\(\int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx\) [90]

Optimal result

Integrand size = 14, antiderivative size = 111 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=-\frac {43 i \log \left (i-3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {43 i \log \left (3 i-\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac {45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))} \] Output:

-43/2048*I*ln(I-3*tanh(1/2*d*x+1/2*c))/d+43/2048*I*ln(3*I-tanh(1/2*d*x+1/2 
*c))/d+5/32*I*cosh(d*x+c)/d/(3+5*I*sinh(d*x+c))^2-45/512*I*cosh(d*x+c)/d/( 
3+5*I*sinh(d*x+c))
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.84 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=\frac {86 \arctan \left (3 \coth \left (\frac {1}{2} (c+d x)\right )\right )+86 \arctan \left (3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )-43 i \log (4-5 \cosh (c+d x))+43 i \log (4+5 \cosh (c+d x))-\frac {80 i}{\left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {80 i}{\left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}+\left (-\frac {120}{3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )}-\frac {360}{\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )}\right ) \sinh \left (\frac {1}{2} (c+d x)\right )}{4096 d} \] Input:

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

Output:

(86*ArcTan[3*Coth[(c + d*x)/2]] + 86*ArcTan[3*Tanh[(c + d*x)/2]] - (43*I)* 
Log[4 - 5*Cosh[c + d*x]] + (43*I)*Log[4 + 5*Cosh[c + d*x]] - (80*I)/(3*Cos 
h[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2 + (80*I)/(Cosh[(c + d*x)/2] + (3*I 
)*Sinh[(c + d*x)/2])^2 + (-120/(3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) 
 - 360/(Cosh[(c + d*x)/2] + (3*I)*Sinh[(c + d*x)/2]))*Sinh[(c + d*x)/2])/( 
4096*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3143, 25, 3042, 3233, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3143

\(\displaystyle \frac {1}{32} \int -\frac {6-5 i \sinh (c+d x)}{(5 i \sinh (c+d x)+3)^2}dx+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac {1}{32} \int \frac {6-5 i \sinh (c+d x)}{(5 i \sinh (c+d x)+3)^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac {1}{32} \int \frac {6-5 \sin (i c+i d x)}{(5 \sin (i c+i d x)+3)^2}dx\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {1}{32} \left (-\frac {1}{16} \int -\frac {43}{5 i \sinh (c+d x)+3}dx-\frac {45 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\right )+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{32} \left (\frac {43}{16} \int \frac {1}{5 i \sinh (c+d x)+3}dx-\frac {45 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\right )+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{32} \left (\frac {43}{16} \int \frac {1}{5 \sin (i c+i d x)+3}dx-\frac {45 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\right )+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {1}{32} \left (-\frac {43 i \int \frac {1}{-3 \tanh ^2\left (\frac {1}{2} (c+d x)\right )+10 i \tanh \left (\frac {1}{2} (c+d x)\right )+3}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {45 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\right )+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 1081

\(\displaystyle \frac {1}{32} \left (-\frac {129 i \int \left (\frac {1}{8 \left (3 i \tanh \left (\frac {1}{2} (c+d x)\right )+1\right )}-\frac {1}{24 \left (i \tanh \left (\frac {1}{2} (c+d x)\right )+3\right )}\right )d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {45 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\right )+\frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}+\frac {1}{32} \left (-\frac {129 i \left (\frac {1}{24} \log \left (1+3 i \tanh \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{24} \log \left (3+i \tanh \left (\frac {1}{2} (c+d x)\right )\right )\right )}{8 d}-\frac {45 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\right )\)

Input:

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

Output:

((((-129*I)/8)*(-1/24*Log[3 + I*Tanh[(c + d*x)/2]] + Log[1 + (3*I)*Tanh[(c 
 + d*x)/2]]/24))/d - (((45*I)/16)*Cosh[c + d*x])/(d*(3 + (5*I)*Sinh[c + d* 
x])))/32 + (((5*I)/32)*Cosh[c + d*x])/(d*(3 + (5*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 1081

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3139

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]
 

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 = 0.57 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {i \left (-387 i {\mathrm e}^{2 d x +2 c}+215 \,{\mathrm e}^{3 d x +3 c}+225 i-325 \,{\mathrm e}^{d x +c}\right )}{256 d \left (5 \,{\mathrm e}^{2 d x +2 c}-5-6 i {\mathrm e}^{d x +c}\right )^{2}}-\frac {43 i \ln \left ({\mathrm e}^{d x +c}-\frac {4}{5}-\frac {3 i}{5}\right )}{2048 d}+\frac {43 i \ln \left ({\mathrm e}^{d x +c}+\frac {4}{5}-\frac {3 i}{5}\right )}{2048 d}\)	\(100\)
derivativedivides	\(\frac {\frac {25 i}{128 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )^{2}}+\frac {43 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{2048}+\frac {15}{512 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}-\frac {43 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{2048}-\frac {25 i}{1152 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {155}{4608 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d}\)	\(110\)
default	\(\frac {\frac {25 i}{128 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )^{2}}+\frac {43 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{2048}+\frac {15}{512 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}-\frac {43 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{2048}-\frac {25 i}{1152 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {155}{4608 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d}\)	\(110\)
parallelrisch	\(\frac {9460-6966 \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-9 i\right )+6966 \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+1935 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {i}{3}\right ) \left (5-5 \cosh \left (2 d x +2 c \right )+12 i \sinh \left (d x +c \right )\right )+1935 \left (-5+5 \cosh \left (2 d x +2 c \right )-12 i \sinh \left (d x +c \right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )+3960 \cosh \left (d x +c \right )-5500 \cosh \left (2 d x +2 c \right )+13200 i \sinh \left (d x +c \right )+8100 i \sinh \left (2 d x +2 c \right )}{18432 \left (43 i-25 i \cosh \left (2 d x +2 c \right )-60 \sinh \left (d x +c \right )\right ) d}\)	\(181\)

Input:

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

Output:

-1/256*I*(-387*I*exp(2*d*x+2*c)+215*exp(3*d*x+3*c)+225*I-325*exp(d*x+c))/d 
/(5*exp(2*d*x+2*c)-5-6*I*exp(d*x+c))^2-43/2048*I/d*ln(exp(d*x+c)-4/5-3/5*I 
)+43/2048*I/d*ln(exp(d*x+c)+4/5-3/5*I)
 

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. \(2 (85) = 170\).

Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.74 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=-\frac {43 \, {\left (-25 i \, e^{\left (4 \, d x + 4 \, c\right )} - 60 \, e^{\left (3 \, d x + 3 \, c\right )} + 86 i \, e^{\left (2 \, d x + 2 \, c\right )} + 60 \, e^{\left (d x + c\right )} - 25 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) + 43 \, {\left (25 i \, e^{\left (4 \, d x + 4 \, c\right )} + 60 \, e^{\left (3 \, d x + 3 \, c\right )} - 86 i \, e^{\left (2 \, d x + 2 \, c\right )} - 60 \, e^{\left (d x + c\right )} + 25 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i - \frac {4}{5}\right ) + 1720 i \, e^{\left (3 \, d x + 3 \, c\right )} + 3096 \, e^{\left (2 \, d x + 2 \, c\right )} - 2600 i \, e^{\left (d x + c\right )} - 1800}{2048 \, {\left (25 \, d e^{\left (4 \, d x + 4 \, c\right )} - 60 i \, d e^{\left (3 \, d x + 3 \, c\right )} - 86 \, d e^{\left (2 \, d x + 2 \, c\right )} + 60 i \, d e^{\left (d x + c\right )} + 25 \, d\right )}} \] Input:

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

Output:

-1/2048*(43*(-25*I*e^(4*d*x + 4*c) - 60*e^(3*d*x + 3*c) + 86*I*e^(2*d*x + 
2*c) + 60*e^(d*x + c) - 25*I)*log(e^(d*x + c) - 3/5*I + 4/5) + 43*(25*I*e^ 
(4*d*x + 4*c) + 60*e^(3*d*x + 3*c) - 86*I*e^(2*d*x + 2*c) - 60*e^(d*x + c) 
 + 25*I)*log(e^(d*x + c) - 3/5*I - 4/5) + 1720*I*e^(3*d*x + 3*c) + 3096*e^ 
(2*d*x + 2*c) - 2600*I*e^(d*x + c) - 1800)/(25*d*e^(4*d*x + 4*c) - 60*I*d* 
e^(3*d*x + 3*c) - 86*d*e^(2*d*x + 2*c) + 60*I*d*e^(d*x + c) + 25*d)
 

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=\frac {- 215 i e^{3 c} e^{3 d x} - 387 e^{2 c} e^{2 d x} + 325 i e^{c} e^{d x} + 225}{6400 d e^{4 c} e^{4 d x} - 15360 i d e^{3 c} e^{3 d x} - 22016 d e^{2 c} e^{2 d x} + 15360 i d e^{c} e^{d x} + 6400 d} + \frac {\operatorname {RootSum} {\left (4194304 z^{2} + 1849, \left ( i \mapsto i \log {\left (\frac {\left (- 8192 i i - 129 i\right ) e^{- c}}{215} + e^{d x} \right )} \right )\right )}}{d} \] Input:

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

Output:

(-215*I*exp(3*c)*exp(3*d*x) - 387*exp(2*c)*exp(2*d*x) + 325*I*exp(c)*exp(d 
*x) + 225)/(6400*d*exp(4*c)*exp(4*d*x) - 15360*I*d*exp(3*c)*exp(3*d*x) - 2 
2016*d*exp(2*c)*exp(2*d*x) + 15360*I*d*exp(c)*exp(d*x) + 6400*d) + RootSum 
(4194304*_z**2 + 1849, Lambda(_i, _i*log((-8192*_i*I - 129*I)*exp(-c)/215 
+ exp(d*x))))/d
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=-\frac {43 i \, \log \left (\frac {5 \, e^{\left (-d x - c\right )} + 3 i - 4}{5 \, e^{\left (-d x - c\right )} + 3 i + 4}\right )}{2048 \, d} - \frac {-325 i \, e^{\left (-d x - c\right )} - 387 \, e^{\left (-2 \, d x - 2 \, c\right )} + 215 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 225}{-256 \, d {\left (60 i \, e^{\left (-d x - c\right )} + 86 \, e^{\left (-2 \, d x - 2 \, c\right )} - 60 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 25 \, e^{\left (-4 \, d x - 4 \, c\right )} - 25\right )}} \] Input:

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

Output:

-43/2048*I*log((5*e^(-d*x - c) + 3*I - 4)/(5*e^(-d*x - c) + 3*I + 4))/d - 
(-325*I*e^(-d*x - c) - 387*e^(-2*d*x - 2*c) + 215*I*e^(-3*d*x - 3*c) + 225 
)/(d*(-15360*I*e^(-d*x - c) - 22016*e^(-2*d*x - 2*c) + 15360*I*e^(-3*d*x - 
 3*c) + 6400*e^(-4*d*x - 4*c) + 6400))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=-\frac {\frac {8 \, {\left (-215 i \, e^{\left (3 \, d x + 3 \, c\right )} - 387 \, e^{\left (2 \, d x + 2 \, c\right )} + 325 i \, e^{\left (d x + c\right )} + 225\right )}}{{\left (-5 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 5 i\right )}^{2}} - 43 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) + 43 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{2048 \, d} \] Input:

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

Output:

-1/2048*(8*(-215*I*e^(3*d*x + 3*c) - 387*e^(2*d*x + 2*c) + 325*I*e^(d*x + 
c) + 225)/(-5*I*e^(2*d*x + 2*c) - 6*e^(d*x + c) + 5*I)^2 - 43*I*log(-(I - 
2)*e^(d*x + c) - 2*I + 1) + 43*I*log(-(2*I - 1)*e^(d*x + c) + I - 2))/d
 

Mupad [B] (verification not implemented)

Time = 2.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx=\frac {\frac {129}{6400\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,43{}\mathrm {i}}{1280\,d}}{1-{\mathrm {e}}^{2\,c+2\,d\,x}+\frac {{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}}{5}}-\frac {\ln \left (-\frac {215}{4}+{\mathrm {e}}^{c+d\,x}\,\left (43-\frac {129}{4}{}\mathrm {i}\right )\right )\,43{}\mathrm {i}}{2048\,d}+\frac {\ln \left (\frac {215}{4}+{\mathrm {e}}^{c+d\,x}\,\left (43+\frac {129}{4}{}\mathrm {i}\right )\right )\,43{}\mathrm {i}}{2048\,d}-\frac {-\frac {3}{200\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,7{}\mathrm {i}}{1000\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-\frac {86\,{\mathrm {e}}^{2\,c+2\,d\,x}}{25}+1+\frac {{\mathrm {e}}^{c+d\,x}\,12{}\mathrm {i}}{5}-\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,12{}\mathrm {i}}{5}} \] Input:

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

Output:

((exp(c + d*x)*43i)/(1280*d) + 129/(6400*d))/((exp(c + d*x)*6i)/5 - exp(2* 
c + 2*d*x) + 1) - (log(exp(c + d*x)*(43 - 129i/4) - 215/4)*43i)/(2048*d) + 
 (log(exp(c + d*x)*(43 + 129i/4) + 215/4)*43i)/(2048*d) - ((exp(c + d*x)*7 
i)/(1000*d) - 3/(200*d))/((exp(c + d*x)*12i)/5 - (86*exp(2*c + 2*d*x))/25 
- (exp(3*c + 3*d*x)*12i)/5 + exp(4*c + 4*d*x) + 1)
 

Reduce [F]

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

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

Output:

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