Integrand size = 29, antiderivative size = 233 \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {3 (e+f x) \arctan \left (e^{c+d x}\right )}{4 a d}-\frac {3 i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{8 a d^2}+\frac {3 i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{8 a d^2}+\frac {3 f \text {sech}(c+d x)}{8 a d^2}+\frac {f \text {sech}^3(c+d x)}{12 a d^2}+\frac {i (e+f x) \text {sech}^4(c+d x)}{4 a d}-\frac {i f \tanh (c+d x)}{4 a d^2}+\frac {3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{8 a d}+\frac {(e+f x) \text {sech}^3(c+d x) \tanh (c+d x)}{4 a d}+\frac {i f \tanh ^3(c+d x)}{12 a d^2} \]
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
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(617\) vs. \(2(233)=466\).
Time = 4.02 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.65 \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {2 (f+6 i d (e+f x))+\frac {6 i d (e+f x)}{\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}-9 (c+d x) (c f-d (2 e+f x)) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-(9-9 i) \left (\frac {1}{2} d^2 f x^2+d e (c+d x)-(1-i) (d e-c f) (c+d x)+(1-i) f (c+d x) \log \left (1+i e^{-c-d x}\right )+(1-i) (d e-c f) \log \left (i+e^{c+d x}\right )-(1-i) f \operatorname {PolyLog}\left (2,-i e^{-c-d x}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-(9+9 i) \left (\frac {1}{2} d^2 f x^2+d e (c+d x)-(1+i) (d e-c f) (c+d x)+(1+i) f (c+d x) \log \left (1-i e^{-c-d x}\right )+(1+i) (d e-c f) \log \left (i-e^{c+d x}\right )-(1+i) f \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-\frac {6 i d (e+f x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}{\left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 i f \sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )}+\frac {12 i f \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )}+28 f \sinh \left (\frac {1}{2} (c+d x)\right ) \left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{48 d^2 (a+i a \sinh (c+d x))} \]
(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]))
Time = 1.15 (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 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )+\frac {f \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 d}\right )+\frac {f \text {sech}^3(c+d x)}{12 d^2}+\frac {(e+f x) \tanh (c+d x) \text {sech}^3(c+d x)}{4 d}}{a}-\frac {i \left (-\frac {(e+f x) \text {sech}^4(c+d x)}{4 d}+\frac {i f \left (\frac {1}{3} i \tanh ^3(c+d x)-i \tanh (c+d x)\right )}{4 d^2}\right )}{a}\) |
((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
3.3.85.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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]
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]
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]
Time = 25.57 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.70
method | result | size |
risch | \(\frac {9 d f x \,{\mathrm e}^{5 d x +5 c}+6 d e \,{\mathrm e}^{3 d x +3 c}-18 i d e \,{\mathrm e}^{4 d x +4 c}+9 d e \,{\mathrm e}^{5 d x +5 c}-18 i d f x \,{\mathrm e}^{4 d x +4 c}-22 i f \,{\mathrm e}^{2 d x +2 c}+9 d e \,{\mathrm e}^{d x +c}-18 i f \,{\mathrm e}^{4 d x +4 c}-4 i f +6 d f x \,{\mathrm e}^{3 d x +3 c}+18 i d e \,{\mathrm e}^{2 d x +2 c}+18 i d f x \,{\mathrm e}^{2 d x +2 c}+9 d f x \,{\mathrm e}^{d x +c}+9 f \,{\mathrm e}^{5 d x +5 c}+8 f \,{\mathrm e}^{3 d x +3 c}-f \,{\mathrm e}^{d x +c}}{12 \left ({\mathrm e}^{d x +c}+i\right )^{2} \left ({\mathrm e}^{d x +c}-i\right )^{4} d^{2} a}+\frac {3 e \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d a}+\frac {3 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) x}{8 d a}+\frac {3 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) c}{8 d^{2} a}+\frac {3 i f \operatorname {polylog}\left (2, i {\mathrm e}^{d x +c}\right )}{8 a \,d^{2}}-\frac {3 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{8 d a}-\frac {3 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{8 d^{2} a}-\frac {3 i f \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{8 a \,d^{2}}-\frac {3 f c \arctan \left ({\mathrm e}^{d x +c}\right )}{4 d^{2} a}\) | \(396\) |
1/12*(9*d*f*x*exp(5*d*x+5*c)+6*d*e*exp(3*d*x+3*c)-18*I*d*e*exp(4*d*x+4*c)+ 9*d*e*exp(5*d*x+5*c)-18*I*d*f*x*exp(4*d*x+4*c)-22*I*f*exp(2*d*x+2*c)+9*d*e *exp(d*x+c)-18*I*f*exp(4*d*x+4*c)-4*I*f+6*d*f*x*exp(3*d*x+3*c)+18*I*d*e*ex p(2*d*x+2*c)+18*I*d*f*x*exp(2*d*x+2*c)+9*d*f*x*exp(d*x+c)+9*f*exp(5*d*x+5* c)+8*f*exp(3*d*x+3*c)-f*exp(d*x+c))/(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))
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 = 0.27 (sec) , antiderivative size = 917, normalized size of antiderivative = 3.94 \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Too large to display} \]
-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...
\[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e \operatorname {sech}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f x \operatorname {sech}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \]
-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
Exception generated. \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )} \operatorname {sech}\left (d x + c\right )^{3}}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {e+f\,x}{{\mathrm {cosh}\left (c+d\,x\right )}^3\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]