\(\int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx\) [163]

Optimal result

Integrand size = 13, antiderivative size = 15 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=-i \sinh (x)+\frac {\sinh ^2(x)}{2} \]

[Out]

-I*sinh(x)+1/2*sinh(x)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2746} \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=\frac {\sinh ^2(x)}{2}-i \sinh (x) \]

[In]

Int[Cosh[x]^3/(I + Sinh[x]),x]

[Out]

(-I)*Sinh[x] + Sinh[x]^2/2

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}(\int (i-x) \, dx,x,\sinh (x)) \\ & = -i \sinh (x)+\frac {\sinh ^2(x)}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=\frac {1}{2} \sinh (x) (-2 i+\sinh (x)) \]

[In]

Integrate[Cosh[x]^3/(I + Sinh[x]),x]

[Out]

(Sinh[x]*(-2*I + Sinh[x]))/2

Maple [A] (verified)

Time = 16.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(-i \sinh \left (x \right )+\frac {\sinh \left (x \right )^{2}}{2}\)	\(13\)
default	\(-i \sinh \left (x \right )+\frac {\sinh \left (x \right )^{2}}{2}\)	\(13\)
risch	\(\frac {{\mathrm e}^{2 x}}{8}-\frac {i {\mathrm e}^{x}}{2}+\frac {i {\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{-2 x}}{8}\)	\(26\)

[In]

int(cosh(x)^3/(I+sinh(x)),x,method=_RETURNVERBOSE)

[Out]

-I*sinh(x)+1/2*sinh(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=\frac {1}{8} \, {\left (e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} + 4 i \, e^{x} + 1\right )} e^{\left (-2 \, x\right )} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (10) = 20\).

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.80 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=\frac {e^{2 x}}{8} - \frac {i e^{x}}{2} + \frac {i e^{- x}}{2} + \frac {e^{- 2 x}}{8} \]

[In]

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

[Out]

exp(2*x)/8 - I*exp(x)/2 + I*exp(-x)/2 + exp(-2*x)/8

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. 27 vs. \(2 (11) = 22\).

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.80 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=\frac {1}{8} \, {\left (-4 i \, e^{\left (-x\right )} + 1\right )} e^{\left (2 \, x\right )} + \frac {1}{2} i \, e^{\left (-x\right )} + \frac {1}{8} \, e^{\left (-2 \, x\right )} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (11) = 22\).

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=-\frac {1}{8} \, {\left (-4 i \, e^{x} - 1\right )} e^{\left (-2 \, x\right )} + \frac {1}{8} \, e^{\left (2 \, x\right )} - \frac {1}{2} i \, e^{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07 \[ \int \frac {\cosh ^3(x)}{i+\sinh (x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,\left ({\mathrm {e}}^{4\,x}+1\right )}{8}-\frac {{\mathrm {e}}^{-2\,x}\,\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )\,1{}\mathrm {i}}{8} \]

[In]

int(cosh(x)^3/(sinh(x) + 1i),x)

[Out]

(exp(-2*x)*(exp(4*x) + 1))/8 - (exp(-2*x)*(4*exp(3*x) - 4*exp(x))*1i)/8