\(\int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx\) [94]

Optimal result

Integrand size = 14, antiderivative size = 95 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=\frac {59 x}{2048}-\frac {59 i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{1024 d}-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac {45 i \cosh (c+d x)}{512 d (5+3 i \sinh (c+d x))} \] Output:

59/2048*x-59/1024*I*arctan(cosh(d*x+c)/(3+I*sinh(d*x+c)))/d-3/32*I*cosh(d* 
x+c)/d/(5+3*I*sinh(d*x+c))^2-45/512*I*cosh(d*x+c)/d/(5+3*I*sinh(d*x+c))
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(95)=190\).

Time = 0.69 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.92 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=\frac {-118 i \arctan \left (\frac {2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-2 \sinh \left (\frac {1}{2} (c+d x)\right )}\right )+118 i \arctan \left (\frac {\cosh \left (\frac {1}{2} (c+d x)\right )+2 \sinh \left (\frac {1}{2} (c+d x)\right )}{2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )}\right )-59 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+59 \log (5 \cosh (c+d x)+4 \sinh (c+d x))+\frac {48}{\left ((1+2 i) \cosh \left (\frac {1}{2} (c+d x)\right )-(2+i) \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48}{\left ((2+i) \cosh \left (\frac {1}{2} (c+d x)\right )+(1+2 i) \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {144 \sinh \left (\frac {1}{2} (c+d x)\right ) \left (-3 i \cosh \left (\frac {1}{2} (c+d x)\right )+5 \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{-5 i+3 \sinh (c+d x)}}{4096 d} \] Input:

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

Output:

((-118*I)*ArcTan[(2*Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2])/(Cosh[(c + d*x) 
/2] - 2*Sinh[(c + d*x)/2])] + (118*I)*ArcTan[(Cosh[(c + d*x)/2] + 2*Sinh[( 
c + d*x)/2])/(2*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])] - 59*Log[5*Cosh[c 
+ d*x] - 4*Sinh[c + d*x]] + 59*Log[5*Cosh[c + d*x] + 4*Sinh[c + d*x]] + 48 
/((1 + 2*I)*Cosh[(c + d*x)/2] - (2 + I)*Sinh[(c + d*x)/2])^2 + 48/((2 + I) 
*Cosh[(c + d*x)/2] + (1 + 2*I)*Sinh[(c + d*x)/2])^2 - (144*Sinh[(c + d*x)/ 
2]*((-3*I)*Cosh[(c + d*x)/2] + 5*Sinh[(c + d*x)/2]))/(-5*I + 3*Sinh[c + d* 
x]))/(4096*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3143, 25, 3042, 3233, 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.

\(\displaystyle \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(5+3 \sin (i c+i d x))^3}dx\)

\(\Big \downarrow \) 3143

\(\displaystyle -\frac {1}{32} \int -\frac {10-3 i \sinh (c+d x)}{(3 i \sinh (c+d x)+5)^2}dx-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{32} \int \frac {10-3 i \sinh (c+d x)}{(3 i \sinh (c+d x)+5)^2}dx-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{32} \int \frac {10-3 \sin (i c+i d x)}{(3 \sin (i c+i d x)+5)^2}dx-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {1}{32} \left (-\frac {1}{16} \int -\frac {59}{3 i \sinh (c+d x)+5}dx-\frac {45 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\right )-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{32} \left (\frac {59}{16} \int \frac {1}{3 i \sinh (c+d x)+5}dx-\frac {45 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\right )-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{32} \left (\frac {59}{16} \int \frac {1}{3 \sin (i c+i d x)+5}dx-\frac {45 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\right )-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 3136

\(\displaystyle \frac {1}{32} \left (\frac {59}{16} \left (\frac {x}{4}-\frac {i \arctan \left (\frac {\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{2 d}\right )-\frac {45 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\right )-\frac {3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}\)

Input:

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

Output:

((59*(x/4 - ((I/2)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d))/16 - ( 
((45*I)/16)*Cosh[c + d*x])/(d*(5 + (3*I)*Sinh[c + d*x])))/32 - (((3*I)/32) 
*Cosh[c + d*x])/(d*(5 + (3*I)*Sinh[c + d*x])^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3136

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]
 

rule 3143

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]
 

rule 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01

method	result	size
risch	\(-\frac {3 i \left (-295 i {\mathrm e}^{2 d x +2 c}+59 \,{\mathrm e}^{3 d x +3 c}+45 i-241 \,{\mathrm e}^{d x +c}\right )}{256 d \left (3 \,{\mathrm e}^{2 d x +2 c}-3-10 i {\mathrm e}^{d x +c}\right )^{2}}-\frac {59 \ln \left ({\mathrm e}^{d x +c}-3 i\right )}{2048 d}+\frac {59 \ln \left (-\frac {i}{3}+{\mathrm e}^{d x +c}\right )}{2048 d}\)	\(96\)
derivativedivides	\(\frac {\frac {\frac {63}{3200}-\frac {27 i}{400}}{\left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )^{2}}+\frac {-\frac {963}{12800}-\frac {123 i}{1600}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {59 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{2048}+\frac {-\frac {63}{3200}-\frac {27 i}{400}}{\left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )^{2}}+\frac {-\frac {963}{12800}+\frac {123 i}{1600}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {59 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{2048}}{d}\)	\(126\)
default	\(\frac {\frac {\frac {63}{3200}-\frac {27 i}{400}}{\left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )^{2}}+\frac {-\frac {963}{12800}-\frac {123 i}{1600}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {59 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{2048}+\frac {-\frac {63}{3200}-\frac {27 i}{400}}{\left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )^{2}}+\frac {-\frac {963}{12800}+\frac {123 i}{1600}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {59 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{2048}}{d}\)	\(126\)
parallelrisch	\(\frac {\left (-87025+13275 \cosh \left (2 d x +2 c \right )-88500 i \sinh \left (d x +c \right )\right ) \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )+\left (88500 i \sinh \left (d x +c \right )-13275 \cosh \left (2 d x +2 c \right )+87025\right ) \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )-54600 i \cosh \left (d x +c \right )+9828 i \cosh \left (2 d x +2 c \right )-64428 i+65520 \sinh \left (d x +c \right )+13500 \sinh \left (2 d x +2 c \right )}{51200 d \left (59-9 \cosh \left (2 d x +2 c \right )+60 i \sinh \left (d x +c \right )\right )}\)	\(151\)

Input:

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

Output:

-3/256*I*(-295*I*exp(2*d*x+2*c)+59*exp(3*d*x+3*c)+45*I-241*exp(d*x+c))/d/( 
3*exp(2*d*x+2*c)-3-10*I*exp(d*x+c))^2-59/2048/d*ln(exp(d*x+c)-3*I)+59/2048 
/d*ln(-1/3*I+exp(d*x+c))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (75) = 150\).

Time = 0.10 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.03 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=\frac {59 \, {\left (9 \, e^{\left (4 \, d x + 4 \, c\right )} - 60 i \, e^{\left (3 \, d x + 3 \, c\right )} - 118 \, e^{\left (2 \, d x + 2 \, c\right )} + 60 i \, e^{\left (d x + c\right )} + 9\right )} \log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - 59 \, {\left (9 \, e^{\left (4 \, d x + 4 \, c\right )} - 60 i \, e^{\left (3 \, d x + 3 \, c\right )} - 118 \, e^{\left (2 \, d x + 2 \, c\right )} + 60 i \, e^{\left (d x + c\right )} + 9\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 1416 i \, e^{\left (3 \, d x + 3 \, c\right )} - 7080 \, e^{\left (2 \, d x + 2 \, c\right )} + 5784 i \, e^{\left (d x + c\right )} + 1080}{2048 \, {\left (9 \, d e^{\left (4 \, d x + 4 \, c\right )} - 60 i \, d e^{\left (3 \, d x + 3 \, c\right )} - 118 \, d e^{\left (2 \, d x + 2 \, c\right )} + 60 i \, d e^{\left (d x + c\right )} + 9 \, d\right )}} \] Input:

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

Output:

1/2048*(59*(9*e^(4*d*x + 4*c) - 60*I*e^(3*d*x + 3*c) - 118*e^(2*d*x + 2*c) 
 + 60*I*e^(d*x + c) + 9)*log(e^(d*x + c) - 1/3*I) - 59*(9*e^(4*d*x + 4*c) 
- 60*I*e^(3*d*x + 3*c) - 118*e^(2*d*x + 2*c) + 60*I*e^(d*x + c) + 9)*log(e 
^(d*x + c) - 3*I) - 1416*I*e^(3*d*x + 3*c) - 7080*e^(2*d*x + 2*c) + 5784*I 
*e^(d*x + c) + 1080)/(9*d*e^(4*d*x + 4*c) - 60*I*d*e^(3*d*x + 3*c) - 118*d 
*e^(2*d*x + 2*c) + 60*I*d*e^(d*x + c) + 9*d)
 

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=\frac {- 177 i e^{3 c} e^{3 d x} - 885 e^{2 c} e^{2 d x} + 723 i e^{c} e^{d x} + 135}{2304 d e^{4 c} e^{4 d x} - 15360 i d e^{3 c} e^{3 d x} - 30208 d e^{2 c} e^{2 d x} + 15360 i d e^{c} e^{d x} + 2304 d} + \frac {- \frac {59 \log {\left (e^{d x} - 3 i e^{- c} \right )}}{2048} + \frac {59 \log {\left (e^{d x} - \frac {i e^{- c}}{3} \right )}}{2048}}{d} \] Input:

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

Output:

(-177*I*exp(3*c)*exp(3*d*x) - 885*exp(2*c)*exp(2*d*x) + 723*I*exp(c)*exp(d 
*x) + 135)/(2304*d*exp(4*c)*exp(4*d*x) - 15360*I*d*exp(3*c)*exp(3*d*x) - 3 
0208*d*exp(2*c)*exp(2*d*x) + 15360*I*d*exp(c)*exp(d*x) + 2304*d) + (-59*lo 
g(exp(d*x) - 3*I*exp(-c))/2048 + 59*log(exp(d*x) - I*exp(-c)/3)/2048)/d
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=-\frac {59 i \, \arctan \left (\frac {3}{4} \, e^{\left (-d x - c\right )} + \frac {5}{4} i\right )}{1024 \, d} + \frac {3 \, {\left (241 i \, e^{\left (-d x - c\right )} + 295 \, e^{\left (-2 \, d x - 2 \, c\right )} - 59 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 45\right )}}{-256 \, d {\left (60 i \, e^{\left (-d x - c\right )} + 118 \, e^{\left (-2 \, d x - 2 \, c\right )} - 60 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} - 9\right )}} \] Input:

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

Output:

-59/1024*I*arctan(3/4*e^(-d*x - c) + 5/4*I)/d + 3*(241*I*e^(-d*x - c) + 29 
5*e^(-2*d*x - 2*c) - 59*I*e^(-3*d*x - 3*c) - 45)/(d*(-15360*I*e^(-d*x - c) 
 - 30208*e^(-2*d*x - 2*c) + 15360*I*e^(-3*d*x - 3*c) + 2304*e^(-4*d*x - 4* 
c) + 2304))
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=-\frac {\frac {24 \, {\left (-59 i \, e^{\left (3 \, d x + 3 \, c\right )} - 295 \, e^{\left (2 \, d x + 2 \, c\right )} + 241 i \, e^{\left (d x + c\right )} + 45\right )}}{{\left (-3 i \, e^{\left (2 \, d x + 2 \, c\right )} - 10 \, e^{\left (d x + c\right )} + 3 i\right )}^{2}} - 59 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right ) + 59 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{2048 \, d} \] Input:

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

Output:

-1/2048*(24*(-59*I*e^(3*d*x + 3*c) - 295*e^(2*d*x + 2*c) + 241*I*e^(d*x + 
c) + 45)/(-3*I*e^(2*d*x + 2*c) - 10*e^(d*x + c) + 3*I)^2 - 59*log(3*e^(d*x 
 + c) - I) + 59*log(e^(d*x + c) - 3*I))/d
 

Mupad [B] (verification not implemented)

Time = 2.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=\frac {\frac {295}{2304\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,59{}\mathrm {i}}{768\,d}}{1-{\mathrm {e}}^{2\,c+2\,d\,x}+\frac {{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}}{3}}-\frac {59\,\ln \left (-\frac {59\,{\mathrm {e}}^{c+d\,x}}{4}+\frac {177}{4}{}\mathrm {i}\right )}{2048\,d}+\frac {59\,\ln \left (\frac {531\,{\mathrm {e}}^{c+d\,x}}{4}-\frac {177}{4}{}\mathrm {i}\right )}{2048\,d}-\frac {\frac {5}{72\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,41{}\mathrm {i}}{216\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-\frac {118\,{\mathrm {e}}^{2\,c+2\,d\,x}}{9}+1+\frac {{\mathrm {e}}^{c+d\,x}\,20{}\mathrm {i}}{3}-\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,20{}\mathrm {i}}{3}} \] Input:

int(1/(sinh(c + d*x)*3i + 5)^3,x)
                                                                                    
                                                                                    
 

Output:

((exp(c + d*x)*59i)/(768*d) + 295/(2304*d))/((exp(c + d*x)*10i)/3 - exp(2* 
c + 2*d*x) + 1) - (59*log(177i/4 - (59*exp(c + d*x))/4))/(2048*d) + (59*lo 
g((531*exp(c + d*x))/4 - 177i/4))/(2048*d) - ((exp(c + d*x)*41i)/(216*d) + 
 5/(72*d))/((exp(c + d*x)*20i)/3 - (118*exp(2*c + 2*d*x))/9 - (exp(3*c + 3 
*d*x)*20i)/3 + exp(4*c + 4*d*x) + 1)
 

Reduce [F]

\[ \int \frac {1}{(5+3 i \sinh (c+d x))^3} \, dx=-\left (\int \frac {1}{27 \sinh \left (d x +c \right )^{3} i +135 \sinh \left (d x +c \right )^{2}-225 \sinh \left (d x +c \right ) i -125}d x \right ) \] Input:

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

Output:

 - int(1/(27*sinh(c + d*x)**3*i + 135*sinh(c + d*x)**2 - 225*sinh(c + d*x) 
*i - 125),x)