3.182 \(\int \frac {\cosh ^2(x)}{(1+i \sinh (x))^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {i \cosh ^3(x)}{3 (1+i \sinh (x))^3} \]

[Out]

1/3*I*cosh(x)^3/(1+I*sinh(x))^3

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2671} \[ \frac {i \cosh ^3(x)}{3 (1+i \sinh (x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(1 + I*Sinh[x])^3,x]

[Out]

((I/3)*Cosh[x]^3)/(1 + I*Sinh[x])^3

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 40, normalized size = 2.00 \[ -\frac {i \left (\cosh \left (\frac {3 x}{2}\right )-3 \cosh \left (\frac {x}{2}\right )\right )}{3 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^2/(1 + I*Sinh[x])^3,x]

[Out]

((-1/3*I)*(-3*Cosh[x/2] + Cosh[(3*x)/2]))/(Cosh[x/2] + I*Sinh[x/2])^3

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fricas [B]  time = 0.48, size = 29, normalized size = 1.45 \[ \frac {-6 i \, e^{\left (2 \, x\right )} + 2 i}{3 \, e^{\left (3 \, x\right )} - 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} + 3 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(1+I*sinh(x))^3,x, algorithm="fricas")

[Out]

(-6*I*e^(2*x) + 2*I)/(3*e^(3*x) - 9*I*e^(2*x) - 9*e^x + 3*I)

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giac [A]  time = 0.21, size = 16, normalized size = 0.80 \[ -\frac {6 i \, e^{\left (2 \, x\right )} - 2 i}{3 \, {\left (e^{x} - i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(1+I*sinh(x))^3,x, algorithm="giac")

[Out]

-1/3*(6*I*e^(2*x) - 2*I)/(e^x - I)^3

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maple [B]  time = 0.08, size = 36, normalized size = 1.80 \[ \frac {2}{\tanh \left (\frac {x}{2}\right )-i}-\frac {8}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}+\frac {4 i}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(1+I*sinh(x))^3,x)

[Out]

2/(tanh(1/2*x)-I)-8/3/(tanh(1/2*x)-I)^3+4*I/(tanh(1/2*x)-I)^2

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maxima [B]  time = 0.33, size = 53, normalized size = 2.65 \[ \frac {6 \, e^{\left (-2 \, x\right )}}{-9 i \, e^{\left (-x\right )} - 9 \, e^{\left (-2 \, x\right )} + 3 i \, e^{\left (-3 \, x\right )} + 3} - \frac {2}{-9 i \, e^{\left (-x\right )} - 9 \, e^{\left (-2 \, x\right )} + 3 i \, e^{\left (-3 \, x\right )} + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(1+I*sinh(x))^3,x, algorithm="maxima")

[Out]

6*e^(-2*x)/(-9*I*e^(-x) - 9*e^(-2*x) + 3*I*e^(-3*x) + 3) - 2/(-9*I*e^(-x) - 9*e^(-2*x) + 3*I*e^(-3*x) + 3)

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mupad [B]  time = 0.66, size = 19, normalized size = 0.95 \[ -\frac {2\,{\mathrm {e}}^{2\,x}-\frac {2}{3}}{{\left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(sinh(x)*1i + 1)^3,x)

[Out]

-(2*exp(2*x) - 2/3)/(exp(x)*1i + 1)^3

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sympy [B]  time = 0.15, size = 31, normalized size = 1.55 \[ \frac {6 e^{2 x} - 2}{3 i e^{3 x} + 9 e^{2 x} - 9 i e^{x} - 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**2/(1+I*sinh(x))**3,x)

[Out]

(6*exp(2*x) - 2)/(3*I*exp(3*x) + 9*exp(2*x) - 9*I*exp(x) - 3)

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