3.43 \(\int \frac{\sinh (x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=14 \[ x-\frac{\cosh (x)}{\sinh (x)+i} \]

[Out]

x - Cosh[x]/(I + Sinh[x])

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Rubi [A]  time = 0.0271906, antiderivative size = 14, 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 = {2735, 2648} \[ x-\frac{\cosh (x)}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(I + Sinh[x]),x]

[Out]

x - Cosh[x]/(I + Sinh[x])

Rule 2735

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

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

Mathematica [B]  time = 0.0538288, size = 43, normalized size = 3.07 \[ i \text{sech}(x) \left (i \sinh (x)+2 \sqrt{\cosh ^2(x)} \sin ^{-1}\left (\frac{\sqrt{1-i \sinh (x)}}{\sqrt{2}}\right )+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(I + Sinh[x]),x]

[Out]

I*Sech[x]*(1 + 2*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[Cosh[x]^2] + I*Sinh[x])

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Maple [B]  time = 0.023, size = 29, normalized size = 2.1 \begin{align*} \ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-1}-\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.15381, size = 16, normalized size = 1.14 \begin{align*} x + \frac{2 i}{e^{\left (-x\right )} - i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*I/(e^(-x) - I)

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Fricas [A]  time = 2.07709, size = 42, normalized size = 3. \begin{align*} \frac{x e^{x} + i \, x + 2 i}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^x + I*x + 2*I)/(e^x + I)

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Sympy [A]  time = 0.157277, size = 8, normalized size = 0.57 \begin{align*} x + \frac{2 i}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*I/(exp(x) + I)

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Giac [A]  time = 1.29781, size = 14, normalized size = 1. \begin{align*} x + \frac{2 i}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*I/(e^x + I)