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

Optimal. Leaf size=31 \[ -\frac{2 i \cosh (x)}{3 (\sinh (x)+i)}-\frac{\cosh (x)}{3 (\sinh (x)+i)^2} \]

[Out]

-Cosh[x]/(3*(I + Sinh[x])^2) - (((2*I)/3)*Cosh[x])/(I + Sinh[x])

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Rubi [A]  time = 0.0295506, antiderivative size = 31, 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 = {2750, 2648} \[ -\frac{2 i \cosh (x)}{3 (\sinh (x)+i)}-\frac{\cosh (x)}{3 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(I + Sinh[x])^2,x]

[Out]

-Cosh[x]/(3*(I + Sinh[x])^2) - (((2*I)/3)*Cosh[x])/(I + Sinh[x])

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{(i+\sinh (x))^2} \, dx &=-\frac{\cosh (x)}{3 (i+\sinh (x))^2}+\frac{2}{3} \int \frac{1}{i+\sinh (x)} \, dx\\ &=-\frac{\cosh (x)}{3 (i+\sinh (x))^2}-\frac{2 i \cosh (x)}{3 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0087043, size = 22, normalized size = 0.71 \[ \frac{(1-2 i \sinh (x)) \cosh (x)}{3 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(I + Sinh[x])^2,x]

[Out]

(Cosh[x]*(1 - (2*I)*Sinh[x]))/(3*(I + Sinh[x])^2)

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Maple [A]  time = 0.033, size = 25, normalized size = 0.8 \begin{align*} 2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-2}-{{\frac{4\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.21631, size = 109, normalized size = 3.52 \begin{align*} -\frac{6 i \, e^{\left (-x\right )}}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} + \frac{6 \, e^{\left (-2 \, x\right )}}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} - \frac{4}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*I*e^(-x)/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I) + 6*e^(-2*x)/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) -
3*I) - 4/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I)

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Fricas [A]  time = 1.98499, size = 95, normalized size = 3.06 \begin{align*} -\frac{2 \,{\left (3 \, e^{\left (2 \, x\right )} + 3 i \, e^{x} - 2\right )}}{3 \, e^{\left (3 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} - 3 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*e^(2*x) + 3*I*e^x - 2)/(3*e^(3*x) + 9*I*e^(2*x) - 9*e^x - 3*I)

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Sympy [A]  time = 0.261966, size = 36, normalized size = 1.16 \begin{align*} \frac{- 2 e^{2 x} - 2 i e^{x} + \frac{4}{3}}{e^{3 x} + 3 i e^{2 x} - 3 e^{x} - i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.30797, size = 27, normalized size = 0.87 \begin{align*} -\frac{2 \,{\left (3 \, e^{\left (2 \, x\right )} + 3 i \, e^{x} - 2\right )}}{3 \,{\left (e^{x} + i\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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