3.202 \(\int \frac{\sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=83 \[ \frac{2 \cosh (c+d x)}{a d}-\frac{\sinh ^2(c+d x) \cosh (c+d x)}{d (a+i a \sinh (c+d x))}-\frac{3 i \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{3 i x}{2 a} \]

[Out]

(((3*I)/2)*x)/a + (2*Cosh[c + d*x])/(a*d) - (((3*I)/2)*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) - (Cosh[c + d*x]*Sin
h[c + d*x]^2)/(d*(a + I*a*Sinh[c + d*x]))

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Rubi [A]  time = 0.0795302, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2767, 2734} \[ \frac{2 \cosh (c+d x)}{a d}-\frac{\sinh ^2(c+d x) \cosh (c+d x)}{d (a+i a \sinh (c+d x))}-\frac{3 i \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{3 i x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]

[Out]

(((3*I)/2)*x)/a + (2*Cosh[c + d*x])/(a*d) - (((3*I)/2)*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) - (Cosh[c + d*x]*Sin
h[c + d*x]^2)/(d*(a + I*a*Sinh[c + d*x]))

Rule 2767

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int \frac{\sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac{\cosh (c+d x) \sinh ^2(c+d x)}{d (a+i a \sinh (c+d x))}+\frac{\int \sinh (c+d x) (2 a-3 i a \sinh (c+d x)) \, dx}{a^2}\\ &=\frac{3 i x}{2 a}+\frac{2 \cosh (c+d x)}{a d}-\frac{3 i \cosh (c+d x) \sinh (c+d x)}{2 a d}-\frac{\cosh (c+d x) \sinh ^2(c+d x)}{d (a+i a \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.163607, size = 109, normalized size = 1.31 \[ \frac{\left (3 \sqrt{1+i \sinh (c+d x)} \sinh ^{-1}(\sinh (c+d x))+\sqrt{1-i \sinh (c+d x)} \left (-i \sinh ^2(c+d x)+\sinh (c+d x)-4 i\right )\right ) \cosh (c+d x)}{2 a d \sqrt{1-i \sinh (c+d x)} (\sinh (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]

[Out]

(Cosh[c + d*x]*(3*ArcSinh[Sinh[c + d*x]]*Sqrt[1 + I*Sinh[c + d*x]] + Sqrt[1 - I*Sinh[c + d*x]]*(-4*I + Sinh[c
+ d*x] - I*Sinh[c + d*x]^2)))/(2*a*d*Sqrt[1 - I*Sinh[c + d*x]]*(-I + Sinh[c + d*x]))

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Maple [B]  time = 0.041, size = 196, normalized size = 2.4 \begin{align*}{\frac{-2\,i}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{{\frac{i}{2}}}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{{\frac{3\,i}{2}}}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{\frac{i}{2}}}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{\frac{3\,i}{2}}}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{\frac{i}{2}}}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{\frac{i}{2}}}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x)

[Out]

-2*I/d/a/(-I+tanh(1/2*d*x+1/2*c))+1/2*I/d/a/(tanh(1/2*d*x+1/2*c)+1)^2+3/2*I/d/a*ln(tanh(1/2*d*x+1/2*c)+1)+1/d/
a/(tanh(1/2*d*x+1/2*c)+1)-1/2*I/d/a/(tanh(1/2*d*x+1/2*c)+1)-3/2*I/d/a*ln(tanh(1/2*d*x+1/2*c)-1)-1/2*I/d/a/(tan
h(1/2*d*x+1/2*c)-1)^2-1/d/a/(tanh(1/2*d*x+1/2*c)-1)-1/2*I/d/a/(tanh(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.02886, size = 132, normalized size = 1.59 \begin{align*} \frac{3 i \,{\left (d x + c\right )}}{2 \, a d} + \frac{3 i \, e^{\left (-d x - c\right )} + 20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{8 \,{\left (i \, a e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-3 \, d x - 3 \, c\right )}\right )} d} + \frac{i \,{\left (-4 i \, e^{\left (-d x - c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}}{8 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

3/2*I*(d*x + c)/(a*d) + 1/8*(3*I*e^(-d*x - c) + 20*e^(-2*d*x - 2*c) + 1)/((I*a*e^(-2*d*x - 2*c) + a*e^(-3*d*x
- 3*c))*d) + 1/8*I*(-4*I*e^(-d*x - c) + e^(-2*d*x - 2*c))/(a*d)

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Fricas [A]  time = 2.59988, size = 243, normalized size = 2.93 \begin{align*} \frac{{\left (12 i \, d x - 4 i\right )} e^{\left (3 \, d x + 3 \, c\right )} + 4 \,{\left (3 \, d x + 5\right )} e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (5 \, d x + 5 \, c\right )} + 3 \, e^{\left (4 \, d x + 4 \, c\right )} - 3 i \, e^{\left (d x + c\right )} + 1}{8 \, a d e^{\left (3 \, d x + 3 \, c\right )} - 8 i \, a d e^{\left (2 \, d x + 2 \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.39073, size = 180, normalized size = 2.17 \begin{align*} \begin{cases} \frac{\left (- 32 i a^{3} d^{3} e^{5 c} e^{2 d x} + 128 a^{3} d^{3} e^{4 c} e^{d x} + 128 a^{3} d^{3} e^{2 c} e^{- d x} + 32 i a^{3} d^{3} e^{c} e^{- 2 d x}\right ) e^{- 3 c}}{256 a^{4} d^{4}} & \text{for}\: 256 a^{4} d^{4} e^{3 c} \neq 0 \\x \left (- \frac{\left (i e^{4 c} - 2 e^{3 c} - 6 i e^{2 c} + 2 e^{c} + i\right ) e^{- 2 c}}{4 a} - \frac{3 i}{2 a}\right ) & \text{otherwise} \end{cases} + \frac{3 i x}{2 a} + \frac{2 e^{c}}{a d \left (i e^{c} + e^{- d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)

[Out]

Piecewise(((-32*I*a**3*d**3*exp(5*c)*exp(2*d*x) + 128*a**3*d**3*exp(4*c)*exp(d*x) + 128*a**3*d**3*exp(2*c)*exp
(-d*x) + 32*I*a**3*d**3*exp(c)*exp(-2*d*x))*exp(-3*c)/(256*a**4*d**4), Ne(256*a**4*d**4*exp(3*c), 0)), (x*(-(I
*exp(4*c) - 2*exp(3*c) - 6*I*exp(2*c) + 2*exp(c) + I)*exp(-2*c)/(4*a) - 3*I/(2*a)), True)) + 3*I*x/(2*a) + 2*e
xp(c)/(a*d*(I*exp(c) + exp(-d*x)))

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Giac [A]  time = 1.47046, size = 127, normalized size = 1.53 \begin{align*} \frac{3 i \,{\left (d x + c\right )}}{2 \, a d} + \frac{{\left (20 \, e^{\left (2 \, d x + 2 \, c\right )} - 3 i \, e^{\left (d x + c\right )} + 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a d{\left (e^{\left (d x + c\right )} - i\right )}} - \frac{i \, a d e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a d e^{\left (d x + c\right )}}{8 \, a^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

3/2*I*(d*x + c)/(a*d) + 1/8*(20*e^(2*d*x + 2*c) - 3*I*e^(d*x + c) + 1)*e^(-2*d*x - 2*c)/(a*d*(e^(d*x + c) - I)
) - 1/8*(I*a*d*e^(2*d*x + 2*c) - 4*a*d*e^(d*x + c))/(a^2*d^2)