+3/8*I*f*polylog(2,I*exp(d*x+c))/a/d^2-3/8*I/d/a*f 
*ln(1+I*exp(d*x+c))*x-3/8*I/d^2/a*f*ln(1+I*exp(d*x+c))*c-3/8*I*f*polylog(2 
,-I*exp(d*x+c))/a/d^2-3/4/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. 917 vs. \(2 (197) = 394\).

Time = 0.11 (sec) , antiderivative size = 917, normalized size of antiderivative = 3.94 \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
 

Output:

-1/24*(9*(-I*f*e^(6*d*x + 6*c) - 2*f*e^(5*d*x + 5*c) - I*f*e^(4*d*x + 4*c) 
 - 4*f*e^(3*d*x + 3*c) + I*f*e^(2*d*x + 2*c) - 2*f*e^(d*x + c) + I*f)*dilo 
g(I*e^(d*x + c)) + 9*(I*f*e^(6*d*x + 6*c) + 2*f*e^(5*d*x + 5*c) + I*f*e^(4 
*d*x + 4*c) + 4*f*e^(3*d*x + 3*c) - I*f*e^(2*d*x + 2*c) + 2*f*e^(d*x + c) 
- I*f)*dilog(-I*e^(d*x + c)) - 18*(d*f*x + d*e + f)*e^(5*d*x + 5*c) + 36*( 
I*d*f*x + I*d*e + I*f)*e^(4*d*x + 4*c) - 4*(3*d*f*x + 3*d*e + 4*f)*e^(3*d* 
x + 3*c) + 4*(-9*I*d*f*x - 9*I*d*e + 11*I*f)*e^(2*d*x + 2*c) - 2*(9*d*f*x 
+ 9*d*e - f)*e^(d*x + c) + 9*(I*d*e - I*c*f + (-I*d*e + I*c*f)*e^(6*d*x + 
6*c) - 2*(d*e - c*f)*e^(5*d*x + 5*c) + (-I*d*e + I*c*f)*e^(4*d*x + 4*c) - 
4*(d*e - c*f)*e^(3*d*x + 3*c) + (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) + 9*(-I*d*e + I*c*f + (I*d*e - I*c 
*f)*e^(6*d*x + 6*c) + 2*(d*e - c*f)*e^(5*d*x + 5*c) + (I*d*e - I*c*f)*e^(4 
*d*x + 4*c) + 4*(d*e - c*f)*e^(3*d*x + 3*c) + (-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) + 9*(-I*d*f*x - I*c 
*f + (I*d*f*x + I*c*f)*e^(6*d*x + 6*c) + 2*(d*f*x + c*f)*e^(5*d*x + 5*c) + 
 (I*d*f*x + I*c*f)*e^(4*d*x + 4*c) + 4*(d*f*x + c*f)*e^(3*d*x + 3*c) + (-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) + 9*(I*d*f*x + I*c*f + (-I*d*f*x - I*c*f)*e^(6*d*x + 6*c) - 2* 
(d*f*x + c*f)*e^(5*d*x + 5*c) + (-I*d*f*x - I*c*f)*e^(4*d*x + 4*c) - 4*(d* 
f*x + c*f)*e^(3*d*x + 3*c) + (I*d*f*x + I*c*f)*e^(2*d*x + 2*c) - 2*(d*f...
 

Sympy [F]

\[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e \operatorname {sech}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f x \operatorname {sech}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \] Input:

integrate((f*x+e)*sech(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)
 

Output:

-I*(Integral(e*sech(c + d*x)**3/(sinh(c + d*x) - I), x) + Integral(f*x*sec 
h(c + d*x)**3/(sinh(c + d*x) - I), x))/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((f*x+e)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

\[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \operatorname {sech}\left (d x + c\right )^{3}}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:

integrate((f*x+e)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)*sech(d*x + c)^3/(I*a*sinh(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {e+f\,x}{{\mathrm {cosh}\left (c+d\,x\right )}^3\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:

int((e + f*x)/(cosh(c + d*x)^3*(a + a*sinh(c + d*x)*1i)),x)
 

Output:

int((e + f*x)/(cosh(c + d*x)^3*(a + a*sinh(c + d*x)*1i)), x)
 

Reduce [F]

\[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (\int \frac {\mathrm {sech}\left (d x +c \right )^{3}}{\sinh \left (d x +c \right ) i +1}d x \right ) e +\left (\int \frac {\mathrm {sech}\left (d x +c \right )^{3} x}{\sinh \left (d x +c \right ) i +1}d x \right ) f}{a} \] Input:

int((f*x+e)*sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x)
 

Output:

(int(sech(c + d*x)**3/(sinh(c + d*x)*i + 1),x)*e + int((sech(c + d*x)**3*x 
)/(sinh(c + d*x)*i + 1),x)*f)/a