3.255 \(\int \frac{(e+f x) \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=73 \[ -\frac{2 i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}-\frac{2 i (e+f x) \log \left (1+i e^{c+d x}\right )}{a d}+\frac{i (e+f x)^2}{2 a f} \]

[Out]

((I/2)*(e + f*x)^2)/(a*f) - ((2*I)*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d) - ((2*I)*f*PolyLog[2, (-I)*E^(c + d
*x)])/(a*d^2)

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Rubi [A]  time = 0.111111, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {5559, 2190, 2279, 2391} \[ -\frac{2 i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}-\frac{2 i (e+f x) \log \left (1+i e^{c+d x}\right )}{a d}+\frac{i (e+f x)^2}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(e + f*x)^2)/(a*f) - ((2*I)*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d) - ((2*I)*f*PolyLog[2, (-I)*E^(c + d
*x)])/(a*d^2)

Rule 5559

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + Dist[2, Int[((e + f*x)^m*E^(c + d*x))/(a + b*E^(c + d*x)), x], x
] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[a^2 + b^2, 0]

Rule 2190

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

Rule 2279

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 2391

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]

Rubi steps

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

Mathematica [A]  time = 0.0258735, size = 66, normalized size = 0.9 \[ \frac{i \left (d (e+f x) \left (d (e+f x)-4 f \log \left (1+i e^{c+d x}\right )\right )-4 f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )\right )}{2 a d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(d*(e + f*x)*(d*(e + f*x) - 4*f*Log[1 + I*E^(c + d*x)]) - 4*f^2*PolyLog[2, (-I)*E^(c + d*x)]))/(a*d^2*f
)

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Maple [B]  time = 0.073, size = 188, normalized size = 2.6 \begin{align*}{\frac{{\frac{i}{2}}f{x}^{2}}{a}}-{\frac{iex}{a}}-{\frac{2\,i\ln \left ({{\rm e}^{dx+c}}-i \right ) e}{da}}+{\frac{2\,i\ln \left ({{\rm e}^{dx+c}} \right ) e}{da}}+{\frac{2\,ifcx}{da}}+{\frac{if{c}^{2}}{a{d}^{2}}}-{\frac{2\,if\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{da}}-{\frac{2\,if\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{2}}}-{\frac{2\,if{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{2\,ifc\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}}-{\frac{2\,ifc\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, f{\left (-\frac{i \, x^{2}}{a} + 4 \, \int \frac{x}{a e^{\left (d x + c\right )} - i \, a}\,{d x}\right )} - \frac{i \, e \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f*(-I*x^2/a + 4*integrate(x/(a*e^(d*x + c) - I*a), x)) - I*e*log(I*a*sinh(d*x + c) + a)/(a*d)

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Fricas [A]  time = 2.18812, size = 252, normalized size = 3.45 \begin{align*} \frac{i \, d^{2} f x^{2} + 2 i \, d^{2} e x + 4 i \, c d e - 2 i \, c^{2} f - 4 i \, f{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) +{\left (-4 i \, d e + 4 i \, c f\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (-4 i \, d f x - 4 i \, c f\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{2 \, a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*d^2*f*x^2 + 2*I*d^2*e*x + 4*I*c*d*e - 2*I*c^2*f - 4*I*f*dilog(-I*e^(d*x + c)) + (-4*I*d*e + 4*I*c*f)*lo
g(e^(d*x + c) - I) + (-4*I*d*f*x - 4*I*c*f)*log(I*e^(d*x + c) + 1))/(a*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e \cosh{\left (c + d x \right )}}{i \sinh{\left (c + d x \right )} + 1}\, dx + \int \frac{f x \cosh{\left (c + d x \right )}}{i \sinh{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e*cosh(c + d*x)/(I*sinh(c + d*x) + 1), x) + Integral(f*x*cosh(c + d*x)/(I*sinh(c + d*x) + 1), x))/a

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )} \cosh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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