\(\int \frac {1}{(1-i \sinh (c+d x))^2} \, dx\) [61]

Optimal result

Integrand size = 14, antiderivative size = 59 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))^2}-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2729, 2727} \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))}-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))^2} \]

[In]

Int[(1 - I*Sinh[c + d*x])^(-2),x]

[Out]

((-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]))

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)}{3 d (1-i \sinh (c+d x))^2}+\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}-\frac {i \cosh (c+d x)}{3 d (1-i \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=-\frac {\cosh \left (\frac {3}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )}{3 d \left (i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

Integrate[(1 - I*Sinh[c + d*x])^(-2),x]

[Out]

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.47

method	result	size
risch	\(\frac {\frac {2 i}{3}+2 \,{\mathrm e}^{d x +c}}{d \left ({\mathrm e}^{d x +c}+i\right )^{3}}\)	\(28\)
derivativedivides	\(\frac {-\frac {4}{3 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{d}\)	\(55\)
default	\(\frac {-\frac {4}{3 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}}{d}\)	\(55\)
parallelrisch	\(\frac {6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4}{3 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+3 i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}\)	\(74\)

[In]

int(1/(1-I*sinh(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

2/3*(I+3*exp(d*x+c))/d/(exp(d*x+c)+I)^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=\frac {2 \, {\left (3 \, e^{\left (d x + c\right )} + i\right )}}{3 \, {\left (d e^{\left (3 \, d x + 3 \, c\right )} + 3 i \, d e^{\left (2 \, d x + 2 \, c\right )} - 3 \, d e^{\left (d x + c\right )} - i \, d\right )}} \]

[In]

integrate(1/(1-I*sinh(d*x+c))^2,x, algorithm="fricas")

[Out]

2/3*(3*e^(d*x + c) + I)/(d*e^(3*d*x + 3*c) + 3*I*d*e^(2*d*x + 2*c) - 3*d*e^(d*x + c) - I*d)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=\frac {6 e^{c} e^{d x} + 2 i}{3 d e^{3 c} e^{3 d x} + 9 i d e^{2 c} e^{2 d x} - 9 d e^{c} e^{d x} - 3 i d} \]

[In]

integrate(1/(1-I*sinh(d*x+c))**2,x)

[Out]

(6*exp(c)*exp(d*x) + 2*I)/(3*d*exp(3*c)*exp(3*d*x) + 9*I*d*exp(2*c)*exp(2*d*x) - 9*d*exp(c)*exp(d*x) - 3*I*d)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=\frac {2 \, e^{\left (-d x - c\right )}}{d {\left (3 \, e^{\left (-d x - c\right )} + 3 i \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-3 \, d x - 3 \, c\right )} - i\right )}} - \frac {2 i}{3 \, d {\left (3 \, e^{\left (-d x - c\right )} + 3 i \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-3 \, d x - 3 \, c\right )} - i\right )}} \]

[In]

integrate(1/(1-I*sinh(d*x+c))^2,x, algorithm="maxima")

[Out]

2*e^(-d*x - c)/(d*(3*e^(-d*x - c) + 3*I*e^(-2*d*x - 2*c) - e^(-3*d*x - 3*c) - I)) - 2/3*I/(d*(3*e^(-d*x - c) +
3*I*e^(-2*d*x - 2*c) - e^(-3*d*x - 3*c) - I))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.42 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=\frac {2 \, {\left (3 \, e^{\left (d x + c\right )} + i\right )}}{3 \, d {\left (e^{\left (d x + c\right )} + i\right )}^{3}} \]

[In]

integrate(1/(1-I*sinh(d*x+c))^2,x, algorithm="giac")

[Out]

2/3*(3*e^(d*x + c) + I)/(d*(e^(d*x + c) + I)^3)

Mupad [B] (verification not implemented)

Time = 1.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(1-i \sinh (c+d x))^2} \, dx=-\frac {2\,\left (-1+{\mathrm {e}}^{c+d\,x}\,3{}\mathrm {i}\right )}{3\,d\,{\left (-1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^3} \]

[In]

int(1/(sinh(c + d*x)*1i - 1)^2,x)

[Out]

-(2*(exp(c + d*x)*3i - 1))/(3*d*(exp(c + d*x)*1i - 1)^3)