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

Optimal. Leaf size=10 \[ -\frac{1}{\sinh (x)+i} \]

[Out]

-(I + Sinh[x])^(-1)

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Rubi [A]  time = 0.0198091, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2667, 32} \[ -\frac{1}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(I + Sinh[x])^2,x]

[Out]

-(I + Sinh[x])^(-1)

Rule 2667

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])

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]

Rubi steps

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

Mathematica [A]  time = 0.0099376, size = 10, normalized size = 1. \[ -\frac{1}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(I + Sinh[x])^2,x]

[Out]

-(I + Sinh[x])^(-1)

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Maple [A]  time = 0.013, size = 10, normalized size = 1. \begin{align*} - \left ( i+\sinh \left ( x \right ) \right ) ^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(I+sinh(x))

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Maxima [A]  time = 1.22747, size = 11, normalized size = 1.1 \begin{align*} -\frac{1}{\sinh \left (x\right ) + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(sinh(x) + I)

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Fricas [A]  time = 1.73557, size = 43, normalized size = 4.3 \begin{align*} -\frac{2 \, e^{x}}{e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^x/(e^(2*x) + 2*I*e^x - 1)

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Sympy [B]  time = 0.211546, size = 19, normalized size = 1.9 \begin{align*} - \frac{2 e^{x}}{e^{2 x} + 2 i e^{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*exp(x)/(exp(2*x) + 2*I*exp(x) - 1)

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Giac [A]  time = 1.29195, size = 14, normalized size = 1.4 \begin{align*} -\frac{2 \, e^{x}}{{\left (e^{x} + i\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^x/(e^x + I)^2