\(\int \frac {\sinh (x)}{i+\sinh (x)} \, dx\) [43]

Optimal result

Integrand size = 11, antiderivative size = 14 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=x-\frac {\cosh (x)}{i+\sinh (x)} \]

[Out]

x-cosh(x)/(I+sinh(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2814, 2727} \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=x-\frac {\cosh (x)}{\sinh (x)+i} \]

[In]

Int[Sinh[x]/(I + Sinh[x]),x]

[Out]

x - Cosh[x]/(I + Sinh[x])

Rule 2727

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]

Rule 2814

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]

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(43\) vs. \(2(14)=28\).

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 3.07 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=i \text {sech}(x) \left (1+2 \arcsin \left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right ) \sqrt {\cosh ^2(x)}+i \sinh (x)\right ) \]

[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])

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
risch	\(x +\frac {2 i}{{\mathrm e}^{x}+i}\)	\(13\)
parallelrisch	\(\frac {-2+i x +x \tanh \left (\frac {x}{2}\right )}{\tanh \left (\frac {x}{2}\right )+i}\)	\(23\)
default	\(-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {2}{\tanh \left (\frac {x}{2}\right )+i}+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )\)	\(29\)

[In]

int(sinh(x)/(I+sinh(x)),x,method=_RETURNVERBOSE)

[Out]

x+2*I/(exp(x)+I)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=\frac {x e^{x} + i \, x + 2 i}{e^{x} + i} \]

[In]

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

[Out]

(x*e^x + I*x + 2*I)/(e^x + I)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=x + \frac {2 i}{e^{x} + i} \]

[In]

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

[Out]

x + 2*I/(exp(x) + I)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=x + \frac {2 i}{e^{\left (-x\right )} - i} \]

[In]

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

[Out]

x + 2*I/(e^(-x) - I)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=x + \frac {2 i}{e^{x} + i} \]

[In]

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

[Out]

x + 2*I/(e^x + I)

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\sinh (x)}{i+\sinh (x)} \, dx=x+\frac {2{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

[In]

int(sinh(x)/(sinh(x) + 1i),x)

[Out]

x + 2i/(exp(x) + 1i)