\(\int \frac {1}{(1+i \sinh (c+d x))^3} \, dx\) [58]

Optimal result

Integrand size = 14, antiderivative size = 88 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=\frac {i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3}+\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))} \]

[Out]

1/5*I*cosh(d*x+c)/d/(1+I*sinh(d*x+c))^3+2/15*I*cosh(d*x+c)/d/(1+I*sinh(d*x+c))^2+2/15*I*cosh(d*x+c)/d/(1+I*sin
h(d*x+c))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2729, 2727} \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))}+\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac {i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3} \]

[In]

Int[(1 + I*Sinh[c + d*x])^(-3),x]

[Out]

((I/5)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x])^3) + (((2*I)/15)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x])^2) + (
((2*I)/15)*Cosh[c + d*x])/(d*(1 + I*Sinh[c + d*x]))

Rule 2727

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 2729

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3}+\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}+\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac {2}{15} \int \frac {1}{1+i \sinh (c+d x)} \, dx \\ & = \frac {i \cosh (c+d x)}{5 d (1+i \sinh (c+d x))^3}+\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))^2}+\frac {2 i \cosh (c+d x)}{15 d (1+i \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=\frac {10-15 \cosh (c+d x)-6 \cosh (2 (c+d x))+\cosh (3 (c+d x))+15 i \sinh (c+d x)-6 i \sinh (2 (c+d x))-i \sinh (3 (c+d x))}{30 d (-i+\sinh (c+d x))^3} \]

[In]

Integrate[(1 + I*Sinh[c + d*x])^(-3),x]

[Out]

(10 - 15*Cosh[c + d*x] - 6*Cosh[2*(c + d*x)] + Cosh[3*(c + d*x)] + (15*I)*Sinh[c + d*x] - (6*I)*Sinh[2*(c + d*
x)] - I*Sinh[3*(c + d*x)])/(30*d*(-I + Sinh[c + d*x])^3)

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45

method	result	size
risch	\(-\frac {4 i \left (-5 i {\mathrm e}^{d x +c}+10 \,{\mathrm e}^{2 d x +2 c}-1\right )}{15 d \left ({\mathrm e}^{d x +c}-i\right )^{5}}\)	\(40\)
derivativedivides	\(\frac {\frac {8}{5 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {4 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16}{3 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(88\)
default	\(\frac {\frac {8}{5 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {4 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16}{3 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(88\)
parallelrisch	\(\frac {\frac {14}{15}-\frac {16 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {8 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}}{d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-5 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-10 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+10 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}\)	\(128\)

[In]

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

[Out]

-4/15*I*(-5*I*exp(d*x+c)+10*exp(2*d*x+2*c)-1)/d/(exp(d*x+c)-I)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=-\frac {4 \, {\left (10 i \, e^{\left (2 \, d x + 2 \, c\right )} + 5 \, e^{\left (d x + c\right )} - i\right )}}{15 \, {\left (d e^{\left (5 \, d x + 5 \, c\right )} - 5 i \, d e^{\left (4 \, d x + 4 \, c\right )} - 10 \, d e^{\left (3 \, d x + 3 \, c\right )} + 10 i \, d e^{\left (2 \, d x + 2 \, c\right )} + 5 \, d e^{\left (d x + c\right )} - i \, d\right )}} \]

[In]

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

[Out]

-4/15*(10*I*e^(2*d*x + 2*c) + 5*e^(d*x + c) - I)/(d*e^(5*d*x + 5*c) - 5*I*d*e^(4*d*x + 4*c) - 10*d*e^(3*d*x +
3*c) + 10*I*d*e^(2*d*x + 2*c) + 5*d*e^(d*x + c) - I*d)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=\frac {- 40 i e^{2 c} e^{2 d x} - 20 e^{c} e^{d x} + 4 i}{15 d e^{5 c} e^{5 d x} - 75 i d e^{4 c} e^{4 d x} - 150 d e^{3 c} e^{3 d x} + 150 i d e^{2 c} e^{2 d x} + 75 d e^{c} e^{d x} - 15 i d} \]

[In]

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

[Out]

(-40*I*exp(2*c)*exp(2*d*x) - 20*exp(c)*exp(d*x) + 4*I)/(15*d*exp(5*c)*exp(5*d*x) - 75*I*d*exp(4*c)*exp(4*d*x)
- 150*d*exp(3*c)*exp(3*d*x) + 150*I*d*exp(2*c)*exp(2*d*x) + 75*d*exp(c)*exp(d*x) - 15*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. 211 vs. \(2 (70) = 140\).

Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.40 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=\frac {20 i \, e^{\left (-d x - c\right )}}{-15 \, d {\left (-5 i \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-5 \, d x - 5 \, c\right )} + 1\right )}} + \frac {40 \, e^{\left (-2 \, d x - 2 \, c\right )}}{-15 \, d {\left (-5 i \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-5 \, d x - 5 \, c\right )} + 1\right )}} - \frac {4}{-15 \, d {\left (-5 i \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-5 \, d x - 5 \, c\right )} + 1\right )}} \]

[In]

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

[Out]

20*I*e^(-d*x - c)/(d*(75*I*e^(-d*x - c) + 150*e^(-2*d*x - 2*c) - 150*I*e^(-3*d*x - 3*c) - 75*e^(-4*d*x - 4*c)
+ 15*I*e^(-5*d*x - 5*c) - 15)) + 40*e^(-2*d*x - 2*c)/(d*(75*I*e^(-d*x - c) + 150*e^(-2*d*x - 2*c) - 150*I*e^(-
3*d*x - 3*c) - 75*e^(-4*d*x - 4*c) + 15*I*e^(-5*d*x - 5*c) - 15)) - 4/(d*(75*I*e^(-d*x - c) + 150*e^(-2*d*x -
2*c) - 150*I*e^(-3*d*x - 3*c) - 75*e^(-4*d*x - 4*c) + 15*I*e^(-5*d*x - 5*c) - 15))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.41 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=-\frac {4 i \, {\left (10 \, e^{\left (2 \, d x + 2 \, c\right )} - 5 i \, e^{\left (d x + c\right )} - 1\right )}}{15 \, d {\left (e^{\left (d x + c\right )} - i\right )}^{5}} \]

[In]

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

[Out]

-4/15*I*(10*e^(2*d*x + 2*c) - 5*I*e^(d*x + c) - 1)/(d*(e^(d*x + c) - I)^5)

Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45 \[ \int \frac {1}{(1+i \sinh (c+d x))^3} \, dx=-\frac {\frac {4}{15}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3}+\frac {{\mathrm {e}}^{c+d\,x}\,4{}\mathrm {i}}{3}}{d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^5} \]

[In]

int(1/(sinh(c + d*x)*1i + 1)^3,x)

[Out]

-((exp(c + d*x)*4i)/3 - (8*exp(2*c + 2*d*x))/3 + 4/15)/(d*(exp(c + d*x)*1i + 1)^5)