3.57 \(\int \frac {1}{(1+i \sinh (c+d x))^2} \, dx\)

Optimal. Leaf size=59 \[ \frac {i \cosh (c+d x)}{3 d (1+i \sinh (c+d x))}+\frac {i \cosh (c+d x)}{3 d (1+i \sinh (c+d x))^2} \]

[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]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2650, 2648} \[ \frac {i \cosh (c+d x)}{3 d (1+i \sinh (c+d x))}+\frac {i \cosh (c+d x)}{3 d (1+i \sinh (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(1 + I*Sinh[c + d*x])^(-2),x]

[Out]

((I/3)*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 2648

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 2650

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*} \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 {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]  time = 0.10, size = 61, normalized size = 1.03 \[ \frac {-4 \sinh (c+d x)+\sinh (2 (c+d x))-4 i \cosh (c+d x)-i \cosh (2 (c+d x))+3 i}{6 d (\sinh (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*I - (4*I)*Cosh[c + d*x] - I*Cosh[2*(c + d*x)] - 4*Sinh[c + d*x] + Sinh[2*(c + d*x)])/(6*d*(-I + Sinh[c + d*
x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 50, normalized size = 0.85 \[ \frac {6 \, e^{\left (d x + c\right )} - 2 i}{3 \, d e^{\left (3 \, d x + 3 \, c\right )} - 9 i \, d e^{\left (2 \, d x + 2 \, c\right )} - 9 \, d e^{\left (d x + c\right )} + 3 i \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.34, size = 25, normalized size = 0.42 \[ \frac {6 \, e^{\left (d x + c\right )} - 2 i}{3 \, d {\left (e^{\left (d x + c\right )} - i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*e^(d*x + c) - 2*I)/(d*(e^(d*x + c) - I)^3)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 55, normalized size = 0.93 \[ \frac {\frac {2 i}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4}{3 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*sinh(d*x+c))^2,x)

[Out]

1/d*(2*I/(-I+tanh(1/2*d*x+1/2*c))^2-4/3/(-I+tanh(1/2*d*x+1/2*c))^3+2/(-I+tanh(1/2*d*x+1/2*c)))

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 94, normalized size = 1.59 \[ \frac {6 \, e^{\left (-d x - c\right )}}{d {\left (9 \, e^{\left (-d x - c\right )} - 9 i \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 i\right )}} + \frac {2 i}{d {\left (9 \, e^{\left (-d x - c\right )} - 9 i \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 i\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.53, size = 29, normalized size = 0.49 \[ -\frac {\frac {2}{3}+{\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}}{d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*1i + 1)^2,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.20, size = 68, normalized size = 1.15 \[ \frac {2 e^{3 c} - 6 i e^{2 c} e^{- d x}}{3 d e^{3 c} - 9 i d e^{2 c} e^{- d x} - 9 d e^{c} e^{- 2 d x} + 3 i d e^{- 3 d x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________