\(\int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx\) [176]

Optimal result

Integrand size = 11, antiderivative size = 10 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{i+\sinh (x)} \]

[Out]

-1/(I+sinh(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2746, 32} \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{\sinh (x)+i} \]

[In]

Int[Cosh[x]/(I + Sinh[x])^2,x]

[Out]

-(I + Sinh[x])^(-1)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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}\left (\int \frac {1}{(i+x)^2} \, dx,x,\sinh (x)\right ) \\ & = -\frac {1}{i+\sinh (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{i+\sinh (x)} \]

[In]

Integrate[Cosh[x]/(I + Sinh[x])^2,x]

[Out]

-(I + Sinh[x])^(-1)

Maple [A] (verified)

Time = 10.89 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(-\frac {1}{i+\sinh \left (x \right )}\)	\(10\)
default	\(-\frac {1}{i+\sinh \left (x \right )}\)	\(10\)
risch	\(-\frac {2 \,{\mathrm e}^{x}}{\left ({\mathrm e}^{x}+i\right )^{2}}\)	\(12\)

[In]

int(cosh(x)/(I+sinh(x))^2,x,method=_RETURNVERBOSE)

[Out]

-1/(I+sinh(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.60 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, e^{x}}{e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1} \]

[In]

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

[Out]

-2*e^x/(e^(2*x) + 2*I*e^x - 1)

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. 19 vs. \(2 (7) = 14\).

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.90 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=- \frac {2 e^{x}}{e^{2 x} + 2 i e^{x} - 1} \]

[In]

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

[Out]

-2*exp(x)/(exp(2*x) + 2*I*exp(x) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{\sinh \left (x\right ) + i} \]

[In]

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

[Out]

-1/(sinh(x) + I)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, e^{x}}{{\left (e^{x} + i\right )}^{2}} \]

[In]

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

[Out]

-2*e^x/(e^x + I)^2

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.20 \[ \int \frac {\cosh (x)}{(i+\sinh (x))^2} \, dx=-\frac {1{}\mathrm {i}}{-1+\mathrm {sinh}\left (x\right )\,1{}\mathrm {i}} \]

[In]

int(cosh(x)/(sinh(x) + 1i)^2,x)

[Out]

-1i/(sinh(x)*1i - 1)