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

Optimal. Leaf size=161 \[ \frac {(e+f x) \text {ArcTan}\left (e^{c+d x}\right )}{a d}-\frac {i f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}+\frac {i f \text {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {f \text {sech}(c+d x)}{2 a d^2}+\frac {i (e+f x) \text {sech}^2(c+d x)}{2 a d}-\frac {i f \tanh (c+d x)}{2 a d^2}+\frac {(e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 a d} \]

[Out]

(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)*sech(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

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Rubi [A]
time = 0.10, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {5690, 4270, 4265, 2317, 2438, 5559, 3852, 8} \begin {gather*} \frac {(e+f x) \text {ArcTan}\left (e^{c+d x}\right )}{a d}-\frac {i f \text {Li}_2\left (-i e^{c+d x}\right )}{2 a d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{2 a d^2}-\frac {i f \tanh (c+d x)}{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 {(e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 3852

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

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 4270

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

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]

Rubi steps

\begin {align*} \int \frac {(e+f x) \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {i \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x) \, dx}{a}+\frac {\int (e+f x) \text {sech}^3(c+d x) \, dx}{a}\\ &=\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 {(e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 a d}+\frac {\int (e+f x) \text {sech}(c+d x) \, dx}{2 a}-\frac {(i f) \int \text {sech}^2(c+d x) \, dx}{2 a d}\\ &=\frac {(e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\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 {(e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 a d}+\frac {f \text {Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{2 a d^2}-\frac {(i f) \int \log \left (1-i e^{c+d x}\right ) \, dx}{2 a d}+\frac {(i f) \int \log \left (1+i e^{c+d x}\right ) \, dx}{2 a d}\\ &=\frac {(e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d}+\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}-\frac {(i f) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{2 a d^2}+\frac {(i f) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{2 a d^2}\\ &=\frac {(e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i f \text {Li}_2\left (-i e^{c+d x}\right )}{2 a d^2}+\frac {i f \text {Li}_2\left (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}\\ \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. \(710\) vs. \(2(161)=322\).
time = 2.21, size = 710, normalized size = 4.41 \begin {gather*} -\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+d e \left (c+d x-2 i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-c f \left (c+d x-2 i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2+d e \left (c+d x+2 i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2-c f \left (c+d x+2 i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\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 {f \left (-2 (-1)^{3/4} (c+d x)^2+\sqrt {2} \left (2 (-2 i c+\pi -2 i d x) \log \left (1+i e^{-c-d x}\right )+\pi \left (3 c+3 d x-4 \log \left (1+e^{c+d x}\right )+4 \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (-\sin \left (\frac {1}{4} (\pi -2 i (c+d x))\right )\right )\right )+4 i \text {PolyLog}\left (2,-i e^{-c-d x}\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}{2 \sqrt {2}}+\frac {f \left (2 \sqrt [4]{-1} (c+d x)^2+\sqrt {2} \left (2 (2 i c+\pi +2 i d x) \log \left (1-i e^{-c-d x}\right )-\pi \left (c+d x-4 \log \left (1+e^{c+d x}\right )+4 \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{4} (\pi +2 i (c+d x))\right )\right )\right )-4 i \text {PolyLog}\left (2,i e^{-c-d x}\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}{2 \sqrt {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))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-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 + d*e*(
c + d*x - (2*I)*Log[Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2 - c*
f*(c + d*x - (2*I)*Log[Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2 +
d*e*(c + d*x + (2*I)*Log[Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^
2 - c*f*(c + d*x + (2*I)*Log[Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2
])^2 + (f*(-2*(-1)^(3/4)*(c + d*x)^2 + Sqrt[2]*(2*((-2*I)*c + Pi - (2*I)*d*x)*Log[1 + I*E^(-c - d*x)] + Pi*(3*
c + 3*d*x - 4*Log[1 + E^(c + d*x)] + 4*Log[Cosh[(c + d*x)/2]] - 2*Log[-Sin[(Pi - (2*I)*(c + d*x))/4]]) + (4*I)
*PolyLog[2, (-I)*E^(-c - d*x)]))*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(2*Sqrt[2]) + (f*(2*(-1)^(1/4)*(
c + d*x)^2 + Sqrt[2]*(2*((2*I)*c + Pi + (2*I)*d*x)*Log[1 - I*E^(-c - d*x)] - Pi*(c + d*x - 4*Log[1 + E^(c + d*
x)] + 4*Log[Cosh[(c + d*x)/2]] + 2*Log[Sin[(Pi + (2*I)*(c + d*x))/4]]) - (4*I)*PolyLog[2, I*E^(-c - d*x)]))*(C
osh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2)/(2*Sqrt[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]))

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Maple [A]
time = 4.04, size = 268, normalized size = 1.66

method	result	size
risch	\(\frac {d f x \,{\mathrm e}^{d x +c}+d e \,{\mathrm e}^{d x +c}+f \,{\mathrm e}^{d x +c}-i f}{\left ({\mathrm e}^{d x +c}-i\right )^{2} d^{2} a}+\frac {i e \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 d a}-\frac {i e \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 d a}+\frac {i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) x}{2 d a}+\frac {i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) c}{2 d^{2} a}+\frac {i f \polylog \left (2, i {\mathrm e}^{d x +c}\right )}{2 a \,d^{2}}-\frac {i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{2 d a}-\frac {i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{2 d^{2} a}-\frac {i f \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{2 a \,d^{2}}-\frac {i f c \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 d^{2} a}+\frac {i f c \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 d^{2} a}\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*(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

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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. 359 vs. \(2 (138) = 276\).
time = 0.40, size = 359, normalized size = 2.23 \begin {gather*} \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 \, c f - i \, d e + {\left (-i \, c f + i \, d e\right )} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, {\left (c f - d e\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + {\left (-i \, c f + i \, d e + {\left (i \, c f - i \, d e\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (c f - d e\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*c*f - I*d*e + (-I*c*f + I*d*e)*e^(2*d*
x + 2*c) - 2*(c*f - d*e)*e^(d*x + c))*log(e^(d*x + c) + I) + (-I*c*f + I*d*e + (I*c*f - I*d*e)*e^(2*d*x + 2*c)
+ 2*(c*f - d*e)*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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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

[Out]

integrate((f*x + e)*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 = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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