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

Optimal. Leaf size=34 \[ -\frac{1}{4 (\sinh (x)+i)}-\frac{i}{4 (\sinh (x)+i)^2}-\frac{1}{4} \tan ^{-1}(\sinh (x)) \]

[Out]

-ArcTan[Sinh[x]]/4 - (I/4)/(I + Sinh[x])^2 - 1/(4*(I + Sinh[x]))

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Rubi [A]  time = 0.0391219, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2667, 44, 203} \[ -\frac{1}{4 (\sinh (x)+i)}-\frac{i}{4 (\sinh (x)+i)^2}-\frac{1}{4} \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(I + Sinh[x])^2,x]

[Out]

-ArcTan[Sinh[x]]/4 - (I/4)/(I + Sinh[x])^2 - 1/(4*(I + Sinh[x]))

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{sech}(x)}{(i+\sinh (x))^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{(i-x) (i+x)^3} \, dx,x,\sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{i}{2 (i+x)^3}-\frac{1}{4 (i+x)^2}+\frac{1}{4 \left (1+x^2\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=-\frac{i}{4 (i+\sinh (x))^2}-\frac{1}{4 (i+\sinh (x))}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{4} \tan ^{-1}(\sinh (x))-\frac{i}{4 (i+\sinh (x))^2}-\frac{1}{4 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.042567, size = 26, normalized size = 0.76 \[ \frac{1}{4} \left (-\tan ^{-1}(\sinh (x))-\frac{\sinh (x)+2 i}{(\sinh (x)+i)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(I + Sinh[x])^2,x]

[Out]

(-ArcTan[Sinh[x]] - (2*I + Sinh[x])/(I + Sinh[x])^2)/4

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Maple [B]  time = 0.04, size = 70, normalized size = 2.1 \begin{align*}{\frac{i}{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) +{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}-{\frac{i}{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) -{{\frac{5\,i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}-2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-3}+{\frac{3}{2} \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)/(I+sinh(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.12828, size = 95, normalized size = 2.79 \begin{align*} -\frac{2 \,{\left (e^{\left (-x\right )} + 4 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )}\right )}}{16 i \, e^{\left (-x\right )} - 24 \, e^{\left (-2 \, x\right )} - 16 i \, e^{\left (-3 \, x\right )} + 4 \, e^{\left (-4 \, x\right )} + 4} - \frac{1}{4} i \, \log \left (i \, e^{\left (-x\right )} + 1\right ) + \frac{1}{4} i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(e^(-x) + 4*I*e^(-2*x) - e^(-3*x))/(16*I*e^(-x) - 24*e^(-2*x) - 16*I*e^(-3*x) + 4*e^(-4*x) + 4) - 1/4*I*log
(I*e^(-x) + 1) + 1/4*I*log(I*e^(-x) - 1)

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Fricas [B]  time = 1.84944, size = 298, normalized size = 8.76 \begin{align*} \frac{{\left (-i \, e^{\left (4 \, x\right )} + 4 \, e^{\left (3 \, x\right )} + 6 i \, e^{\left (2 \, x\right )} - 4 \, e^{x} - i\right )} \log \left (e^{x} + i\right ) +{\left (i \, e^{\left (4 \, x\right )} - 4 \, e^{\left (3 \, x\right )} - 6 i \, e^{\left (2 \, x\right )} + 4 \, e^{x} + i\right )} \log \left (e^{x} - i\right ) - 2 \, e^{\left (3 \, x\right )} - 8 i \, e^{\left (2 \, x\right )} + 2 \, e^{x}}{4 \, e^{\left (4 \, x\right )} + 16 i \, e^{\left (3 \, x\right )} - 24 \, e^{\left (2 \, x\right )} - 16 i \, e^{x} + 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-I*e^(4*x) + 4*e^(3*x) + 6*I*e^(2*x) - 4*e^x - I)*log(e^x + I) + (I*e^(4*x) - 4*e^(3*x) - 6*I*e^(2*x) + 4*e^
x + I)*log(e^x - I) - 2*e^(3*x) - 8*I*e^(2*x) + 2*e^x)/(4*e^(4*x) + 16*I*e^(3*x) - 24*e^(2*x) - 16*I*e^x + 4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}{\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)/(I+sinh(x))**2,x)

[Out]

Integral(sech(x)/(sinh(x) + I)**2, x)

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Giac [B]  time = 1.27089, size = 95, normalized size = 2.79 \begin{align*} \frac{3 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 20 \, e^{\left (-x\right )} - 20 \, e^{x} - 44 i}{16 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{2}} - \frac{1}{8} i \, \log \left (i \, e^{\left (-x\right )} - i \, e^{x} + 2\right ) + \frac{1}{8} i \, \log \left (i \, e^{\left (-x\right )} - i \, e^{x} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*I*(e^(-x) - e^x)^2 + 20*e^(-x) - 20*e^x - 44*I)/(e^(-x) - e^x - 2*I)^2 - 1/8*I*log(I*e^(-x) - I*e^x +
2) + 1/8*I*log(I*e^(-x) - I*e^x - 2)