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

Optimal. Leaf size=20 \[ -\frac{2}{1-\cosh (x)}-\log (1-\cosh (x)) \]

[Out]

-2/(1 - Cosh[x]) - Log[1 - Cosh[x]]

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Rubi [A]  time = 0.0404257, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2667, 43} \[ -\frac{2}{1-\cosh (x)}-\log (1-\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(1 - Cosh[x])^3,x]

[Out]

-2/(1 - Cosh[x]) - Log[1 - Cosh[x]]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sinh ^3(x)}{(1-\cosh (x))^3} \, dx &=\operatorname{Subst}\left (\int \frac{1-x}{(1+x)^2} \, dx,x,-\cosh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{-1-x}+\frac{2}{(1+x)^2}\right ) \, dx,x,-\cosh (x)\right )\\ &=-\frac{2}{1-\cosh (x)}-\log (1-\cosh (x))\\ \end{align*}

Mathematica [A]  time = 0.0124578, size = 27, normalized size = 1.35 \[ \coth ^2\left (\frac{x}{2}\right )-2 \log \left (\tanh \left (\frac{x}{2}\right )\right )-2 \log \left (\cosh \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Coth[x/2]^2 - 2*Log[Cosh[x/2]] - 2*Log[Tanh[x/2]]

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Maple [A]  time = 0.018, size = 17, normalized size = 0.9 \begin{align*} 2\, \left ( -1+\cosh \left ( x \right ) \right ) ^{-1}-\ln \left ( -1+\cosh \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(-1+cosh(x))-ln(-1+cosh(x))

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Maxima [A]  time = 1.15768, size = 47, normalized size = 2.35 \begin{align*} -x - \frac{4 \, e^{\left (-x\right )}}{2 \, e^{\left (-x\right )} - e^{\left (-2 \, x\right )} - 1} - 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 4*e^(-x)/(2*e^(-x) - e^(-2*x) - 1) - 2*log(e^(-x) - 1)

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Fricas [B]  time = 1.88948, size = 335, normalized size = 16.75 \begin{align*} \frac{x \cosh \left (x\right )^{2} + x \sinh \left (x\right )^{2} - 2 \,{\left (x - 2\right )} \cosh \left (x\right ) - 2 \,{\left (\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) - 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 2 \,{\left (x \cosh \left (x\right ) - x + 2\right )} \sinh \left (x\right ) + x}{\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) - 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.949815, size = 194, normalized size = 9.7 \begin{align*} - \frac{2 \log{\left (\cosh{\left (x \right )} - 1 \right )} \cosh ^{2}{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} + \frac{4 \log{\left (\cosh{\left (x \right )} - 1 \right )} \cosh{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} - \frac{2 \log{\left (\cosh{\left (x \right )} - 1 \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} - \frac{\sinh ^{2}{\left (x \right )} \cosh ^{2}{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} + \frac{2 \sinh ^{2}{\left (x \right )} \cosh{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} + \frac{\cosh ^{4}{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} - \frac{2 \cosh ^{3}{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} + \frac{4 \cosh{\left (x \right )}}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} - \frac{3}{2 \cosh ^{2}{\left (x \right )} - 4 \cosh{\left (x \right )} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(cosh(x) - 1)*cosh(x)**2/(2*cosh(x)**2 - 4*cosh(x) + 2) + 4*log(cosh(x) - 1)*cosh(x)/(2*cosh(x)**2 - 4*c
osh(x) + 2) - 2*log(cosh(x) - 1)/(2*cosh(x)**2 - 4*cosh(x) + 2) - sinh(x)**2*cosh(x)**2/(2*cosh(x)**2 - 4*cosh
(x) + 2) + 2*sinh(x)**2*cosh(x)/(2*cosh(x)**2 - 4*cosh(x) + 2) + cosh(x)**4/(2*cosh(x)**2 - 4*cosh(x) + 2) - 2
*cosh(x)**3/(2*cosh(x)**2 - 4*cosh(x) + 2) + 4*cosh(x)/(2*cosh(x)**2 - 4*cosh(x) + 2) - 3/(2*cosh(x)**2 - 4*co
sh(x) + 2)

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Giac [A]  time = 1.13683, size = 27, normalized size = 1.35 \begin{align*} x + \frac{4 \, e^{x}}{{\left (e^{x} - 1\right )}^{2}} - 2 \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4*e^x/(e^x - 1)^2 - 2*log(abs(e^x - 1))