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

Optimal. Leaf size=10 \[ 2 \tanh ^{-1}(\sinh (x))-\sinh (x) \]

[Out]

2*arctanh(sinh(x))-sinh(x)

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Rubi [A]  time = 0.04, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3190, 388, 206} \[ 2 \tanh ^{-1}(\sinh (x))-\sinh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(1 - Sinh[x]^2),x]

[Out]

2*ArcTanh[Sinh[x]] - Sinh[x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.40 \[ -2 \left (\frac {\sinh (x)}{2}-\tanh ^{-1}(\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*(-ArcTanh[Sinh[x]] + Sinh[x]/2)

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fricas [B]  time = 1.01, size = 71, normalized size = 7.10 \[ -\frac {\cosh \relax (x)^{2} - 2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (\sinh \relax (x) + 1\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (\sinh \relax (x) - 1\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1}{2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.14, size = 37, normalized size = 3.70 \[ \frac {1}{2} \, e^{\left (-x\right )} - \frac {1}{2} \, e^{x} + \log \left ({\left | -e^{\left (-x\right )} + e^{x} + 2 \right |}\right ) - \log \left ({\left | -e^{\left (-x\right )} + e^{x} - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 50, normalized size = 5.00 \[ \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )-2 \tanh \left (\frac {x}{2}\right )-1\right )+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+2 \tanh \left (\frac {x}{2}\right )-1\right )+\frac {1}{\tanh \left (\frac {x}{2}\right )-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(tanh(1/2*x)^2-2*tanh(1/2*x)-1)+1/(tanh(1/2*x)+1)-ln(tanh(1/2*x)^2+2*tanh(1/2*x)-1)+1/(tanh(1/2*x)-1)

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maxima [B]  time = 0.46, size = 39, normalized size = 3.90 \[ \frac {1}{2} \, e^{\left (-x\right )} - \frac {1}{2} \, e^{x} - \log \left (2 \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1\right ) + \log \left (-2 \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 39, normalized size = 3.90 \[ \frac {{\mathrm {e}}^{-x}}{2}-\ln \left (32\,{\mathrm {e}}^{2\,x}-64\,{\mathrm {e}}^x-32\right )+\ln \left (32\,{\mathrm {e}}^{2\,x}+64\,{\mathrm {e}}^x-32\right )-\frac {{\mathrm {e}}^x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-cosh(x)^3/(sinh(x)^2 - 1),x)

[Out]

exp(-x)/2 - log(32*exp(2*x) - 64*exp(x) - 32) + log(32*exp(2*x) + 64*exp(x) - 32) - exp(x)/2

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sympy [B]  time = 1.61, size = 129, normalized size = 12.90 \[ \frac {\log {\left (\tanh ^{2}{\left (\frac {x}{2} \right )} - 2 \tanh {\left (\frac {x}{2} \right )} - 1 \right )} \tanh ^{2}{\left (\frac {x}{2} \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} - 1} - \frac {\log {\left (\tanh ^{2}{\left (\frac {x}{2} \right )} - 2 \tanh {\left (\frac {x}{2} \right )} - 1 \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} - 1} - \frac {\log {\left (\tanh ^{2}{\left (\frac {x}{2} \right )} + 2 \tanh {\left (\frac {x}{2} \right )} - 1 \right )} \tanh ^{2}{\left (\frac {x}{2} \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} - 1} + \frac {\log {\left (\tanh ^{2}{\left (\frac {x}{2} \right )} + 2 \tanh {\left (\frac {x}{2} \right )} - 1 \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} - 1} + \frac {2 \tanh {\left (\frac {x}{2} \right )}}{\tanh ^{2}{\left (\frac {x}{2} \right )} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tanh(x/2)**2 - 2*tanh(x/2) - 1)*tanh(x/2)**2/(tanh(x/2)**2 - 1) - log(tanh(x/2)**2 - 2*tanh(x/2) - 1)/(tan
h(x/2)**2 - 1) - log(tanh(x/2)**2 + 2*tanh(x/2) - 1)*tanh(x/2)**2/(tanh(x/2)**2 - 1) + log(tanh(x/2)**2 + 2*ta
nh(x/2) - 1)/(tanh(x/2)**2 - 1) + 2*tanh(x/2)/(tanh(x/2)**2 - 1)

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