3.58 \(\int \frac{1}{(1+i \sinh (c+d x))^3} \, dx\)

Optimal. Leaf size=88 \[ \frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))}+\frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac{i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3} \]

[Out]

((I/5)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x])^3) + (((2*I)/15)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x])^2) + (
((2*I)/15)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x]))

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Rubi [A]  time = 0.0416327, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2650, 2648} \[ \frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))}+\frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac{i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(1 + I*Sinh[c + d*x])^(-3),x]

[Out]

((I/5)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x])^3) + (((2*I)/15)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x])^2) + (
((2*I)/15)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x]))

Rule 2650

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(1+i \sinh (c+d x))^3} \, dx &=\frac{i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3}+\frac{2}{5} \int \frac{1}{(1+i \sinh (c+d x))^2} \, dx\\ &=\frac{i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3}+\frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac{2}{15} \int \frac{1}{1+i \sinh (c+d x)} \, dx\\ &=\frac{i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3}+\frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac{2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.119885, size = 81, normalized size = 0.92 \[ \frac{15 i \sinh (c+d x)-6 i \sinh (2 (c+d x))-i \sinh (3 (c+d x))-15 \cosh (c+d x)-6 \cosh (2 (c+d x))+\cosh (3 (c+d x))+10}{30 d (\sinh (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + I*Sinh[c + d*x])^(-3),x]

[Out]

(10 - 15*Cosh[c + d*x] - 6*Cosh[2*(c + d*x)] + Cosh[3*(c + d*x)] + (15*I)*Sinh[c + d*x] - (6*I)*Sinh[2*(c + d*
x)] - I*Sinh[3*(c + d*x)])/(30*d*(-I + Sinh[c + d*x])^3)

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Maple [A]  time = 0.046, size = 88, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( 2\, \left ( -i+\tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{-1}+{4\,i \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{8}{5} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}-{\frac{16}{3} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{4\,i \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*sinh(d*x+c))^3,x)

[Out]

1/d*(2/(-I+tanh(1/2*d*x+1/2*c))+4*I/(-I+tanh(1/2*d*x+1/2*c))^2+8/5/(-I+tanh(1/2*d*x+1/2*c))^5-16/3/(-I+tanh(1/
2*d*x+1/2*c))^3-4*I/(-I+tanh(1/2*d*x+1/2*c))^4)

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Maxima [B]  time = 1.06219, size = 285, normalized size = 3.24 \begin{align*} \frac{20 i \, e^{\left (-d x - c\right )}}{d{\left (75 i \, e^{\left (-d x - c\right )} + 150 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 75 \, e^{\left (-4 \, d x - 4 \, c\right )} + 15 i \, e^{\left (-5 \, d x - 5 \, c\right )} - 15\right )}} + \frac{40 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (75 i \, e^{\left (-d x - c\right )} + 150 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 75 \, e^{\left (-4 \, d x - 4 \, c\right )} + 15 i \, e^{\left (-5 \, d x - 5 \, c\right )} - 15\right )}} - \frac{4}{d{\left (75 i \, e^{\left (-d x - c\right )} + 150 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 75 \, e^{\left (-4 \, d x - 4 \, c\right )} + 15 i \, e^{\left (-5 \, d x - 5 \, c\right )} - 15\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*sinh(d*x+c))^3,x, algorithm="maxima")

[Out]

20*I*e^(-d*x - c)/(d*(75*I*e^(-d*x - c) + 150*e^(-2*d*x - 2*c) - 150*I*e^(-3*d*x - 3*c) - 75*e^(-4*d*x - 4*c)
+ 15*I*e^(-5*d*x - 5*c) - 15)) + 40*e^(-2*d*x - 2*c)/(d*(75*I*e^(-d*x - c) + 150*e^(-2*d*x - 2*c) - 150*I*e^(-
3*d*x - 3*c) - 75*e^(-4*d*x - 4*c) + 15*I*e^(-5*d*x - 5*c) - 15)) - 4/(d*(75*I*e^(-d*x - c) + 150*e^(-2*d*x -
2*c) - 150*I*e^(-3*d*x - 3*c) - 75*e^(-4*d*x - 4*c) + 15*I*e^(-5*d*x - 5*c) - 15))

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Fricas [A]  time = 1.98127, size = 235, normalized size = 2.67 \begin{align*} \frac{-40 i \, e^{\left (2 \, d x + 2 \, c\right )} - 20 \, e^{\left (d x + c\right )} + 4 i}{15 \, d e^{\left (5 \, d x + 5 \, c\right )} - 75 i \, d e^{\left (4 \, d x + 4 \, c\right )} - 150 \, d e^{\left (3 \, d x + 3 \, c\right )} + 150 i \, d e^{\left (2 \, d x + 2 \, c\right )} + 75 \, d e^{\left (d x + c\right )} - 15 i \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*sinh(d*x+c))^3,x, algorithm="fricas")

[Out]

(-40*I*e^(2*d*x + 2*c) - 20*e^(d*x + c) + 4*I)/(15*d*e^(5*d*x + 5*c) - 75*I*d*e^(4*d*x + 4*c) - 150*d*e^(3*d*x
+ 3*c) + 150*I*d*e^(2*d*x + 2*c) + 75*d*e^(d*x + c) - 15*I*d)

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Sympy [A]  time = 2.8668, size = 114, normalized size = 1.3 \begin{align*} \frac{\frac{4 i e^{5 c}}{15 d} + \frac{4 e^{4 c} e^{- d x}}{3 d} - \frac{8 i e^{3 c} e^{- 2 d x}}{3 d}}{i e^{5 c} + 5 e^{4 c} e^{- d x} - 10 i e^{3 c} e^{- 2 d x} - 10 e^{2 c} e^{- 3 d x} + 5 i e^{c} e^{- 4 d x} + e^{- 5 d x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*sinh(d*x+c))**3,x)

[Out]

(4*I*exp(5*c)/(15*d) + 4*exp(4*c)*exp(-d*x)/(3*d) - 8*I*exp(3*c)*exp(-2*d*x)/(3*d))/(I*exp(5*c) + 5*exp(4*c)*e
xp(-d*x) - 10*I*exp(3*c)*exp(-2*d*x) - 10*exp(2*c)*exp(-3*d*x) + 5*I*exp(c)*exp(-4*d*x) + exp(-5*d*x))

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Giac [A]  time = 1.37125, size = 49, normalized size = 0.56 \begin{align*} -\frac{40 i \, e^{\left (2 \, d x + 2 \, c\right )} + 20 \, e^{\left (d x + c\right )} - 4 i}{15 \, d{\left (e^{\left (d x + c\right )} - i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*sinh(d*x+c))^3,x, algorithm="giac")

[Out]

-1/15*(40*I*e^(2*d*x + 2*c) + 20*e^(d*x + c) - 4*I)/(d*(e^(d*x + c) - I)^5)