\(\int \frac {1}{1+i \sinh (c+d x)} \, dx\) [56]

Optimal result

Integrand size = 14, antiderivative size = 27 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {i \cosh (c+d x)}{d (1+i \sinh (c+d x))} \]

[Out]

I*cosh(d*x+c)/d/(1+I*sinh(d*x+c))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2727} \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {i \cosh (c+d x)}{d (1+i \sinh (c+d x))} \]

[In]

Int[(1 + I*Sinh[c + d*x])^(-1),x]

[Out]

(I*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*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]

Rubi steps \begin{align*} \text {integral}& = \frac {i \cosh (c+d x)}{d (1+i \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {2 \sinh \left (\frac {1}{2} (c+d x)\right )}{d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[(1 + I*Sinh[c + d*x])^(-1),x]

[Out]

(2*Sinh[(c + d*x)/2])/(d*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {2 i}{d \left ({\mathrm e}^{d x +c}-i\right )}\)	\(18\)
derivativedivides	\(\frac {2}{d \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(20\)
default	\(\frac {2}{d \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(20\)
parallelrisch	\(-\frac {2}{d \left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(22\)

[In]

int(1/(1+I*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2*I/d/(exp(d*x+c)-I)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {2 i}{d e^{\left (d x + c\right )} - i \, d} \]

[In]

integrate(1/(1+I*sinh(d*x+c)),x, algorithm="fricas")

[Out]

2*I/(d*e^(d*x + c) - I*d)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {2 i}{d e^{c} e^{d x} - i d} \]

[In]

integrate(1/(1+I*sinh(d*x+c)),x)

[Out]

2*I/(d*exp(c)*exp(d*x) - I*d)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=-\frac {2}{d {\left (i \, e^{\left (-d x - c\right )} - 1\right )}} \]

[In]

integrate(1/(1+I*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-2/(d*(I*e^(-d*x - c) - 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {2 i}{d {\left (e^{\left (d x + c\right )} - i\right )}} \]

[In]

integrate(1/(1+I*sinh(d*x+c)),x, algorithm="giac")

[Out]

2*I/(d*(e^(d*x + c) - I))

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {1}{1+i \sinh (c+d x)} \, dx=\frac {2{}\mathrm {i}}{d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \]

[In]

int(1/(sinh(c + d*x)*1i + 1),x)

[Out]

2i/(d*(exp(c + d*x) - 1i))