3.3.72 \(\int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [272]

Optimal. Leaf size=268 \[ \frac {(e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right )}{a d}-\frac {f^2 \text {ArcTan}(\sinh (c+d x))}{a d^3}+\frac {i f^2 \log (\cosh (c+d x))}{a d^3}-\frac {i f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {i f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )}{a d^2}+\frac {i f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {i f^2 \text {PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d} \]

[Out]

(f*x+e)^2*arctan(exp(d*x+c))/a/d-f^2*arctan(sinh(d*x+c))/a/d^3+I*f^2*ln(cosh(d*x+c))/a/d^3-I*f*(f*x+e)*polylog
(2,-I*exp(d*x+c))/a/d^2+I*f*(f*x+e)*polylog(2,I*exp(d*x+c))/a/d^2+I*f^2*polylog(3,-I*exp(d*x+c))/a/d^3-I*f^2*p
olylog(3,I*exp(d*x+c))/a/d^3+f*(f*x+e)*sech(d*x+c)/a/d^2+1/2*I*(f*x+e)^2*sech(d*x+c)^2/a/d-I*f*(f*x+e)*tanh(d*
x+c)/a/d^2+1/2*(f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/a/d

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Rubi [A]
time = 0.19, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {5690, 4271, 3855, 4265, 2611, 2320, 6724, 5559, 4269, 3556} \begin {gather*} -\frac {f^2 \text {ArcTan}(\sinh (c+d x))}{a d^3}+\frac {(e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right )}{a d}+\frac {i f^2 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^3}-\frac {i f^2 \text {Li}_3\left (i e^{c+d x}\right )}{a d^3}+\frac {i f^2 \log (\cosh (c+d x))}{a d^3}-\frac {i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right )}{a d^2}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \tanh (c+d x) \text {sech}(c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)^2*ArcTan[E^(c + d*x)])/(a*d) - (f^2*ArcTan[Sinh[c + d*x]])/(a*d^3) + (I*f^2*Log[Cosh[c + d*x]])/(a*
d^3) - (I*f*(e + f*x)*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^2) + (I*f*(e + f*x)*PolyLog[2, I*E^(c + d*x)])/(a*d^2
) + (I*f^2*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^3) - (I*f^2*PolyLog[3, I*E^(c + d*x)])/(a*d^3) + (f*(e + f*x)*Se
ch[c + d*x])/(a*d^2) + ((I/2)*(e + f*x)^2*Sech[c + d*x]^2)/(a*d) - (I*f*(e + f*x)*Tanh[c + d*x])/(a*d^2) + ((e
+ f*x)^2*Sech[c + d*x]*Tanh[c + d*x])/(2*a*d)

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3556

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

Rule 3855

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

Rule 4265

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 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[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 4271

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-b^2)*(c + d*x)^m*Cot[e
+ f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))), Int[(c +
d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)^m*(b*Csc[e + f*x])^
(n - 2), x], x] - Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 5559

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Sim
p[(-(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 5690

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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {i \int (e+f x)^2 \text {sech}^2(c+d x) \tanh (c+d x) \, dx}{a}+\frac {\int (e+f x)^2 \text {sech}^3(c+d x) \, dx}{a}\\ &=\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d}+\frac {\int (e+f x)^2 \text {sech}(c+d x) \, dx}{2 a}-\frac {(i f) \int (e+f x) \text {sech}^2(c+d x) \, dx}{a d}-\frac {f^2 \int \text {sech}(c+d x) \, dx}{a d^2}\\ &=\frac {(e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {f^2 \tan ^{-1}(\sinh (c+d x))}{a d^3}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d}-\frac {(i f) \int (e+f x) \log \left (1-i e^{c+d x}\right ) \, dx}{a d}+\frac {(i f) \int (e+f x) \log \left (1+i e^{c+d x}\right ) \, dx}{a d}+\frac {\left (i f^2\right ) \int \tanh (c+d x) \, dx}{a d^2}\\ &=\frac {(e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {f^2 \tan ^{-1}(\sinh (c+d x))}{a d^3}+\frac {i f^2 \log (\cosh (c+d x))}{a d^3}-\frac {i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right )}{a d^2}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d}+\frac {\left (i f^2\right ) \int \text {Li}_2\left (-i e^{c+d x}\right ) \, dx}{a d^2}-\frac {\left (i f^2\right ) \int \text {Li}_2\left (i e^{c+d x}\right ) \, dx}{a d^2}\\ &=\frac {(e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {f^2 \tan ^{-1}(\sinh (c+d x))}{a d^3}+\frac {i f^2 \log (\cosh (c+d x))}{a d^3}-\frac {i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right )}{a d^2}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d}+\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}-\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}\\ &=\frac {(e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {f^2 \tan ^{-1}(\sinh (c+d x))}{a d^3}+\frac {i f^2 \log (\cosh (c+d x))}{a d^3}-\frac {i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^2}+\frac {i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right )}{a d^2}+\frac {i f^2 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^3}-\frac {i f^2 \text {Li}_3\left (i e^{c+d x}\right )}{a d^3}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(599\) vs. \(2(268)=536\).
time = 7.25, size = 599, normalized size = 2.24 \begin {gather*} -\frac {\frac {-3 i d^3 e^2 e^c x+12 i d e^c f^2 x-3 i d^3 e e^c f x^2-i d^3 e^c f^2 x^3+3 d^2 e^2 \log \left (i-e^{c+d x}\right )+3 i d^2 e^2 e^c \log \left (i-e^{c+d x}\right )-12 f^2 \log \left (i-e^{c+d x}\right )-12 i e^c f^2 \log \left (i-e^{c+d x}\right )+6 d^2 e f x \log \left (1+i e^{c+d x}\right )+6 i d^2 e e^c f x \log \left (1+i e^{c+d x}\right )+3 d^2 f^2 x^2 \log \left (1+i e^{c+d x}\right )+3 i d^2 e^c f^2 x^2 \log \left (1+i e^{c+d x}\right )+6 d \left (1+i e^c\right ) f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )-6 \left (1+i e^c\right ) f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{d^3 \left (-i+e^c\right )}+\frac {d^2 \left (i d e^c x \left (3 e^2+3 e f x+f^2 x^2\right )+3 \left (1-i e^c\right ) (e+f x)^2 \log \left (1-i e^{c+d x}\right )\right )+6 d \left (1-i e^c\right ) f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )+6 i \left (i+e^c\right ) f^2 \text {PolyLog}\left (3,i e^{c+d x}\right )}{d^3 \left (i+e^c\right )}-x \left (3 e^2+3 e f x+f^2 x^2\right ) \text {sech}(c)-\frac {3 i (e+f x)^2}{d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 i f (e+f x) \sinh \left (\frac {d x}{2}\right )}{d^2 \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}}{6 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(((-3*I)*d^3*e^2*E^c*x + (12*I)*d*E^c*f^2*x - (3*I)*d^3*e*E^c*f*x^2 - I*d^3*E^c*f^2*x^3 + 3*d^2*e^2*Log[I
- E^(c + d*x)] + (3*I)*d^2*e^2*E^c*Log[I - E^(c + d*x)] - 12*f^2*Log[I - E^(c + d*x)] - (12*I)*E^c*f^2*Log[I
- E^(c + d*x)] + 6*d^2*e*f*x*Log[1 + I*E^(c + d*x)] + (6*I)*d^2*e*E^c*f*x*Log[1 + I*E^(c + d*x)] + 3*d^2*f^2*x
^2*Log[1 + I*E^(c + d*x)] + (3*I)*d^2*E^c*f^2*x^2*Log[1 + I*E^(c + d*x)] + 6*d*(1 + I*E^c)*f*(e + f*x)*PolyLog
[2, (-I)*E^(c + d*x)] - 6*(1 + I*E^c)*f^2*PolyLog[3, (-I)*E^(c + d*x)])/(d^3*(-I + E^c)) + (d^2*(I*d*E^c*x*(3*
e^2 + 3*e*f*x + f^2*x^2) + 3*(1 - I*E^c)*(e + f*x)^2*Log[1 - I*E^(c + d*x)]) + 6*d*(1 - I*E^c)*f*(e + f*x)*Pol
yLog[2, I*E^(c + d*x)] + (6*I)*(I + E^c)*f^2*PolyLog[3, I*E^(c + d*x)])/(d^3*(I + E^c)) - x*(3*e^2 + 3*e*f*x +
f^2*x^2)*Sech[c] - ((3*I)*(e + f*x)^2)/(d*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2) + ((12*I)*f*(e + f*x)*
Sinh[(d*x)/2])/(d^2*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))/a

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (248 ) = 496\).
time = 3.21, size = 613, normalized size = 2.29

method	result	size
risch	\(\frac {d \,x^{2} f^{2} {\mathrm e}^{d x +c}+2 d e f x \,{\mathrm e}^{d x +c}+d \,e^{2} {\mathrm e}^{d x +c}-2 i f^{2} x +2 f^{2} x \,{\mathrm e}^{d x +c}-2 i e f +2 e f \,{\mathrm e}^{d x +c}}{\left ({\mathrm e}^{d x +c}-i\right )^{2} d^{2} a}+\frac {i e^{2} \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a d}+\frac {i \polylog \left (2, i {\mathrm e}^{d x +c}\right ) f^{2} x}{a \,d^{2}}-\frac {i e f c \ln \left ({\mathrm e}^{d x +c}+i\right )}{a \,d^{2}}+\frac {i f c e \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{2} a}+\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) c e f}{a \,d^{2}}+\frac {i f^{2} \polylog \left (3, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) f^{2} x^{2}}{2 a d}-\frac {i e f \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {2 i f^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) e f x}{a d}-\frac {i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{2 d a}-\frac {i \ln \left (1+i {\mathrm e}^{d x +c}\right ) c e f}{a \,d^{2}}+\frac {2 i f^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}-\frac {i c^{2} f^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 a \,d^{3}}-\frac {i f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right ) x}{d^{2} a}-\frac {i \ln \left (1+i {\mathrm e}^{d x +c}\right ) e f x}{a d}-\frac {i f^{2} \polylog \left (3, i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {i e f \polylog \left (2, i {\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) c^{2} f^{2}}{2 a \,d^{3}}-\frac {i e^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 a d}+\frac {i \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2} f^{2}}{2 a \,d^{3}}+\frac {i c^{2} f^{2} \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 a \,d^{3}}\)	\(613\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x^2*f^2*exp(d*x+c)+2*d*e*f*x*exp(d*x+c)+d*e^2*exp(d*x+c)-2*I*f^2*x+2*f^2*x*exp(d*x+c)-2*I*e*f+2*e*f*exp(d*x
+c))/(exp(d*x+c)-I)^2/d^2/a+1/2*I/a/d*e^2*ln(exp(d*x+c)+I)-I/a/d^2*e*f*c*ln(exp(d*x+c)+I)+I/a/d^2*e*f*c*ln(exp
(d*x+c)-I)+I/a/d^2*ln(1-I*exp(d*x+c))*c*e*f+I*f^2*polylog(3,-I*exp(d*x+c))/a/d^3+1/2*I/a/d*ln(1-I*exp(d*x+c))*
f^2*x^2-I/a/d^2*e*f*polylog(2,-I*exp(d*x+c))+I/a/d^2*polylog(2,I*exp(d*x+c))*f^2*x-2*I/a/d^3*f^2*ln(exp(d*x+c)
)+I/a/d*ln(1-I*exp(d*x+c))*e*f*x-1/2*I/a/d*ln(1+I*exp(d*x+c))*f^2*x^2-I*f^2*polylog(3,I*exp(d*x+c))/a/d^3-I/a/
d^2*ln(1+I*exp(d*x+c))*c*e*f+2*I/a/d^3*f^2*ln(exp(d*x+c)-I)-1/2*I/a/d^3*c^2*f^2*ln(exp(d*x+c)-I)-I/a/d*ln(1+I*
exp(d*x+c))*e*f*x-I/a/d^2*polylog(2,-I*exp(d*x+c))*f^2*x-1/2*I/a/d^3*ln(1-I*exp(d*x+c))*c^2*f^2+I/a/d^2*e*f*po
lylog(2,I*exp(d*x+c))-1/2*I/a/d*e^2*ln(exp(d*x+c)-I)+1/2*I/a/d^3*ln(1+I*exp(d*x+c))*c^2*f^2+1/2*I/a/d^3*c^2*f^
2*ln(exp(d*x+c)+I)

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Maxima [A]
time = 0.44, size = 394, normalized size = 1.47 \begin {gather*} -\frac {1}{2} \, {\left (\frac {4 \, e^{\left (-d x - c\right )}}{-2 \, {\left (-2 i \, a e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} + 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 )} e^{2} + \frac {-2 i \, f^{2} x - 2 i \, f e + {\left (d f^{2} x^{2} e^{c} + 2 \, {\left (d f e^{\left (c + 1\right )} + f^{2} e^{c}\right )} x + 2 \, f e^{\left (c + 1\right )}\right )} e^{\left (d x\right )}}{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}} - \frac {2 i \, f^{2} x}{a d^{2}} - \frac {i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} f e}{a d^{2}} + \frac {i \, {\left (d x \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right )\right )} f e}{a d^{2}} - \frac {i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{2}}{2 \, a d^{3}} + \frac {i \, {\left (d^{2} x^{2} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(i \, e^{\left (d x + c\right )})\right )} f^{2}}{2 \, a d^{3}} + \frac {2 i \, f^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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))*e^2 + (-2*I*f^2*x - 2*I*f*e + (d*f^2*x^2*e^c + 2*(d*f*e^(c + 1) + f^2*e^c)*x +
2*f*e^(c + 1))*e^(d*x))/(a*d^2*e^(2*d*x + 2*c) - 2*I*a*d^2*e^(d*x + c) - a*d^2) - 2*I*f^2*x/(a*d^2) - I*(d*x*
log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*f*e/(a*d^2) + I*(d*x*log(-I*e^(d*x + c) + 1) + dilog(I*e^(d*x
+ c)))*f*e/(a*d^2) - 1/2*I*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dilog(-I*e^(d*x + c)) - 2*polylog(3, -I*e^(
d*x + c)))*f^2/(a*d^3) + 1/2*I*(d^2*x^2*log(-I*e^(d*x + c) + 1) + 2*d*x*dilog(I*e^(d*x + c)) - 2*polylog(3, I*
e^(d*x + c)))*f^2/(a*d^3) + 2*I*f^2*log(I*e^(d*x + c) + 1)/(a*d^3)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 827 vs. \(2 (242) = 484\).
time = 0.40, size = 827, normalized size = 3.09 \begin {gather*} \frac {4 i \, c f^{2} - 4 i \, d f e - 2 \, {\left (i \, d f^{2} x + i \, d f e + {\left (-i \, d f^{2} x - i \, d f e\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, {\left (d f^{2} x + d f e\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) - 2 \, {\left (-i \, d f^{2} x - i \, d f e + {\left (i \, d f^{2} x + i \, d f e\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d f^{2} x + d f e\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 4 \, {\left (i \, d f^{2} x + i \, c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d^{2} f^{2} x^{2} - 2 \, d f^{2} x - 4 \, c f^{2} + d^{2} e^{2} + 2 \, {\left (d^{2} f x + d f\right )} e\right )} e^{\left (d x + c\right )} + {\left (-i \, c^{2} f^{2} + 2 i \, c d f e - i \, d^{2} e^{2} + {\left (i \, c^{2} f^{2} - 2 i \, c d f e + i \, d^{2} e^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + {\left (-2 i \, c d f e + {\left (i \, c^{2} - 4 i\right )} f^{2} + i \, d^{2} e^{2} + {\left (2 i \, c d f e + {\left (-i \, c^{2} + 4 i\right )} f^{2} - i \, d^{2} e^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (2 \, c d f e - {\left (c^{2} - 4\right )} f^{2} - d^{2} e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (i \, d^{2} f^{2} x^{2} - i \, c^{2} f^{2} - 2 \, {\left (-i \, d^{2} f x - i \, c d f\right )} e + {\left (-i \, d^{2} f^{2} x^{2} + i \, c^{2} f^{2} - 2 \, {\left (i \, d^{2} f x + i \, c d f\right )} e\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\left (-i \, d^{2} f^{2} x^{2} + i \, c^{2} f^{2} - 2 \, {\left (i \, d^{2} f x + i \, c d f\right )} e + {\left (i \, d^{2} f^{2} x^{2} - i \, c^{2} f^{2} - 2 \, {\left (-i \, d^{2} f x - i \, c d f\right )} e\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} e^{\left (d x + c\right )}\right )} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) - 2 \, {\left (i \, f^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, f^{2} e^{\left (d x + c\right )} - i \, f^{2}\right )} {\rm polylog}\left (3, i \, e^{\left (d x + c\right )}\right ) - 2 \, {\left (-i \, f^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, f^{2} e^{\left (d x + c\right )} + i \, f^{2}\right )} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{2 \, {\left (a d^{3} e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d^{3} e^{\left (d x + c\right )} - a d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*I*c*f^2 - 4*I*d*f*e - 2*(I*d*f^2*x + I*d*f*e + (-I*d*f^2*x - I*d*f*e)*e^(2*d*x + 2*c) - 2*(d*f^2*x + d*
f*e)*e^(d*x + c))*dilog(I*e^(d*x + c)) - 2*(-I*d*f^2*x - I*d*f*e + (I*d*f^2*x + I*d*f*e)*e^(2*d*x + 2*c) + 2*(
d*f^2*x + d*f*e)*e^(d*x + c))*dilog(-I*e^(d*x + c)) - 4*(I*d*f^2*x + I*c*f^2)*e^(2*d*x + 2*c) + 2*(d^2*f^2*x^2
- 2*d*f^2*x - 4*c*f^2 + d^2*e^2 + 2*(d^2*f*x + d*f)*e)*e^(d*x + c) + (-I*c^2*f^2 + 2*I*c*d*f*e - I*d^2*e^2 +
(I*c^2*f^2 - 2*I*c*d*f*e + I*d^2*e^2)*e^(2*d*x + 2*c) + 2*(c^2*f^2 - 2*c*d*f*e + d^2*e^2)*e^(d*x + c))*log(e^(
d*x + c) + I) + (-2*I*c*d*f*e + (I*c^2 - 4*I)*f^2 + I*d^2*e^2 + (2*I*c*d*f*e + (-I*c^2 + 4*I)*f^2 - I*d^2*e^2)
*e^(2*d*x + 2*c) + 2*(2*c*d*f*e - (c^2 - 4)*f^2 - d^2*e^2)*e^(d*x + c))*log(e^(d*x + c) - I) + (I*d^2*f^2*x^2
- I*c^2*f^2 - 2*(-I*d^2*f*x - I*c*d*f)*e + (-I*d^2*f^2*x^2 + I*c^2*f^2 - 2*(I*d^2*f*x + I*c*d*f)*e)*e^(2*d*x +
2*c) - 2*(d^2*f^2*x^2 - c^2*f^2 + 2*(d^2*f*x + c*d*f)*e)*e^(d*x + c))*log(I*e^(d*x + c) + 1) + (-I*d^2*f^2*x^
2 + I*c^2*f^2 - 2*(I*d^2*f*x + I*c*d*f)*e + (I*d^2*f^2*x^2 - I*c^2*f^2 - 2*(-I*d^2*f*x - I*c*d*f)*e)*e^(2*d*x
+ 2*c) + 2*(d^2*f^2*x^2 - c^2*f^2 + 2*(d^2*f*x + c*d*f)*e)*e^(d*x + c))*log(-I*e^(d*x + c) + 1) - 2*(I*f^2*e^(
2*d*x + 2*c) + 2*f^2*e^(d*x + c) - I*f^2)*polylog(3, I*e^(d*x + c)) - 2*(-I*f^2*e^(2*d*x + 2*c) - 2*f^2*e^(d*x
+ c) + I*f^2)*polylog(3, -I*e^(d*x + c)))/(a*d^3*e^(2*d*x + 2*c) - 2*I*a*d^3*e^(d*x + c) - a*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \left (\int \frac {e^{2} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{2} x^{2} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {2 e f x \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^2}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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