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

Optimal. Leaf size=12 \[ \cosh (x)+2 \log (1-\cosh (x)) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 12, normalized size of antiderivative = 1.00, 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} \[ \cosh (x)+2 \log (1-\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(1 - Cosh[x])^2,x]

[Out]

Cosh[x] + 2*Log[1 - Cosh[x]]

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])

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])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 13, normalized size = 1.08 \[ \cosh (x)+4 \log \left (\sinh \left (\frac {x}{2}\right )\right )-1 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1 + Cosh[x] + 4*Log[Sinh[x/2]]

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fricas [B]  time = 1.09, size = 54, normalized size = 4.50 \[ -\frac {4 \, x \cosh \relax (x) - \cosh \relax (x)^{2} - 8 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) + 2 \, {\left (2 \, x - \cosh \relax (x)\right )} \sinh \relax (x) - \sinh \relax (x)^{2} - 1}{2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*x*cosh(x) - cosh(x)^2 - 8*(cosh(x) + sinh(x))*log(cosh(x) + sinh(x) - 1) + 2*(2*x - cosh(x))*sinh(x) -
sinh(x)^2 - 1)/(cosh(x) + sinh(x))

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giac [A]  time = 0.13, size = 22, normalized size = 1.83 \[ -2 \, x + \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} + 4 \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 11, normalized size = 0.92 \[ \cosh \relax (x )+2 \ln \left (-1+\cosh \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 23, normalized size = 1.92 \[ 2 \, x + \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} + 4 \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.94, size = 10, normalized size = 0.83 \[ 2\,\ln \left (\mathrm {cosh}\relax (x)-1\right )+\mathrm {cosh}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(cosh(x) - 1)^2,x)

[Out]

2*log(cosh(x) - 1) + cosh(x)

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sympy [B]  time = 0.53, size = 58, normalized size = 4.83 \[ \frac {2 \log {\left (\cosh {\relax (x )} - 1 \right )} \cosh {\relax (x )}}{\cosh {\relax (x )} - 1} - \frac {2 \log {\left (\cosh {\relax (x )} - 1 \right )}}{\cosh {\relax (x )} - 1} - \frac {\sinh ^{2}{\relax (x )}}{\cosh {\relax (x )} - 1} + \frac {2 \cosh ^{2}{\relax (x )}}{\cosh {\relax (x )} - 1} - \frac {2}{\cosh {\relax (x )} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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