\(\int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx\) [185]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=-\frac {i \cosh ^3(x)}{3 (1-i \sinh (x))^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2750} \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=-\frac {i \cosh ^3(x)}{3 (1-i \sinh (x))^3} \]

[In]

Int[Cosh[x]^2/(1 - I*Sinh[x])^3,x]

[Out]

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

Rule 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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*} \text {integral}& = -\frac {i \cosh ^3(x)}{3 (1-i \sinh (x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=-\frac {\cosh ^3(x)}{3 (i+\sinh (x))^3} \]

[In]

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

[Out]

-1/3*Cosh[x]^3/(I + Sinh[x])^3

Maple [A] (verified)

Time = 57.56 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {2 i \left (3 \,{\mathrm e}^{2 x}-1\right )}{3 \left ({\mathrm e}^{x}+i\right )^{3}}\)	\(19\)
default	\(\frac {2}{\tanh \left (\frac {x}{2}\right )+i}-\frac {4 i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {8}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}\)	\(36\)

[In]

int(cosh(x)^2/(1-I*sinh(x))^3,x,method=_RETURNVERBOSE)

[Out]

2/3*I*(3*exp(2*x)-1)/(exp(x)+I)^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=-\frac {2 \, {\left (-3 i \, e^{\left (2 \, x\right )} + i\right )}}{3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=\frac {6 i e^{2 x} - 2 i}{3 e^{3 x} + 9 i e^{2 x} - 9 e^{x} - 3 i} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (14) = 28\).

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.65 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=-\frac {6 \, e^{\left (-2 \, x\right )}}{-9 i \, e^{\left (-x\right )} + 9 \, e^{\left (-2 \, x\right )} + 3 i \, e^{\left (-3 \, x\right )} - 3} + \frac {2}{-9 i \, e^{\left (-x\right )} + 9 \, e^{\left (-2 \, x\right )} + 3 i \, e^{\left (-3 \, x\right )} - 3} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=-\frac {2 \, {\left (-3 i \, e^{\left (2 \, x\right )} + i\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} \]

[In]

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

[Out]

-2/3*(-3*I*e^(2*x) + I)/(e^x + I)^3

Mupad [B] (verification not implemented)

Time = 1.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\cosh ^2(x)}{(1-i \sinh (x))^3} \, dx=\frac {2\,\left (3\,{\mathrm {e}}^{2\,x}-1\right )}{3\,{\left (-1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^3} \]

[In]

int(-cosh(x)^2/(sinh(x)*1i - 1)^3,x)

[Out]

(2*(3*exp(2*x) - 1))/(3*(exp(x)*1i - 1)^3)