3.166 \(\int \frac{\text{sech}(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=24 \[ -\frac{i}{2 (\sinh (x)+i)}-\frac{1}{2} i \tan ^{-1}(\sinh (x)) \]

[Out]

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

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Rubi [A]  time = 0.034876, antiderivative size = 24, 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} \[ -\frac{i}{2 (\sinh (x)+i)}-\frac{1}{2} i \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(I + Sinh[x]),x]

[Out]

(-I/2)*ArcTan[Sinh[x]] - (I/2)/(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)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{(i-x) (i+x)^2} \, dx,x,\sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{i}{2 (i+x)^2}+\frac{i}{2 \left (1+x^2\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=-\frac{i}{2 (i+\sinh (x))}-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{2} i \tan ^{-1}(\sinh (x))-\frac{i}{2 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.021417, size = 18, normalized size = 0.75 \[ -\frac{1}{2} i \left (\tan ^{-1}(\sinh (x))+\frac{1}{\sinh (x)+i}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(I + Sinh[x]),x]

[Out]

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

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Maple [B]  time = 0.029, size = 43, normalized size = 1.8 \begin{align*} -{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) }+{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}+{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.19936, size = 55, normalized size = 2.29 \begin{align*} \frac{2 i \, e^{\left (-x\right )}}{-4 i \, e^{\left (-x\right )} + 2 \, e^{\left (-2 \, x\right )} - 2} - \frac{1}{2} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac{1}{2} \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.78173, size = 158, normalized size = 6.58 \begin{align*} \frac{{\left (e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )} \log \left (e^{x} + i\right ) -{\left (e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )} \log \left (e^{x} - i\right ) - 2 i \, e^{x}}{2 \,{\left (e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((e^(2*x) + 2*I*e^x - 1)*log(e^x + I) - (e^(2*x) + 2*I*e^x - 1)*log(e^x - I) - 2*I*e^x)/(e^(2*x) + 2*I*e^x
- 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)/(sinh(x) + I), x)

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Giac [B]  time = 1.29249, size = 69, normalized size = 2.88 \begin{align*} -\frac{e^{\left (-x\right )} - e^{x} - 6 i}{4 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}} + \frac{1}{4} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{1}{4} \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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