\(\int \frac {1}{(1-i \sinh (c+d x))^4} \, dx\) [63]

Optimal result

Integrand size = 14, antiderivative size = 117 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}-\frac {3 i \cosh (c+d x)}{35 d (1-i \sinh (c+d x))^3}-\frac {2 i \cosh (c+d x)}{35 d (1-i \sinh (c+d x))^2}-\frac {2 i \cosh (c+d x)}{35 d (1-i \sinh (c+d x))} \] Output:

-1/7*I*cosh(d*x+c)/d/(1-I*sinh(d*x+c))^4-3/35*I*cosh(d*x+c)/d/(1-I*sinh(d* 
x+c))^3-2/35*I*cosh(d*x+c)/d/(1-I*sinh(d*x+c))^2-2/35*I*cosh(d*x+c)/d/(1-I 
*sinh(d*x+c))
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=\frac {-21 i \cosh \left (\frac {3}{2} (c+d x)\right )+i \cosh \left (\frac {7}{2} (c+d x)\right )+35 \sinh \left (\frac {1}{2} (c+d x)\right )-7 \sinh \left (\frac {5}{2} (c+d x)\right )}{70 d \left (\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^7} \] Input:

Integrate[(1 - I*Sinh[c + d*x])^(-4),x]
 

Output:

((-21*I)*Cosh[(3*(c + d*x))/2] + I*Cosh[(7*(c + d*x))/2] + 35*Sinh[(c + d* 
x)/2] - 7*Sinh[(5*(c + d*x))/2])/(70*d*(Cosh[(c + d*x)/2] - I*Sinh[(c + d* 
x)/2])^7)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3129, 3042, 3129, 3042, 3129, 3042, 3127}

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.

\(\displaystyle \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(1-\sin (i c+i d x))^4}dx\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {3}{7} \int \frac {1}{(1-i \sinh (c+d x))^3}dx-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{7} \int \frac {1}{(1-\sin (i c+i d x))^3}dx-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {3}{7} \left (\frac {2}{5} \int \frac {1}{(1-i \sinh (c+d x))^2}dx-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}\right )-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{7} \left (\frac {2}{5} \int \frac {1}{(1-\sin (i c+i d x))^2}dx-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}\right )-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {3}{7} \left (\frac {2}{5} \left (\frac {1}{3} \int \frac {1}{1-i \sinh (c+d x)}dx-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))^2}\right )-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}\right )-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{7} \left (\frac {2}{5} \left (\frac {1}{3} \int \frac {1}{1-\sin (i c+i d x)}dx-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))^2}\right )-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}\right )-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

\(\Big \downarrow \) 3127

\(\displaystyle \frac {3}{7} \left (\frac {2}{5} \left (-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))}-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))^2}\right )-\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}\right )-\frac {i \cosh (c+d x)}{7 d (1-i \sinh (c+d x))^4}\)

Input:

Int[(1 - I*Sinh[c + d*x])^(-4),x]
 

Output:

(3*((2*(((-1/3*I)*Cosh[c + d*x])/(d*(1 - I*Sinh[c + d*x])^2) - ((I/3)*Cosh 
[c + d*x])/(d*(1 - I*Sinh[c + d*x]))))/5 - ((I/5)*Cosh[c + d*x])/(d*(1 - I 
*Sinh[c + d*x])^3)))/7 - ((I/7)*Cosh[c + d*x])/(d*(1 - I*Sinh[c + d*x])^4)
 

Defintions of rubi rules used

rule 3042

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

rule 3127

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

rule 3129

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c 
+ d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n 
+ 1))   Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] 
&& EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.44

method	result	size
risch	\(-\frac {4 \left (-7 \,{\mathrm e}^{d x +c}+21 i {\mathrm e}^{2 d x +2 c}+35 \,{\mathrm e}^{3 d x +3 c}-i\right )}{35 d \left ({\mathrm e}^{d x +c}+i\right )^{7}}\)	\(51\)
derivativedivides	\(\frac {-\frac {6 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {12}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {16}{7 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {72}{5 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {16 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}}{d}\)	\(121\)
default	\(\frac {-\frac {6 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {12}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {16}{7 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {72}{5 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {16 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {8 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}}{d}\)	\(121\)
parallelrisch	\(\frac {-\frac {12}{35}-\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}+\frac {12 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35}-\frac {6 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5}+\frac {2 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{d \left (7 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-35 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-21 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+21 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+35 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-i-7 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(169\)

Input:

int(1/(-I*sinh(d*x+c)+1)^4,x,method=_RETURNVERBOSE)
 

Output:

-4/35*(-7*exp(d*x+c)+21*I*exp(2*d*x+2*c)+35*exp(3*d*x+3*c)-I)/d/(exp(d*x+c 
)+I)^7
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=-\frac {4 \, {\left (35 \, e^{\left (3 \, d x + 3 \, c\right )} + 21 i \, e^{\left (2 \, d x + 2 \, c\right )} - 7 \, e^{\left (d x + c\right )} - i\right )}}{35 \, {\left (d e^{\left (7 \, d x + 7 \, c\right )} + 7 i \, d e^{\left (6 \, d x + 6 \, c\right )} - 21 \, d e^{\left (5 \, d x + 5 \, c\right )} - 35 i \, d e^{\left (4 \, d x + 4 \, c\right )} + 35 \, d e^{\left (3 \, d x + 3 \, c\right )} + 21 i \, d e^{\left (2 \, d x + 2 \, c\right )} - 7 \, d e^{\left (d x + c\right )} - i \, d\right )}} \] Input:

integrate(1/(1-I*sinh(d*x+c))^4,x, algorithm="fricas")
 

Output:

-4/35*(35*e^(3*d*x + 3*c) + 21*I*e^(2*d*x + 2*c) - 7*e^(d*x + c) - I)/(d*e 
^(7*d*x + 7*c) + 7*I*d*e^(6*d*x + 6*c) - 21*d*e^(5*d*x + 5*c) - 35*I*d*e^( 
4*d*x + 4*c) + 35*d*e^(3*d*x + 3*c) + 21*I*d*e^(2*d*x + 2*c) - 7*d*e^(d*x 
+ c) - I*d)
 

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=\frac {- 140 e^{3 c} e^{3 d x} - 84 i e^{2 c} e^{2 d x} + 28 e^{c} e^{d x} + 4 i}{35 d e^{7 c} e^{7 d x} + 245 i d e^{6 c} e^{6 d x} - 735 d e^{5 c} e^{5 d x} - 1225 i d e^{4 c} e^{4 d x} + 1225 d e^{3 c} e^{3 d x} + 735 i d e^{2 c} e^{2 d x} - 245 d e^{c} e^{d x} - 35 i d} \] Input:

integrate(1/(1-I*sinh(d*x+c))**4,x)
 

Output:

(-140*exp(3*c)*exp(3*d*x) - 84*I*exp(2*c)*exp(2*d*x) + 28*exp(c)*exp(d*x) 
+ 4*I)/(35*d*exp(7*c)*exp(7*d*x) + 245*I*d*exp(6*c)*exp(6*d*x) - 735*d*exp 
(5*c)*exp(5*d*x) - 1225*I*d*exp(4*c)*exp(4*d*x) + 1225*d*exp(3*c)*exp(3*d* 
x) + 735*I*d*exp(2*c)*exp(2*d*x) - 245*d*exp(c)*exp(d*x) - 35*I*d)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (93) = 186\).

Time = 0.05 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.18 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=\frac {4 \, e^{\left (-d x - c\right )}}{5 \, d {\left (7 \, e^{\left (-d x - c\right )} + 21 i \, e^{\left (-2 \, d x - 2 \, c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} - 35 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 21 \, e^{\left (-5 \, d x - 5 \, c\right )} + 7 i \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-7 \, d x - 7 \, c\right )} - i\right )}} + \frac {12 i \, e^{\left (-2 \, d x - 2 \, c\right )}}{5 \, d {\left (7 \, e^{\left (-d x - c\right )} + 21 i \, e^{\left (-2 \, d x - 2 \, c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} - 35 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 21 \, e^{\left (-5 \, d x - 5 \, c\right )} + 7 i \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-7 \, d x - 7 \, c\right )} - i\right )}} - \frac {4 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (7 \, e^{\left (-d x - c\right )} + 21 i \, e^{\left (-2 \, d x - 2 \, c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} - 35 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 21 \, e^{\left (-5 \, d x - 5 \, c\right )} + 7 i \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-7 \, d x - 7 \, c\right )} - i\right )}} - \frac {4 i}{35 \, d {\left (7 \, e^{\left (-d x - c\right )} + 21 i \, e^{\left (-2 \, d x - 2 \, c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} - 35 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 21 \, e^{\left (-5 \, d x - 5 \, c\right )} + 7 i \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-7 \, d x - 7 \, c\right )} - i\right )}} \] Input:

integrate(1/(1-I*sinh(d*x+c))^4,x, algorithm="maxima")
 

Output:

4/5*e^(-d*x - c)/(d*(7*e^(-d*x - c) + 21*I*e^(-2*d*x - 2*c) - 35*e^(-3*d*x 
 - 3*c) - 35*I*e^(-4*d*x - 4*c) + 21*e^(-5*d*x - 5*c) + 7*I*e^(-6*d*x - 6* 
c) - e^(-7*d*x - 7*c) - I)) + 12/5*I*e^(-2*d*x - 2*c)/(d*(7*e^(-d*x - c) + 
 21*I*e^(-2*d*x - 2*c) - 35*e^(-3*d*x - 3*c) - 35*I*e^(-4*d*x - 4*c) + 21* 
e^(-5*d*x - 5*c) + 7*I*e^(-6*d*x - 6*c) - e^(-7*d*x - 7*c) - I)) - 4*e^(-3 
*d*x - 3*c)/(d*(7*e^(-d*x - c) + 21*I*e^(-2*d*x - 2*c) - 35*e^(-3*d*x - 3* 
c) - 35*I*e^(-4*d*x - 4*c) + 21*e^(-5*d*x - 5*c) + 7*I*e^(-6*d*x - 6*c) - 
e^(-7*d*x - 7*c) - I)) - 4/35*I/(d*(7*e^(-d*x - c) + 21*I*e^(-2*d*x - 2*c) 
 - 35*e^(-3*d*x - 3*c) - 35*I*e^(-4*d*x - 4*c) + 21*e^(-5*d*x - 5*c) + 7*I 
*e^(-6*d*x - 6*c) - e^(-7*d*x - 7*c) - I))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.40 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=-\frac {4 \, {\left (35 \, e^{\left (3 \, d x + 3 \, c\right )} + 21 i \, e^{\left (2 \, d x + 2 \, c\right )} - 7 \, e^{\left (d x + c\right )} - i\right )}}{35 \, d {\left (e^{\left (d x + c\right )} + i\right )}^{7}} \] Input:

integrate(1/(1-I*sinh(d*x+c))^4,x, algorithm="giac")
 

Output:

-4/35*(35*e^(3*d*x + 3*c) + 21*I*e^(2*d*x + 2*c) - 7*e^(d*x + c) - I)/(d*( 
e^(d*x + c) + I)^7)
 

Mupad [B] (verification not implemented)

Time = 2.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=-\frac {4\,\left (21\,{\mathrm {e}}^{2\,c+2\,d\,x}-1+{\mathrm {e}}^{c+d\,x}\,7{}\mathrm {i}-{\mathrm {e}}^{3\,c+3\,d\,x}\,35{}\mathrm {i}\right )}{35\,d\,{\left (-1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^7} \] Input:

int(1/(sinh(c + d*x)*1i - 1)^4,x)
 

Output:

-(4*(exp(c + d*x)*7i + 21*exp(2*c + 2*d*x) - exp(3*c + 3*d*x)*35i - 1))/(3 
5*d*(exp(c + d*x)*1i - 1)^7)
 

Reduce [F]

\[ \int \frac {1}{(1-i \sinh (c+d x))^4} \, dx=\int \frac {1}{\sinh \left (d x +c \right )^{4}+4 \sinh \left (d x +c \right )^{3} i -6 \sinh \left (d x +c \right )^{2}-4 \sinh \left (d x +c \right ) i +1}d x \] Input:

int(1/(1-I*sinh(d*x+c))^4,x)
 

Output:

int(1/(sinh(c + d*x)**4 + 4*sinh(c + d*x)**3*i - 6*sinh(c + d*x)**2 - 4*si 
nh(c + d*x)*i + 1),x)