3.38 \(\int \frac {1}{(1-\cosh (c+d x))^3} \, dx\)

Optimal. Leaf size=76 \[ -\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2650, 2648} \[ -\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cosh[c + d*x])^(-3),x]

[Out]

-Sinh[c + d*x]/(5*d*(1 - Cosh[c + d*x])^3) - (2*Sinh[c + d*x])/(15*d*(1 - Cosh[c + d*x])^2) - (2*Sinh[c + d*x]
)/(15*d*(1 - Cosh[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-\cosh (c+d x))^3} \, dx &=-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}+\frac {2}{5} \int \frac {1}{(1-\cosh (c+d x))^2} \, dx\\ &=-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}+\frac {2}{15} \int \frac {1}{1-\cosh (c+d x)} \, dx\\ &=-\frac {\sinh (c+d x)}{5 d (1-\cosh (c+d x))^3}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))^2}-\frac {2 \sinh (c+d x)}{15 d (1-\cosh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 41, normalized size = 0.54 \[ \frac {\sinh (c+d x) (-6 \cosh (c+d x)+\cosh (2 (c+d x))+8)}{15 d (\cosh (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cosh[c + d*x])^(-3),x]

[Out]

((8 - 6*Cosh[c + d*x] + Cosh[2*(c + d*x)])*Sinh[c + d*x])/(15*d*(-1 + Cosh[c + d*x])^3)

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fricas [B]  time = 0.48, size = 174, normalized size = 2.29 \[ \frac {4 \, {\left (11 \, \cosh \left (d x + c\right ) + 9 \, \sinh \left (d x + c\right ) - 5\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{4} + d \sinh \left (d x + c\right )^{4} - 5 \, d \cosh \left (d x + c\right )^{3} + {\left (4 \, d \cosh \left (d x + c\right ) - 5 \, d\right )} \sinh \left (d x + c\right )^{3} + 10 \, d \cosh \left (d x + c\right )^{2} + {\left (6 \, d \cosh \left (d x + c\right )^{2} - 15 \, d \cosh \left (d x + c\right ) + 10 \, d\right )} \sinh \left (d x + c\right )^{2} - 11 \, d \cosh \left (d x + c\right ) + {\left (4 \, d \cosh \left (d x + c\right )^{3} - 15 \, d \cosh \left (d x + c\right )^{2} + 20 \, d \cosh \left (d x + c\right ) - 9 \, d\right )} \sinh \left (d x + c\right ) + 5 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(11*cosh(d*x + c) + 9*sinh(d*x + c) - 5)/(d*cosh(d*x + c)^4 + d*sinh(d*x + c)^4 - 5*d*cosh(d*x + c)^3 + (
4*d*cosh(d*x + c) - 5*d)*sinh(d*x + c)^3 + 10*d*cosh(d*x + c)^2 + (6*d*cosh(d*x + c)^2 - 15*d*cosh(d*x + c) +
10*d)*sinh(d*x + c)^2 - 11*d*cosh(d*x + c) + (4*d*cosh(d*x + c)^3 - 15*d*cosh(d*x + c)^2 + 20*d*cosh(d*x + c)
- 9*d)*sinh(d*x + c) + 5*d)

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giac [A]  time = 0.13, size = 36, normalized size = 0.47 \[ \frac {4 \, {\left (10 \, e^{\left (2 \, d x + 2 \, c\right )} - 5 \, e^{\left (d x + c\right )} + 1\right )}}{15 \, d {\left (e^{\left (d x + c\right )} - 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 45, normalized size = 0.59 \[ \frac {-\frac {1}{6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {1}{20 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cosh(d*x+c))^3,x)

[Out]

1/d*(-1/6/tanh(1/2*d*x+1/2*c)^3+1/4/tanh(1/2*d*x+1/2*c)+1/20/tanh(1/2*d*x+1/2*c)^5)

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maxima [B]  time = 0.41, size = 205, normalized size = 2.70 \[ \frac {4 \, e^{\left (-d x - c\right )}}{3 \, d {\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{3 \, d {\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} - \frac {4}{15 \, d {\left (5 \, e^{\left (-d x - c\right )} - 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} - 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*e^(-d*x - c)/(d*(5*e^(-d*x - c) - 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) - 5*e^(-4*d*x - 4*c) + e^(-5*d
*x - 5*c) - 1)) - 8/3*e^(-2*d*x - 2*c)/(d*(5*e^(-d*x - c) - 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) - 5*e^(-
4*d*x - 4*c) + e^(-5*d*x - 5*c) - 1)) - 4/15/(d*(5*e^(-d*x - c) - 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) -
5*e^(-4*d*x - 4*c) + e^(-5*d*x - 5*c) - 1))

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mupad [B]  time = 0.91, size = 36, normalized size = 0.47 \[ \frac {4\,\left (10\,{\mathrm {e}}^{2\,c+2\,d\,x}-5\,{\mathrm {e}}^{c+d\,x}+1\right )}{15\,d\,{\left ({\mathrm {e}}^{c+d\,x}-1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(cosh(c + d*x) - 1)^3,x)

[Out]

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

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sympy [A]  time = 2.52, size = 70, normalized size = 0.92 \[ \begin {cases} \tilde {\infty } x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\\frac {x}{\left (1 - \cosh {\relax (c )}\right )^{3}} & \text {for}\: d = 0 \\\frac {1}{4 d \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}} - \frac {1}{6 d \tanh ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}} + \frac {1}{20 d \tanh ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(d*x+c))**3,x)

[Out]

Piecewise((zoo*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x/(1 - cosh(c))**3, Eq(d, 0)), (1/(4*
d*tanh(c/2 + d*x/2)) - 1/(6*d*tanh(c/2 + d*x/2)**3) + 1/(20*d*tanh(c/2 + d*x/2)**5), True))

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