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

Optimal. Leaf size=63 \[ \frac{(c+d x) \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a f}-\frac{2 d \log \left (\cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )\right )}{a f^2} \]

[Out]

(-2*d*Log[Cosh[e/2 + (I/4)*Pi + (f*x)/2]])/(a*f^2) + ((c + d*x)*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(a*f)

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Rubi [A]  time = 0.0739995, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3318, 4184, 3475} \[ \frac{(c+d x) \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a f}-\frac{2 d \log \left (\cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )\right )}{a f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*Log[Cosh[e/2 + (I/4)*Pi + (f*x)/2]])/(a*f^2) + ((c + d*x)*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(a*f)

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4184

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 3475

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

Rubi steps

\begin{align*} \int \frac{c+d x}{a+i a \sinh (e+f x)} \, dx &=\frac{\int (c+d x) \csc ^2\left (\frac{1}{2} \left (i e+\frac{\pi }{2}\right )+\frac{i f x}{2}\right ) \, dx}{2 a}\\ &=\frac{(c+d x) \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}-\frac{d \int \coth \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=-\frac{2 d \log \left (\cosh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )\right )}{a f^2}+\frac{(c+d x) \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}\\ \end{align*}

Mathematica [B]  time = 0.450552, size = 185, normalized size = 2.94 \[ \frac{2 c f \sinh \left (\frac{f x}{2}\right )+i d f x \cosh \left (e+\frac{f x}{2}\right )-i d \sinh \left (e+\frac{f x}{2}\right ) \log (\cosh (e+f x))+2 d \sinh \left (e+\frac{f x}{2}\right ) \tan ^{-1}\left (\sinh \left (\frac{f x}{2}\right ) \text{sech}\left (e+\frac{f x}{2}\right )\right )+\cosh \left (\frac{f x}{2}\right ) \left (-d \log (\cosh (e+f x))-2 i d \tan ^{-1}\left (\sinh \left (\frac{f x}{2}\right ) \text{sech}\left (e+\frac{f x}{2}\right )\right )\right )+d f x \sinh \left (\frac{f x}{2}\right )}{a f^2 \left (\cosh \left (\frac{e}{2}\right )+i \sinh \left (\frac{e}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*d*f*x*Cosh[e + (f*x)/2] + Cosh[(f*x)/2]*((-2*I)*d*ArcTan[Sech[e + (f*x)/2]*Sinh[(f*x)/2]] - d*Log[Cosh[e +
f*x]]) + 2*c*f*Sinh[(f*x)/2] + d*f*x*Sinh[(f*x)/2] + 2*d*ArcTan[Sech[e + (f*x)/2]*Sinh[(f*x)/2]]*Sinh[e + (f*x
)/2] - I*d*Log[Cosh[e + f*x]]*Sinh[e + (f*x)/2])/(a*f^2*(Cosh[e/2] + I*Sinh[e/2])*(Cosh[(e + f*x)/2] + I*Sinh[
(e + f*x)/2]))

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Maple [A]  time = 0.056, size = 66, normalized size = 1.1 \begin{align*} 2\,{\frac{dx}{af}}+2\,{\frac{de}{a{f}^{2}}}+{\frac{2\,i \left ( dx+c \right ) }{af \left ({{\rm e}^{fx+e}}-i \right ) }}-2\,{\frac{d\ln \left ({{\rm e}^{fx+e}}-i \right ) }{a{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06809, size = 101, normalized size = 1.6 \begin{align*} 2 \, d{\left (\frac{x e^{\left (f x + e\right )}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac{\log \left ({\left (e^{\left (f x + e\right )} - i\right )} e^{\left (-e\right )}\right )}{a f^{2}}\right )} - \frac{2 \, c}{{\left (i \, a e^{\left (-f x - e\right )} - a\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.1936, size = 151, normalized size = 2.4 \begin{align*} \frac{2 \, d f x e^{\left (f x + e\right )} + 2 i \, c f -{\left (2 \, d e^{\left (f x + e\right )} - 2 i \, d\right )} \log \left (e^{\left (f x + e\right )} - i\right )}{a f^{2} e^{\left (f x + e\right )} - i \, a f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.689824, size = 56, normalized size = 0.89 \begin{align*} \frac{2 d x}{a f} - \frac{2 d \log{\left (e^{f x} - i e^{- e} \right )}}{a f^{2}} + \frac{\left (2 i c + 2 i d x\right ) e^{- e}}{a f \left (e^{f x} - i e^{- e}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*x/(a*f) - 2*d*log(exp(f*x) - I*exp(-e))/(a*f**2) + (2*I*c + 2*I*d*x)*exp(-e)/(a*f*(exp(f*x) - I*exp(-e)))

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Giac [A]  time = 1.2167, size = 97, normalized size = 1.54 \begin{align*} \frac{2 \, d f x e^{\left (f x + e\right )} - 2 \, d e^{\left (f x + e\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + 2 i \, c f + 2 i \, d \log \left (e^{\left (f x + e\right )} - i\right )}{a f^{2} e^{\left (f x + e\right )} - i \, a f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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