\(\int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) [190]

Optimal result

Integrand size = 22, antiderivative size = 35 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i x}{a}-\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \]

[Out]

-I*x/a-cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2814, 2727} \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {i x}{a} \]

[In]

Int[Sinh[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

((-I)*x)/a - Cosh[c + d*x]/(d*(a + I*a*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]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.74 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \cosh (c+d x) \left (1-\frac {\text {arcsinh}(\sinh (c+d x)) (-i+\sinh (c+d x))}{\sqrt {\cosh ^2(c+d x)}}\right )}{a d (-i+\sinh (c+d x))} \]

[In]

Integrate[Sinh[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

(I*Cosh[c + d*x]*(1 - (ArcSinh[Sinh[c + d*x]]*(-I + Sinh[c + d*x]))/Sqrt[Cosh[c + d*x]^2]))/(a*d*(-I + Sinh[c
+ d*x]))

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {i x}{a}-\frac {2}{d a \left ({\mathrm e}^{d x +c}-i\right )}\)	\(28\)
parallelrisch	\(\frac {i x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) d +d x -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(53\)
derivativedivides	\(\frac {i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 i}{-2 i+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}\)	\(57\)
default	\(\frac {i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 i}{-2 i+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {-i \, d x e^{\left (d x + c\right )} - d x - 2}{a d e^{\left (d x + c\right )} - i \, a d} \]

[In]

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

[Out]

(-I*d*x*e^(d*x + c) - d*x - 2)/(a*d*e^(d*x + c) - I*a*d)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.69 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {2}{a d e^{c} e^{d x} - i a d} - \frac {i x}{a} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \, {\left (d x + c\right )}}{a d} - \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d} \]

[In]

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

[Out]

-I*(d*x + c)/(a*d) - 2/((a*e^(-d*x - c) + I*a)*d)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\frac {i \, {\left (d x + c\right )}}{a} + \frac {2 i}{a {\left (i \, e^{\left (d x + c\right )} + 1\right )}}}{d} \]

[In]

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

[Out]

-(I*(d*x + c)/a + 2*I/(a*(I*e^(d*x + c) + 1)))/d

Mupad [B] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {x\,1{}\mathrm {i}}{a}-\frac {2}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \]

[In]

int(sinh(c + d*x)/(a + a*sinh(c + d*x)*1i),x)

[Out]

- (x*1i)/a - 2/(a*d*(exp(c + d*x) - 1i))