3.1.62 \(\int \frac {1}{(1-i \sinh (c+d x))^3} \, dx\) [62]

Optimal. Leaf size=88 \[ -\frac {i \cosh (c+d x)}{5 d (1-i \sinh (c+d x))^3}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))^2}-\frac {2 i \cosh (c+d x)}{15 d (1-i \sinh (c+d x))} \]

[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*si
nh(d*x+c))

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Rubi [A]
time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2729, 2727} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - I*Sinh[c + d*x])^(-3),x]

[Out]

((-1/5*I)*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*} \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}{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*}

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Mathematica [A]
time = 0.09, size = 81, normalized size = 0.92 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [A]
time = 1.17, size = 88, normalized size = 1.00

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	\(\frac {\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8}{5 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{3 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{d}\)	\(88\)
default	\(\frac {\frac {2}{i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8}{5 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {4 i}{\left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{3 \left (i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{d}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(2/(I+tanh(1/2*d*x+1/2*c))+4*I/(I+tanh(1/2*d*x+1/2*c))^4+8/5/(I+tanh(1/2*d*x+1/2*c))^5-4*I/(I+tanh(1/2*d*x
+1/2*c))^2-16/3/(I+tanh(1/2*d*x+1/2*c))^3)

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Maxima [B] 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 = 0.32, size = 211, normalized size = 2.40 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]
time = 0.42, size = 85, normalized size = 0.97 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]
time = 0.18, size = 109, normalized size = 1.24 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 0.41, size = 36, normalized size = 0.41 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 0.63, size = 40, normalized size = 0.45 \begin {gather*} -\frac {4\,\left (10\,{\mathrm {e}}^{2\,c+2\,d\,x}-1+{\mathrm {e}}^{c+d\,x}\,5{}\mathrm {i}\right )}{15\,d\,{\left (-1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(sinh(c + d*x)*1i - 1)^3,x)

[Out]

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

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