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

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

[Out]

-1/(I+sinh(x))

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ -\frac {1}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(I + Sinh[x])^2,x]

[Out]

-(I + Sinh[x])^(-1)

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fricas [A]  time = 0.56, size = 16, normalized size = 1.60 \[ -\frac {2 \, e^{x}}{e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1} \]

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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giac [A]  time = 0.14, size = 10, normalized size = 1.00 \[ -\frac {2 \, e^{x}}{{\left (e^{x} + i\right )}^{2}} \]

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

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maple [A]  time = 0.03, size = 10, normalized size = 1.00 \[ -\frac {1}{i+\sinh \relax (x )} \]

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 = 0.31, size = 8, normalized size = 0.80 \[ -\frac {1}{\sinh \relax (x) + i} \]

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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mupad [B]  time = 0.58, size = 12, normalized size = 1.20 \[ -\frac {1{}\mathrm {i}}{-1+\mathrm {sinh}\relax (x)\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(sinh(x) + 1i)^2,x)

[Out]

-1i/(sinh(x)*1i - 1)

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sympy [B]  time = 0.12, size = 17, normalized size = 1.70 \[ \frac {2 e^{x}}{- e^{2 x} - 2 i e^{x} + 1} \]

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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