Integrand size = 14, antiderivative size = 82 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {3 i \log \left (i-3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}-\frac {3 i \log \left (3 i-\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))} \] Output:
3/64*I*ln(I-3*tanh(1/2*d*x+1/2*c))/d-3/64*I*ln(3*I-tanh(1/2*d*x+1/2*c))/d+ 5/16*I*cosh(d*x+c)/d/(3+5*I*sinh(d*x+c))
Time = 0.36 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.73 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {-9 \left (2 \arctan \left (3 \coth \left (\frac {1}{2} (c+d x)\right )\right )+2 \arctan \left (3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )-i \log (4-5 \cosh (c+d x))+i \log (4+5 \cosh (c+d x))\right )+40 \left (\frac {1}{3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )}+\frac {3}{\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )}\right ) \sinh \left (\frac {1}{2} (c+d x)\right )}{384 d} \] Input:
Integrate[(3 + (5*I)*Sinh[c + d*x])^(-2),x]
Output:
(-9*(2*ArcTan[3*Coth[(c + d*x)/2]] + 2*ArcTan[3*Tanh[(c + d*x)/2]] - I*Log [4 - 5*Cosh[c + d*x]] + I*Log[4 + 5*Cosh[c + d*x]]) + 40*((3*Cosh[(c + d*x )/2] + I*Sinh[(c + d*x)/2])^(-1) + 3/(Cosh[(c + d*x)/2] + (3*I)*Sinh[(c + d*x)/2]))*Sinh[(c + d*x)/2])/(384*d)
Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3143, 27, 3042, 3139, 1081, 2009}
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.
\(\displaystyle \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(3+5 \sin (i c+i d x))^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {1}{16} \int -\frac {3}{5 i \sinh (c+d x)+3}dx+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac {3}{16} \int \frac {1}{5 i \sinh (c+d x)+3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac {3}{16} \int \frac {1}{5 \sin (i c+i d x)+3}dx\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {3 i \int \frac {1}{-3 \tanh ^2\left (\frac {1}{2} (c+d x)\right )+10 i \tanh \left (\frac {1}{2} (c+d x)\right )+3}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\) |
\(\Big \downarrow \) 1081 |
\(\displaystyle \frac {9 i \int \left (\frac {1}{8 \left (3 i \tanh \left (\frac {1}{2} (c+d x)\right )+1\right )}-\frac {1}{24 \left (i \tanh \left (\frac {1}{2} (c+d x)\right )+3\right )}\right )d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 i \left (\frac {1}{24} \log \left (1+3 i \tanh \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{24} \log \left (3+i \tanh \left (\frac {1}{2} (c+d x)\right )\right )\right )}{8 d}+\frac {5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\) |
Input:
Int[(3 + (5*I)*Sinh[c + d*x])^(-2),x]
Output:
(((9*I)/8)*(-1/24*Log[3 + I*Tanh[(c + d*x)/2]] + Log[1 + (3*I)*Tanh[(c + d *x)/2]]/24))/d + (((5*I)/16)*Cosh[c + d*x])/(d*(3 + (5*I)*Sinh[c + d*x]))
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_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[c Int[ExpandIntegrand[1/((b/2 - q/2 + c*x)*(b/2 + q/2 + c*x)), x], x], x]] /; FreeQ[{a, b, c}, x] && NiceSqrtQ[b^2 - 4*a*c]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
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 = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\frac {3 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{64}+\frac {5}{48 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}-\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{64}+\frac {5}{16 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}}{d}\) | \(74\) |
default | \(\frac {\frac {3 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{64}+\frac {5}{48 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}-\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{64}+\frac {5}{16 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}}{d}\) | \(74\) |
risch | \(\frac {i \left (3 \,{\mathrm e}^{d x +c}-5 i\right )}{8 d \left (5 \,{\mathrm e}^{2 d x +2 c}-5-6 i {\mathrm e}^{d x +c}\right )}-\frac {3 i \ln \left ({\mathrm e}^{d x +c}+\frac {4}{5}-\frac {3 i}{5}\right )}{64 d}+\frac {3 i \ln \left ({\mathrm e}^{d x +c}-\frac {4}{5}-\frac {3 i}{5}\right )}{64 d}\) | \(77\) |
parallelrisch | \(\frac {\left (-9-15 i \sinh \left (d x +c \right )\right ) \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+15 i \sinh \left (d x +c \right ) \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-9 i\right )+\ln \left (19683\right )-20 \cosh \left (d x +c \right )+9 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{64 d \left (3 i-5 \sinh \left (d x +c \right )\right )}\) | \(96\) |
Input:
int(1/(3+5*I*sinh(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(3/64*I*ln(3*tanh(1/2*d*x+1/2*c)-I)+5/48/(3*tanh(1/2*d*x+1/2*c)-I)-3/6 4*I*ln(tanh(1/2*d*x+1/2*c)-3*I)+5/16/(tanh(1/2*d*x+1/2*c)-3*I))
Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {3 \, {\left (5 i \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} - 5 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) + 3 \, {\left (-5 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 5 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i - \frac {4}{5}\right ) - 24 i \, e^{\left (d x + c\right )} - 40}{64 \, {\left (5 \, d e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, d e^{\left (d x + c\right )} - 5 \, d\right )}} \] Input:
integrate(1/(3+5*I*sinh(d*x+c))^2,x, algorithm="fricas")
Output:
-1/64*(3*(5*I*e^(2*d*x + 2*c) + 6*e^(d*x + c) - 5*I)*log(e^(d*x + c) - 3/5 *I + 4/5) + 3*(-5*I*e^(2*d*x + 2*c) - 6*e^(d*x + c) + 5*I)*log(e^(d*x + c) - 3/5*I - 4/5) - 24*I*e^(d*x + c) - 40)/(5*d*e^(2*d*x + 2*c) - 6*I*d*e^(d *x + c) - 5*d)
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {3 i e^{c} e^{d x} + 5}{40 d e^{2 c} e^{2 d x} - 48 i d e^{c} e^{d x} - 40 d} + \frac {\operatorname {RootSum} {\left (4096 z^{2} + 9, \left ( i \mapsto i \log {\left (\frac {\left (256 i i - 9 i\right ) e^{- c}}{15} + e^{d x} \right )} \right )\right )}}{d} \] Input:
integrate(1/(3+5*I*sinh(d*x+c))**2,x)
Output:
(3*I*exp(c)*exp(d*x) + 5)/(40*d*exp(2*c)*exp(2*d*x) - 48*I*d*exp(c)*exp(d* x) - 40*d) + RootSum(4096*_z**2 + 9, Lambda(_i, _i*log((256*_i*I - 9*I)*ex p(-c)/15 + exp(d*x))))/d
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=\frac {3 i \, \log \left (\frac {5 \, e^{\left (-d x - c\right )} + 3 i - 4}{5 \, e^{\left (-d x - c\right )} + 3 i + 4}\right )}{64 \, d} + \frac {3 i \, e^{\left (-d x - c\right )} - 5}{-8 \, d {\left (-6 i \, e^{\left (-d x - c\right )} - 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5\right )}} \] Input:
integrate(1/(3+5*I*sinh(d*x+c))^2,x, algorithm="maxima")
Output:
3/64*I*log((5*e^(-d*x - c) + 3*I - 4)/(5*e^(-d*x - c) + 3*I + 4))/d + (3*I *e^(-d*x - c) - 5)/(d*(48*I*e^(-d*x - c) + 40*e^(-2*d*x - 2*c) - 40))
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {\frac {8 \, {\left (-3 i \, e^{\left (d x + c\right )} - 5\right )}}{5 \, e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, e^{\left (d x + c\right )} - 5} + 3 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) - 3 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{64 \, d} \] Input:
integrate(1/(3+5*I*sinh(d*x+c))^2,x, algorithm="giac")
Output:
-1/64*(8*(-3*I*e^(d*x + c) - 5)/(5*e^(2*d*x + 2*c) - 6*I*e^(d*x + c) - 5) + 3*I*log(-(I - 2)*e^(d*x + c) - 2*I + 1) - 3*I*log(-(2*I - 1)*e^(d*x + c) + I - 2))/d
Time = 2.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\frac {5}{8\,\left (5\,d-5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}\right )}-\frac {\ln \left (-\frac {15}{4}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (-3-\frac {9}{4}{}\mathrm {i}\right )\right )\,3{}\mathrm {i}}{64\,d}+\frac {\ln \left (\frac {15}{4}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (-3+\frac {9}{4}{}\mathrm {i}\right )\right )\,3{}\mathrm {i}}{64\,d}-\frac {{\mathrm {e}}^{c+d\,x}\,3{}\mathrm {i}}{8\,\left (5\,d-5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}\right )} \] Input:
int(1/(sinh(c + d*x)*5i + 3)^2,x)
Output:
(log(15/4 - exp(d*x)*exp(c)*(3 - 9i/4))*3i)/(64*d) - (log(- exp(d*x)*exp(c )*(3 + 9i/4) - 15/4)*3i)/(64*d) - 5/(8*(5*d + d*exp(c + d*x)*6i - 5*d*exp( 2*c + 2*d*x))) - (exp(c + d*x)*3i)/(8*(5*d + d*exp(c + d*x)*6i - 5*d*exp(2 *c + 2*d*x)))
\[ \int \frac {1}{(3+5 i \sinh (c+d x))^2} \, dx=-\left (\int \frac {1}{25 \sinh \left (d x +c \right )^{2}-30 \sinh \left (d x +c \right ) i -9}d x \right ) \] Input:
int(1/(3+5*I*sinh(d*x+c))^2,x)
Output:
- int(1/(25*sinh(c + d*x)**2 - 30*sinh(c + d*x)*i - 9),x)