Integrand size = 14, antiderivative size = 66 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {5 x}{64}-\frac {5 i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))} \]
5/64*x-5/32*I*arctan(cosh(d*x+c)/(3+I*sinh(d*x+c)))/d-3/16*I*cosh(d*x+c)/d /(5+3*I*sinh(d*x+c))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(183\) vs. \(2(66)=132\).
Time = 0.37 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.77 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {24 i-50 i \arctan \left (\frac {2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-2 \sinh \left (\frac {1}{2} (c+d x)\right )}\right )+50 i \arctan \left (\frac {\cosh \left (\frac {1}{2} (c+d x)\right )+2 \sinh \left (\frac {1}{2} (c+d x)\right )}{2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )}\right )-25 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+25 \log (5 \cosh (c+d x)+4 \sinh (c+d x))-\frac {120 \cosh (c+d x)}{-5 i+3 \sinh (c+d x)}}{640 d} \]
(24*I - (50*I)*ArcTan[(2*Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2])/(Cosh[(c + d*x)/2] - 2*Sinh[(c + d*x)/2])] + (50*I)*ArcTan[(Cosh[(c + d*x)/2] + 2*Si nh[(c + d*x)/2])/(2*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])] - 25*Log[5*Cos h[c + d*x] - 4*Sinh[c + d*x]] + 25*Log[5*Cosh[c + d*x] + 4*Sinh[c + d*x]] - (120*Cosh[c + d*x])/(-5*I + 3*Sinh[c + d*x]))/(640*d)
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3143, 27, 3042, 3136}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(5+3 \sin (i c+i d x))^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle -\frac {1}{16} \int -\frac {5}{3 i \sinh (c+d x)+5}dx-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{16} \int \frac {1}{3 i \sinh (c+d x)+5}dx-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{16} \int \frac {1}{3 \sin (i c+i d x)+5}dx-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\) |
\(\Big \downarrow \) 3136 |
\(\displaystyle \frac {5}{16} \left (\frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d}\right )-\frac {3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\) |
(5*(x/4 - ((I/2)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d))/16 - ((( 3*I)/16)*Cosh[c + d*x])/(d*(5 + (3*I)*Sinh[c + d*x]))
3.1.93.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Time = 1.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11
method | result | size |
risch | \(-\frac {i \left (5 \,{\mathrm e}^{d x +c}-3 i\right )}{8 d \left (3 \,{\mathrm e}^{2 d x +2 c}-3-10 i {\mathrm e}^{d x +c}\right )}-\frac {5 \ln \left ({\mathrm e}^{d x +c}-3 i\right )}{64 d}+\frac {5 \ln \left (-\frac {i}{3}+{\mathrm e}^{d x +c}\right )}{64 d}\) | \(73\) |
derivativedivides | \(\frac {\frac {-\frac {9}{80}-\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64}+\frac {-\frac {9}{80}+\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64}}{d}\) | \(84\) |
default | \(\frac {\frac {-\frac {9}{80}-\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64}+\frac {-\frac {9}{80}+\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64}}{d}\) | \(84\) |
parallelrisch | \(\frac {125 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )-125 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )-60 i+75 i \sinh \left (d x +c \right ) \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )-75 i \sinh \left (d x +c \right ) \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )-60 i \cosh \left (d x +c \right )+36 \sinh \left (d x +c \right )}{960 i d \sinh \left (d x +c \right )+1600 d}\) | \(124\) |
-1/8*I*(5*exp(d*x+c)-3*I)/d/(3*exp(2*d*x+2*c)-3-10*I*exp(d*x+c))-5/64/d*ln (exp(d*x+c)-3*I)+5/64/d*ln(-1/3*I+exp(d*x+c))
Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.56 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {5 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - 5 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 40 i \, e^{\left (d x + c\right )} - 24}{64 \, {\left (3 \, d e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, d e^{\left (d x + c\right )} - 3 \, d\right )}} \]
1/64*(5*(3*e^(2*d*x + 2*c) - 10*I*e^(d*x + c) - 3)*log(e^(d*x + c) - 1/3*I ) - 5*(3*e^(2*d*x + 2*c) - 10*I*e^(d*x + c) - 3)*log(e^(d*x + c) - 3*I) - 40*I*e^(d*x + c) - 24)/(3*d*e^(2*d*x + 2*c) - 10*I*d*e^(d*x + c) - 3*d)
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {- 5 i e^{c} e^{d x} - 3}{24 d e^{2 c} e^{2 d x} - 80 i d e^{c} e^{d x} - 24 d} + \frac {- \frac {5 \log {\left (e^{d x} - 3 i e^{- c} \right )}}{64} + \frac {5 \log {\left (e^{d x} - \frac {i e^{- c}}{3} \right )}}{64}}{d} \]
(-5*I*exp(c)*exp(d*x) - 3)/(24*d*exp(2*c)*exp(2*d*x) - 80*I*d*exp(c)*exp(d *x) - 24*d) + (-5*log(exp(d*x) - 3*I*exp(-c))/64 + 5*log(exp(d*x) - I*exp( -c)/3)/64)/d
Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=-\frac {5 i \, \arctan \left (\frac {3}{4} \, e^{\left (-d x - c\right )} + \frac {5}{4} i\right )}{32 \, d} - \frac {5 i \, e^{\left (-d x - c\right )} - 3}{-8 \, d {\left (-10 i \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3\right )}} \]
-5/32*I*arctan(3/4*e^(-d*x - c) + 5/4*I)/d - (5*I*e^(-d*x - c) - 3)/(d*(80 *I*e^(-d*x - c) + 24*e^(-2*d*x - 2*c) - 24))
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=-\frac {\frac {8 \, {\left (5 i \, e^{\left (d x + c\right )} + 3\right )}}{3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3} - 5 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right ) + 5 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{64 \, d} \]
-1/64*(8*(5*I*e^(d*x + c) + 3)/(3*e^(2*d*x + 2*c) - 10*I*e^(d*x + c) - 3) - 5*log(3*e^(d*x + c) - I) + 5*log(e^(d*x + c) - 3*I))/d
Time = 1.78 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.55 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx=\frac {3}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )}-\frac {5\,\ln \left (-\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}+\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {5\,\ln \left (\frac {45\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}-\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,5{}\mathrm {i}}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )} \]