3.183 \(\int \frac {\cosh (x)}{(1+i \sinh (x))^3} \, dx\)

Optimal. Leaf size=16 \[ \frac {i}{2 (1+i \sinh (x))^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2667, 32} \[ \frac {i}{2 (1+i \sinh (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(1 + I*Sinh[x])^3,x]

[Out]

(I/2)/(1 + I*Sinh[x])^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]

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)}{(1+i \sinh (x))^3} \, dx &=-\left (i \operatorname {Subst}\left (\int \frac {1}{(1+x)^3} \, dx,x,i \sinh (x)\right )\right )\\ &=\frac {i}{2 (1+i \sinh (x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 14, normalized size = 0.88 \[ -\frac {i}{2 (\sinh (x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/2*I)/(-I + Sinh[x])^2

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fricas [B]  time = 0.60, size = 30, normalized size = 1.88 \[ -\frac {2 i \, e^{\left (2 \, x\right )}}{e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - 6 \, e^{\left (2 \, x\right )} + 4 i \, e^{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*e^(2*x)/(e^(4*x) - 4*I*e^(3*x) - 6*e^(2*x) + 4*I*e^x + 1)

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giac [A]  time = 0.37, size = 12, normalized size = 0.75 \[ -\frac {2 i \, e^{\left (2 \, x\right )}}{{\left (e^{x} - i\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*e^(2*x)/(e^x - I)^4

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maple [A]  time = 0.02, size = 13, normalized size = 0.81 \[ \frac {i}{2 \left (1+i \sinh \relax (x )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(1+I*sinh(x))^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 10, normalized size = 0.62 \[ \frac {i}{2 \, {\left (i \, \sinh \relax (x) + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(1+I*sinh(x))^3,x, algorithm="maxima")

[Out]

1/2*I/(I*sinh(x) + 1)^2

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mupad [B]  time = 0.65, size = 16, normalized size = 1.00 \[ -\frac {{\mathrm {e}}^{2\,x}\,2{}\mathrm {i}}{{\left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(sinh(x)*1i + 1)^3,x)

[Out]

-(exp(2*x)*2i)/(exp(x)*1i + 1)^4

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sympy [B]  time = 0.15, size = 37, normalized size = 2.31 \[ - \frac {2 i e^{2 x}}{e^{4 x} - 4 i e^{3 x} - 6 e^{2 x} + 4 i e^{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*exp(2*x)/(exp(4*x) - 4*I*exp(3*x) - 6*exp(2*x) + 4*I*exp(x) + 1)

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