3.178 \(\int \frac{\text{sech}^2(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=37 \[ -\frac{2 \tanh (x)}{5}-\frac{\text{sech}(x)}{5 (\sinh (x)+i)}-\frac{i \text{sech}(x)}{5 (\sinh (x)+i)^2} \]

[Out]

((-I/5)*Sech[x])/(I + Sinh[x])^2 - Sech[x]/(5*(I + Sinh[x])) - (2*Tanh[x])/5

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Rubi [A]  time = 0.0708574, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2672, 3767, 8} \[ -\frac{2 \tanh (x)}{5}-\frac{\text{sech}(x)}{5 (\sinh (x)+i)}-\frac{i \text{sech}(x)}{5 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(I + Sinh[x])^2,x]

[Out]

((-I/5)*Sech[x])/(I + Sinh[x])^2 - Sech[x]/(5*(I + Sinh[x])) - (2*Tanh[x])/5

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(x)}{(i+\sinh (x))^2} \, dx &=-\frac{i \text{sech}(x)}{5 (i+\sinh (x))^2}-\frac{3}{5} i \int \frac{\text{sech}^2(x)}{i+\sinh (x)} \, dx\\ &=-\frac{i \text{sech}(x)}{5 (i+\sinh (x))^2}-\frac{\text{sech}(x)}{5 (i+\sinh (x))}-\frac{2}{5} \int \text{sech}^2(x) \, dx\\ &=-\frac{i \text{sech}(x)}{5 (i+\sinh (x))^2}-\frac{\text{sech}(x)}{5 (i+\sinh (x))}-\frac{2}{5} i \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (x))\\ &=-\frac{i \text{sech}(x)}{5 (i+\sinh (x))^2}-\frac{\text{sech}(x)}{5 (i+\sinh (x))}-\frac{2 \tanh (x)}{5}\\ \end{align*}

Mathematica [A]  time = 0.0218572, size = 31, normalized size = 0.84 \[ -\frac{\text{sech}(x) (-5 \sinh (x)+\sinh (3 x)+4 i \cosh (2 x))}{10 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(I + Sinh[x])^2,x]

[Out]

-(Sech[x]*((4*I)*Cosh[2*x] - 5*Sinh[x] + Sinh[3*x]))/(10*(I + Sinh[x])^2)

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Maple [B]  time = 0.039, size = 70, normalized size = 1.9 \begin{align*} -{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+{{\frac{5\,i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}-{\frac{4}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}+3\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-3}-{\frac{7}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(tanh(1/2*x)-I)-2*I/(tanh(1/2*x)+I)^4+5/2*I/(tanh(1/2*x)+I)^2-4/5/(tanh(1/2*x)+I)^5+3/(tanh(1/2*x)+I)^3-7
/4/(tanh(1/2*x)+I)

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Maxima [B]  time = 1.31682, size = 158, normalized size = 4.27 \begin{align*} -\frac{16 i \, e^{\left (-x\right )}}{20 i \, e^{\left (-x\right )} - 25 \, e^{\left (-2 \, x\right )} - 25 \, e^{\left (-4 \, x\right )} - 20 i \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-6 \, x\right )} + 5} + \frac{20 \, e^{\left (-2 \, x\right )}}{20 i \, e^{\left (-x\right )} - 25 \, e^{\left (-2 \, x\right )} - 25 \, e^{\left (-4 \, x\right )} - 20 i \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-6 \, x\right )} + 5} - \frac{4}{20 i \, e^{\left (-x\right )} - 25 \, e^{\left (-2 \, x\right )} - 25 \, e^{\left (-4 \, x\right )} - 20 i \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-6 \, x\right )} + 5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*I*e^(-x)/(20*I*e^(-x) - 25*e^(-2*x) - 25*e^(-4*x) - 20*I*e^(-5*x) + 5*e^(-6*x) + 5) + 20*e^(-2*x)/(20*I*e^
(-x) - 25*e^(-2*x) - 25*e^(-4*x) - 20*I*e^(-5*x) + 5*e^(-6*x) + 5) - 4/(20*I*e^(-x) - 25*e^(-2*x) - 25*e^(-4*x
) - 20*I*e^(-5*x) + 5*e^(-6*x) + 5)

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Fricas [A]  time = 1.73613, size = 132, normalized size = 3.57 \begin{align*} -\frac{4 \,{\left (5 \, e^{\left (2 \, x\right )} + 4 i \, e^{x} - 1\right )}}{5 \, e^{\left (6 \, x\right )} + 20 i \, e^{\left (5 \, x\right )} - 25 \, e^{\left (4 \, x\right )} - 25 \, e^{\left (2 \, x\right )} - 20 i \, e^{x} + 5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(5*e^(2*x) + 4*I*e^x - 1)/(5*e^(6*x) + 20*I*e^(5*x) - 25*e^(4*x) - 25*e^(2*x) - 20*I*e^x + 5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (x \right )}}{\left (\sinh{\left (x \right )} + i\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**2/(sinh(x) + I)**2, x)

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Giac [A]  time = 1.23532, size = 55, normalized size = 1.49 \begin{align*} -\frac{i}{4 \,{\left (e^{x} - i\right )}} - \frac{-5 i \, e^{\left (4 \, x\right )} + 30 \, e^{\left (3 \, x\right )} + 80 i \, e^{\left (2 \, x\right )} - 50 \, e^{x} - 11 i}{20 \,{\left (e^{x} + i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I/(e^x - I) - 1/20*(-5*I*e^(4*x) + 30*e^(3*x) + 80*I*e^(2*x) - 50*e^x - 11*I)/(e^x + I)^5