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

Optimal. Leaf size=90 \[ -\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}+\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i e x}{a}-\frac{i f x^2}{2 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.112382, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {5557, 3318, 4184, 3475} \[ -\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}+\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i e x}{a}-\frac{i f x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5557

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

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

Mathematica [B]  time = 0.637151, size = 239, normalized size = 2.66 \[ \frac{-i \cosh \left (\frac{d x}{2}\right ) \left (2 f \log (\cosh (c+d x))+4 i f \tan ^{-1}\left (\sinh \left (\frac{d x}{2}\right ) \text{sech}\left (c+\frac{d x}{2}\right )\right )+d^2 x (2 e+f x)\right )+2 d^2 e x \sinh \left (c+\frac{d x}{2}\right )+d^2 f x^2 \sinh \left (c+\frac{d x}{2}\right )-2 d f x \cosh \left (c+\frac{d x}{2}\right )+2 f \sinh \left (c+\frac{d x}{2}\right ) \log (\cosh (c+d x))+4 i f \sinh \left (c+\frac{d x}{2}\right ) \tan ^{-1}\left (\sinh \left (\frac{d x}{2}\right ) \text{sech}\left (c+\frac{d x}{2}\right )\right )+4 i d e \sinh \left (\frac{d x}{2}\right )+2 i d f x \sinh \left (\frac{d x}{2}\right )}{2 a d^2 \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*f*x*Cosh[c + (d*x)/2] - I*Cosh[(d*x)/2]*(d^2*x*(2*e + f*x) + (4*I)*f*ArcTan[Sech[c + (d*x)/2]*Sinh[(d*x)
/2]] + 2*f*Log[Cosh[c + d*x]]) + (4*I)*d*e*Sinh[(d*x)/2] + (2*I)*d*f*x*Sinh[(d*x)/2] + 2*d^2*e*x*Sinh[c + (d*x
)/2] + d^2*f*x^2*Sinh[c + (d*x)/2] + (4*I)*f*ArcTan[Sech[c + (d*x)/2]*Sinh[(d*x)/2]]*Sinh[c + (d*x)/2] + 2*f*L
og[Cosh[c + d*x]]*Sinh[c + (d*x)/2])/(2*a*d^2*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/
2]))

________________________________________________________________________________________

Maple [A]  time = 0.079, size = 86, normalized size = 1. \begin{align*}{\frac{-{\frac{i}{2}}f{x}^{2}}{a}}-{\frac{iex}{a}}+{\frac{2\,ifx}{da}}+{\frac{2\,ifc}{a{d}^{2}}}-2\,{\frac{fx+e}{da \left ({{\rm e}^{dx+c}}-i \right ) }}-{\frac{2\,if\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Maxima [A]  time = 1.26702, size = 146, normalized size = 1.62 \begin{align*} \frac{1}{2} \, f{\left (\frac{-i \, d x^{2} +{\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac{4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} - e{\left (\frac{i \,{\left (d x + c\right )}}{a d} + \frac{2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.53768, size = 235, normalized size = 2.61 \begin{align*} -\frac{d^{2} f x^{2} + 2 \, d^{2} e x + 4 \, d e -{\left (-i \, d^{2} f x^{2} +{\left (-2 i \, d^{2} e + 4 i \, d f\right )} x\right )} e^{\left (d x + c\right )} -{\left (-4 i \, f e^{\left (d x + c\right )} - 4 \, f\right )} \log \left (e^{\left (d x + c\right )} - i\right )}{2 \, a d^{2} e^{\left (d x + c\right )} - 2 i \, a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.21546, size = 180, normalized size = 2. \begin{align*} -\frac{i \, d^{2} f x^{2} e^{\left (d x + 2 \, c\right )} + d^{2} f x^{2} e^{c} + 2 i \, d^{2} x e^{\left (d x + 2 \, c + 1\right )} - 4 i \, d f x e^{\left (d x + 2 \, c\right )} + 2 \, d^{2} x e^{\left (c + 1\right )} + 4 i \, f e^{\left (d x + 2 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 4 \, f e^{c} \log \left (e^{\left (d x + c\right )} - i\right ) + 4 \, d e^{\left (c + 1\right )}}{2 \,{\left (a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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