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]

1/5*I*cosh(d*x+c)/d/(1+I*sinh(d*x+c))^3+2/15*I*cosh(d*x+c)/d/(1+I*sinh(d*x+c))^2+2/15*I*cosh(d*x+c)/d/(1+I*sin
h(d*x+c))

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [A]  time = 0.54, size = 85, normalized size = 0.97 \[ \frac {-40 i \, e^{\left (2 \, d x + 2 \, c\right )} - 20 \, e^{\left (d x + c\right )} + 4 i}{15 \, d e^{\left (5 \, d x + 5 \, c\right )} - 75 i \, d e^{\left (4 \, d x + 4 \, c\right )} - 150 \, d e^{\left (3 \, d x + 3 \, c\right )} + 150 i \, d e^{\left (2 \, d x + 2 \, c\right )} + 75 \, d e^{\left (d x + c\right )} - 15 i \, d} \]

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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giac [A]  time = 0.23, size = 36, normalized size = 0.41 \[ -\frac {i \, {\left (40 \, e^{\left (2 \, d x + 2 \, c\right )} - 20 i \, e^{\left (d x + c\right )} - 4\right )}}{15 \, d {\left (e^{\left (d x + c\right )} - i\right )}^{5}} \]

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

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maple [A]  time = 0.09, size = 88, normalized size = 1.00 \[ \frac {\frac {4 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16}{3 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {8}{5 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {2}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 211, normalized size = 2.40 \[ \frac {20 i \, e^{\left (-d x - c\right )}}{d {\left (75 i \, e^{\left (-d x - c\right )} + 150 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 75 \, e^{\left (-4 \, d x - 4 \, c\right )} + 15 i \, e^{\left (-5 \, d x - 5 \, c\right )} - 15\right )}} + \frac {40 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (75 i \, e^{\left (-d x - c\right )} + 150 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 75 \, e^{\left (-4 \, d x - 4 \, c\right )} + 15 i \, e^{\left (-5 \, d x - 5 \, c\right )} - 15\right )}} - \frac {4}{d {\left (75 i \, e^{\left (-d x - c\right )} + 150 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 75 \, e^{\left (-4 \, d x - 4 \, c\right )} + 15 i \, e^{\left (-5 \, d x - 5 \, c\right )} - 15\right )}} \]

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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mupad [B]  time = 0.69, size = 40, normalized size = 0.45 \[ -\frac {\frac {4}{15}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3}+\frac {{\mathrm {e}}^{c+d\,x}\,4{}\mathrm {i}}{3}}{d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*1i + 1)^3,x)

[Out]

-((exp(c + d*x)*4i)/3 - (8*exp(2*c + 2*d*x))/3 + 4/15)/(d*(exp(c + d*x)*1i + 1)^5)

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sympy [A]  time = 0.31, size = 114, normalized size = 1.30 \[ \frac {- 4 e^{5 c} + 20 i e^{4 c} e^{- d x} + 40 e^{3 c} e^{- 2 d x}}{- 15 d e^{5 c} + 75 i d e^{4 c} e^{- d x} + 150 d e^{3 c} e^{- 2 d x} - 150 i d e^{2 c} e^{- 3 d x} - 75 d e^{c} e^{- 4 d x} + 15 i d e^{- 5 d x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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