3.195 \(\int \frac{(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=119 \[ \frac{i f \sinh (c+d x)}{a d^2}+\frac{2 f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}-\frac{i (e+f x) \cosh (c+d x)}{a d}-\frac{(e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}+\frac{e x}{a}+\frac{f x^2}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.177117, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {5557, 3296, 2637, 3318, 4184, 3475} \[ \frac{i f \sinh (c+d x)}{a d^2}+\frac{2 f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}-\frac{i (e+f x) \cosh (c+d x)}{a d}-\frac{(e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}+\frac{e x}{a}+\frac{f x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x)/a + (f*x^2)/(2*a) - (I*(e + f*x)*Cosh[c + d*x])/(a*d) + (2*f*Log[Cosh[c/2 + (I/4)*Pi + (d*x)/2]])/(a*d^2
) + (I*f*Sinh[c + d*x])/(a*d^2) - ((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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A]  time = 1.08925, size = 238, normalized size = 2. \[ \frac{\left (\sinh \left (\frac{1}{2} (c+d x)\right )-i \cosh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right ) \left (c^2 (-f)-2 i d (e+f x) \cosh (c+d x)+2 c d e+2 i f \sinh (c+d x)+4 i f \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+2 f \log (\cosh (c+d x))-2 i c f+2 d^2 e x+d^2 f x^2-2 i d f x\right )+\sinh \left (\frac{1}{2} (c+d x)\right ) \left (i (c+d x+2 i) (-c f+2 d e+d f x)+2 d (e+f x) \cosh (c+d x)-2 f \sinh (c+d x)-4 f \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+2 i f \log (\cosh (c+d x))\right )\right )}{2 a d^2 (\sinh (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.106, size = 134, normalized size = 1.1 \begin{align*}{\frac{f{x}^{2}}{2\,a}}+{\frac{ex}{a}}-{\frac{{\frac{i}{2}} \left ( dfx+de-f \right ){{\rm e}^{dx+c}}}{a{d}^{2}}}-{\frac{{\frac{i}{2}} \left ( dfx+de+f \right ){{\rm e}^{-dx-c}}}{a{d}^{2}}}-2\,{\frac{fx}{da}}-2\,{\frac{cf}{a{d}^{2}}}-{\frac{2\,i \left ( fx+e \right ) }{da \left ({{\rm e}^{dx+c}}-i \right ) }}+2\,{\frac{f\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)^2/(a+I*a*sinh(d*x+c)),x)

[Out]

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

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Maxima [B]  time = 1.3415, size = 325, normalized size = 2.73 \begin{align*} -\frac{1}{4} \, f{\left (\frac{4 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac{-2 i \, d^{2} x^{2} e^{c} - 2 i \, d x e^{c} -{\left (2 i \, d x e^{\left (3 \, c\right )} - 2 i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \,{\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - 2 \,{\left (d x + 1\right )} e^{\left (-d x\right )} - 2 i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac{8 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac{1}{4} \, e{\left (\frac{4 \,{\left (d x + c\right )}}{a d} + \frac{2 \,{\left (-5 i \, e^{\left (-d x - c\right )} + 1\right )}}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac{2 i \, e^{\left (-d x - c\right )}}{a d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*f*(4*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) - (-2*I*d^2*x^2*e^c - 2*I*d*x*e^c - (2*I*d*x*e^(3*c) - 2*I*e
^(3*c))*e^(2*d*x) + 2*(d^2*x^2*e^(2*c) - 3*d*x*e^(2*c) + e^(2*c))*e^(d*x) - 2*(d*x + 1)*e^(-d*x) - 2*I*e^c)/(a
*d^2*e^(d*x + 2*c) - I*a*d^2*e^c) - 8*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) + 1/4*e*(4*(d*x + c)/(a*d) + 2*(-
5*I*e^(-d*x - c) + 1)/((I*a*e^(-d*x - c) + a*e^(-2*d*x - 2*c))*d) - 2*I*e^(-d*x - c)/(a*d))

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Fricas [A]  time = 2.66552, size = 416, normalized size = 3.5 \begin{align*} -\frac{d f x + d e -{\left (-i \, d f x - i \, d e + i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} -{\left (d^{2} f x^{2} - d e +{\left (2 \, d^{2} e - 5 \, d f\right )} x + f\right )} e^{\left (2 \, d x + 2 \, c\right )} -{\left (-i \, d^{2} f x^{2} - 5 i \, d e +{\left (-2 i \, d^{2} e - i \, d f\right )} x - i \, f\right )} e^{\left (d x + c\right )} - 4 \,{\left (f e^{\left (2 \, d x + 2 \, c\right )} - i \, f e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + f}{2 \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*f*x + d*e - (-I*d*f*x - I*d*e + I*f)*e^(3*d*x + 3*c) - (d^2*f*x^2 - d*e + (2*d^2*e - 5*d*f)*x + f)*e^(2*d*
x + 2*c) - (-I*d^2*f*x^2 - 5*I*d*e + (-2*I*d^2*e - I*d*f)*x - I*f)*e^(d*x + c) - 4*(f*e^(2*d*x + 2*c) - I*f*e^
(d*x + c))*log(e^(d*x + c) - I) + f)/(2*a*d^2*e^(2*d*x + 2*c) - 2*I*a*d^2*e^(d*x + c))

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Sympy [A]  time = 2.36023, size = 449, normalized size = 3.77 \begin{align*} \begin{cases} \frac{\left (\left (- 2 i a^{3} d^{5} e e^{c} - 2 i a^{3} d^{5} f x e^{c} - 2 i a^{3} d^{4} f e^{c}\right ) e^{- d x} + \left (- 2 i a^{3} d^{5} e e^{3 c} - 2 i a^{3} d^{5} f x e^{3 c} + 2 i a^{3} d^{4} f e^{3 c}\right ) e^{d x}\right ) e^{- 2 c}}{4 a^{4} d^{6}} & \text{for}\: 4 a^{4} d^{6} e^{2 c} \neq 0 \\\frac{x^{2} \left (i f e^{6 c} + 4 f e^{5 c} - 7 i f e^{4 c} - 8 f e^{3 c} + 7 i f e^{2 c} + 4 f e^{c} - i f\right )}{- 4 a e^{5 c} + 16 i a e^{4 c} + 24 a e^{3 c} - 16 i a e^{2 c} - 4 a e^{c}} + \frac{x \left (i e e^{8 c} + 6 e e^{7 c} - 16 i e e^{6 c} - 26 e e^{5 c} + 30 i e e^{4 c} + 26 e e^{3 c} - 16 i e e^{2 c} - 6 e e^{c} + i e\right )}{- 2 a e^{7 c} + 12 i a e^{6 c} + 30 a e^{5 c} - 40 i a e^{4 c} - 30 a e^{3 c} + 12 i a e^{2 c} + 2 a e^{c}} & \text{otherwise} \end{cases} + \frac{f x^{2}}{2 a} + \frac{x \left (d e - 2 f\right )}{a d} - \frac{\left (2 i e + 2 i f x\right ) e^{- c}}{a d \left (e^{d x} - i e^{- c}\right )} + \frac{2 f \log{\left (e^{d x} - i e^{- c} \right )}}{a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((-2*I*a**3*d**5*e*exp(c) - 2*I*a**3*d**5*f*x*exp(c) - 2*I*a**3*d**4*f*exp(c))*exp(-d*x) + (-2*I*a*
*3*d**5*e*exp(3*c) - 2*I*a**3*d**5*f*x*exp(3*c) + 2*I*a**3*d**4*f*exp(3*c))*exp(d*x))*exp(-2*c)/(4*a**4*d**6),
Ne(4*a**4*d**6*exp(2*c), 0)), (x**2*(I*f*exp(6*c) + 4*f*exp(5*c) - 7*I*f*exp(4*c) - 8*f*exp(3*c) + 7*I*f*exp(
2*c) + 4*f*exp(c) - I*f)/(-4*a*exp(5*c) + 16*I*a*exp(4*c) + 24*a*exp(3*c) - 16*I*a*exp(2*c) - 4*a*exp(c)) + x*
(I*e*exp(8*c) + 6*e*exp(7*c) - 16*I*e*exp(6*c) - 26*e*exp(5*c) + 30*I*e*exp(4*c) + 26*e*exp(3*c) - 16*I*e*exp(
2*c) - 6*e*exp(c) + I*e)/(-2*a*exp(7*c) + 12*I*a*exp(6*c) + 30*a*exp(5*c) - 40*I*a*exp(4*c) - 30*a*exp(3*c) +
12*I*a*exp(2*c) + 2*a*exp(c)), True)) + f*x**2/(2*a) + x*(d*e - 2*f)/(a*d) - (2*I*e + 2*I*f*x)*exp(-c)/(a*d*(e
xp(d*x) - I*exp(-c))) + 2*f*log(exp(d*x) - I*exp(-c))/(a*d**2)

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Giac [B]  time = 1.52809, size = 367, normalized size = 3.08 \begin{align*} \frac{d^{2} f x^{2} e^{\left (2 \, d x + 3 \, c\right )} - i \, d^{2} f x^{2} e^{\left (d x + 2 \, c\right )} - i \, d f x e^{\left (3 \, d x + 4 \, c\right )} + 2 \, d^{2} x e^{\left (2 \, d x + 3 \, c + 1\right )} - 5 \, d f x e^{\left (2 \, d x + 3 \, c\right )} - 2 i \, d^{2} x e^{\left (d x + 2 \, c + 1\right )} - i \, d f x e^{\left (d x + 2 \, c\right )} - d f x e^{c} + 4 \, f e^{\left (2 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 4 i \, f e^{\left (d x + 2 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - i \, d e^{\left (3 \, d x + 4 \, c + 1\right )} + i \, f e^{\left (3 \, d x + 4 \, c\right )} - d e^{\left (2 \, d x + 3 \, c + 1\right )} + f e^{\left (2 \, d x + 3 \, c\right )} - 5 i \, d e^{\left (d x + 2 \, c + 1\right )} - i \, f e^{\left (d x + 2 \, c\right )} - d e^{\left (c + 1\right )} - f e^{c}}{2 \,{\left (a d^{2} e^{\left (2 \, d x + 3 \, c\right )} - i \, a d^{2} e^{\left (d x + 2 \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d^2*f*x^2*e^(2*d*x + 3*c) - I*d^2*f*x^2*e^(d*x + 2*c) - I*d*f*x*e^(3*d*x + 4*c) + 2*d^2*x*e^(2*d*x + 3*c
+ 1) - 5*d*f*x*e^(2*d*x + 3*c) - 2*I*d^2*x*e^(d*x + 2*c + 1) - I*d*f*x*e^(d*x + 2*c) - d*f*x*e^c + 4*f*e^(2*d*
x + 3*c)*log(e^(d*x + c) - I) - 4*I*f*e^(d*x + 2*c)*log(e^(d*x + c) - I) - I*d*e^(3*d*x + 4*c + 1) + I*f*e^(3*
d*x + 4*c) - d*e^(2*d*x + 3*c + 1) + f*e^(2*d*x + 3*c) - 5*I*d*e^(d*x + 2*c + 1) - I*f*e^(d*x + 2*c) - d*e^(c
+ 1) - f*e^c)/(a*d^2*e^(2*d*x + 3*c) - I*a*d^2*e^(d*x + 2*c))