\(\int \frac {1}{5+3 i \sinh (c+d x)} \, dx\) [92]

Optimal result

Integrand size = 14, antiderivative size = 37 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d} \]

[Out]

1/4*x-1/2*I*arctan(cosh(d*x+c)/(3+I*sinh(d*x+c)))/d

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, 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 = {2736} \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d} \]

[In]

Int[(5 + (3*I)*Sinh[c + d*x])^(-1),x]

[Out]

x/4 - ((I/2)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d

Rule 2736

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(171\) vs. \(2(37)=74\).

Time = 0.13 (sec) , antiderivative size = 171, normalized size of antiderivative = 4.62 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=-\frac {i \arctan \left (\frac {2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-2 \sinh \left (\frac {1}{2} (c+d x)\right )}\right )}{4 d}+\frac {i \arctan \left (\frac {\cosh \left (\frac {1}{2} (c+d x)\right )+2 \sinh \left (\frac {1}{2} (c+d x)\right )}{2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )}\right )}{4 d}-\frac {\log (5 \cosh (c+d x)-4 \sinh (c+d x))}{8 d}+\frac {\log (5 \cosh (c+d x)+4 \sinh (c+d x))}{8 d} \]

[In]

Integrate[(5 + (3*I)*Sinh[c + d*x])^(-1),x]

[Out]

((-1/4*I)*ArcTan[(2*Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2])/(Cosh[(c + d*x)/2] - 2*Sinh[(c + d*x)/2])])/d + ((I
/4)*ArcTan[(Cosh[(c + d*x)/2] + 2*Sinh[(c + d*x)/2])/(2*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])])/d - Log[5*Cos
h[c + d*x] - 4*Sinh[c + d*x]]/(8*d) + Log[5*Cosh[c + d*x] + 4*Sinh[c + d*x]]/(8*d)

Maple [A] (verified)

Time = 1.75 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {\ln \left (-\frac {i}{3}+{\mathrm e}^{d x +c}\right )}{4 d}-\frac {\ln \left ({\mathrm e}^{d x +c}-3 i\right )}{4 d}\)	\(32\)
parallelrisch	\(\frac {-\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )+\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{4 d}\)	\(41\)
derivativedivides	\(\frac {-\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{4}+\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{4}}{d}\)	\(42\)
default	\(\frac {-\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{4}+\frac {\ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{4}}{d}\)	\(42\)

[In]

int(1/(5+3*I*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/4/d*ln(-1/3*I+exp(d*x+c))-1/4/d*ln(exp(d*x+c)-3*I)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {\log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - \log \left (e^{\left (d x + c\right )} - 3 i\right )}{4 \, d} \]

[In]

integrate(1/(5+3*I*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(log(e^(d*x + c) - 1/3*I) - log(e^(d*x + c) - 3*I))/d

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {- \frac {\log {\left (e^{d x} - 3 i e^{- c} \right )}}{4} + \frac {\log {\left (e^{d x} - \frac {i e^{- c}}{3} \right )}}{4}}{d} \]

[In]

integrate(1/(5+3*I*sinh(d*x+c)),x)

[Out]

(-log(exp(d*x) - 3*I*exp(-c))/4 + log(exp(d*x) - I*exp(-c)/3)/4)/d

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {\log \left (-\frac {6 \, {\left (-i \, e^{\left (-d x - c\right )} + 3\right )}}{6 i \, e^{\left (-d x - c\right )} - 2}\right )}{4 \, d} \]

[In]

integrate(1/(5+3*I*sinh(d*x+c)),x, algorithm="maxima")

[Out]

1/4*log(-6*(-I*e^(-d*x - c) + 3)/(6*I*e^(-d*x - c) - 2))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=\frac {\log \left (3 \, e^{\left (d x + c\right )} - i\right ) - \log \left (e^{\left (d x + c\right )} - 3 i\right )}{4 \, d} \]

[In]

integrate(1/(5+3*I*sinh(d*x+c)),x, algorithm="giac")

[Out]

1/4*(log(3*e^(d*x + c) - I) - log(e^(d*x + c) - 3*I))/d

Mupad [B] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {1}{5+3 i \sinh (c+d x)} \, dx=-\frac {\ln \left (-\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{2}+\frac {3}{2}{}\mathrm {i}\right )-\ln \left (\frac {9\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{2}-\frac {3}{2}{}\mathrm {i}\right )}{4\,d} \]

[In]

int(1/(sinh(c + d*x)*3i + 5),x)

[Out]

-(log(3i/2 - (exp(d*x)*exp(c))/2) - log((9*exp(d*x)*exp(c))/2 - 3i/2))/(4*d)