3.149 \(\int \frac {\sinh ^2(x)}{(1-\cosh (x))^3} \, dx\)

Optimal. Leaf size=16 \[ -\frac {\sinh ^3(x)}{3 (1-\cosh (x))^3} \]

[Out]

-1/3*sinh(x)^3/(1-cosh(x))^3

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Rubi [A]  time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2671} \[ -\frac {\sinh ^3(x)}{3 (1-\cosh (x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(1 - Cosh[x])^3,x]

[Out]

-Sinh[x]^3/(3*(1 - Cosh[x])^3)

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\sinh ^2(x)}{(1-\cosh (x))^3} \, dx &=-\frac {\sinh ^3(x)}{3 (1-\cosh (x))^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 12, normalized size = 0.75 \[ \frac {1}{3} \coth ^3\left (\frac {x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(1 - Cosh[x])^3,x]

[Out]

Coth[x/2]^3/3

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fricas [B]  time = 3.35, size = 33, normalized size = 2.06 \[ \frac {4 \, {\left (2 \, \cosh \relax (x) + \sinh \relax (x)\right )}}{3 \, {\left (\cosh \relax (x)^{2} + 2 \, {\left (\cosh \relax (x) - 1\right )} \sinh \relax (x) + \sinh \relax (x)^{2} - 4 \, \cosh \relax (x) + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1-cosh(x))^3,x, algorithm="fricas")

[Out]

4/3*(2*cosh(x) + sinh(x))/(cosh(x)^2 + 2*(cosh(x) - 1)*sinh(x) + sinh(x)^2 - 4*cosh(x) + 3)

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giac [A]  time = 0.14, size = 16, normalized size = 1.00 \[ \frac {2 \, {\left (3 \, e^{\left (2 \, x\right )} + 1\right )}}{3 \, {\left (e^{x} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1-cosh(x))^3,x, algorithm="giac")

[Out]

2/3*(3*e^(2*x) + 1)/(e^x - 1)^3

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maple [A]  time = 0.06, size = 9, normalized size = 0.56 \[ \frac {1}{3 \tanh \left (\frac {x}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(1-cosh(x))^3,x)

[Out]

1/3/tanh(1/2*x)^3

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maxima [B]  time = 0.31, size = 49, normalized size = 3.06 \[ -\frac {2 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-x\right )} - 3 \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )} - 1} - \frac {2}{3 \, {\left (3 \, e^{\left (-x\right )} - 3 \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(1-cosh(x))^3,x, algorithm="maxima")

[Out]

-2*e^(-2*x)/(3*e^(-x) - 3*e^(-2*x) + e^(-3*x) - 1) - 2/3/(3*e^(-x) - 3*e^(-2*x) + e^(-3*x) - 1)

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mupad [B]  time = 0.92, size = 16, normalized size = 1.00 \[ \frac {2\,\left (3\,{\mathrm {e}}^{2\,x}+1\right )}{3\,{\left ({\mathrm {e}}^x-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-sinh(x)^2/(cosh(x) - 1)^3,x)

[Out]

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

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sympy [A]  time = 1.49, size = 8, normalized size = 0.50 \[ \frac {1}{3 \tanh ^{3}{\left (\frac {x}{2} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**2/(1-cosh(x))**3,x)

[Out]

1/(3*tanh(x/2)**3)

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