3.1.92 \(\int \frac {\cosh ^3(x)}{1+\tanh (x)} \, dx\) [92]

Optimal. Leaf size=29 \[ \frac {4 \sinh (x)}{5}+\frac {4 \sinh ^3(x)}{15}-\frac {\cosh ^3(x)}{5 (1+\tanh (x))} \]

[Out]

4/5*sinh(x)+4/15*sinh(x)^3-1/5*cosh(x)^3/(1+tanh(x))

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3583, 2713} \begin {gather*} \frac {4 \sinh ^3(x)}{15}+\frac {4 \sinh (x)}{5}-\frac {\cosh ^3(x)}{5 (\tanh (x)+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(1 + Tanh[x]),x]

[Out]

(4*Sinh[x])/5 + (4*Sinh[x]^3)/15 - Cosh[x]^3/(5*(1 + Tanh[x]))

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {\cosh ^3(x)}{1+\tanh (x)} \, dx &=-\frac {\cosh ^3(x)}{5 (1+\tanh (x))}+\frac {4}{5} \int \cosh ^3(x) \, dx\\ &=-\frac {\cosh ^3(x)}{5 (1+\tanh (x))}+\frac {4}{5} i \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )\\ &=\frac {4 \sinh (x)}{5}+\frac {4 \sinh ^3(x)}{15}-\frac {\cosh ^3(x)}{5 (1+\tanh (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 36, normalized size = 1.24 \begin {gather*} \frac {\text {sech}(x) (-45+20 \cosh (2 x)+\cosh (4 x)+40 \sinh (2 x)+4 \sinh (4 x))}{120 (1+\tanh (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^3/(1 + Tanh[x]),x]

[Out]

(Sech[x]*(-45 + 20*Cosh[2*x] + Cosh[4*x] + 40*Sinh[2*x] + 4*Sinh[4*x]))/(120*(1 + Tanh[x]))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(79\) vs. \(2(23)=46\).
time = 0.46, size = 80, normalized size = 2.76

method result size
risch \(\frac {{\mathrm e}^{3 x}}{48}+\frac {{\mathrm e}^{x}}{4}-\frac {3 \,{\mathrm e}^{-x}}{8}-\frac {{\mathrm e}^{-3 x}}{12}-\frac {{\mathrm e}^{-5 x}}{80}\) \(30\)
default \(-\frac {1}{6 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {5}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2}{5 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {5}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {3}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {11}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )}\) \(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-1/6/(tanh(1/2*x)-1)^3-1/4/(tanh(1/2*x)-1)^2-5/8/(tanh(1/2*x)-1)-2/5/(tanh(1/2*x)+1)^5+1/(tanh(1/2*x)+1)^4-5/3
/(tanh(1/2*x)+1)^3+3/2/(tanh(1/2*x)+1)^2-11/8/(tanh(1/2*x)+1)

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 33, normalized size = 1.14 \begin {gather*} \frac {1}{48} \, {\left (12 \, e^{\left (-2 \, x\right )} + 1\right )} e^{\left (3 \, x\right )} - \frac {3}{8} \, e^{\left (-x\right )} - \frac {1}{12} \, e^{\left (-3 \, x\right )} - \frac {1}{80} \, e^{\left (-5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(1+tanh(x)),x, algorithm="maxima")

[Out]

1/48*(12*e^(-2*x) + 1)*e^(3*x) - 3/8*e^(-x) - 1/12*e^(-3*x) - 1/80*e^(-5*x)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (23) = 46\).
time = 0.42, size = 60, normalized size = 2.07 \begin {gather*} \frac {\cosh \left (x\right )^{4} + 16 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} + 10\right )} \sinh \left (x\right )^{2} + 20 \, \cosh \left (x\right )^{2} + 16 \, {\left (\cosh \left (x\right )^{3} + 5 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) - 45}{120 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(1+tanh(x)),x, algorithm="fricas")

[Out]

1/120*(cosh(x)^4 + 16*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 10)*sinh(x)^2 + 20*cosh(x)^2 + 16*(cosh
(x)^3 + 5*cosh(x))*sinh(x) - 45)/(cosh(x) + sinh(x))

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (26) = 52\).
time = 0.35, size = 134, normalized size = 4.62 \begin {gather*} - \frac {8 \sinh ^{3}{\left (x \right )} \tanh {\left (x \right )}}{15 \tanh {\left (x \right )} + 15} - \frac {2 \sinh ^{3}{\left (x \right )}}{15 \tanh {\left (x \right )} + 15} - \frac {6 \sinh ^{2}{\left (x \right )} \cosh {\left (x \right )} \tanh {\left (x \right )}}{15 \tanh {\left (x \right )} + 15} + \frac {6 \sinh ^{2}{\left (x \right )} \cosh {\left (x \right )}}{15 \tanh {\left (x \right )} + 15} + \frac {6 \sinh {\left (x \right )} \cosh ^{2}{\left (x \right )} \tanh {\left (x \right )}}{15 \tanh {\left (x \right )} + 15} + \frac {9 \sinh {\left (x \right )} \cosh ^{2}{\left (x \right )}}{15 \tanh {\left (x \right )} + 15} + \frac {3 \cosh ^{3}{\left (x \right )} \tanh {\left (x \right )}}{15 \tanh {\left (x \right )} + 15} - \frac {3 \cosh ^{3}{\left (x \right )}}{15 \tanh {\left (x \right )} + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**3/(1+tanh(x)),x)

[Out]

-8*sinh(x)**3*tanh(x)/(15*tanh(x) + 15) - 2*sinh(x)**3/(15*tanh(x) + 15) - 6*sinh(x)**2*cosh(x)*tanh(x)/(15*ta
nh(x) + 15) + 6*sinh(x)**2*cosh(x)/(15*tanh(x) + 15) + 6*sinh(x)*cosh(x)**2*tanh(x)/(15*tanh(x) + 15) + 9*sinh
(x)*cosh(x)**2/(15*tanh(x) + 15) + 3*cosh(x)**3*tanh(x)/(15*tanh(x) + 15) - 3*cosh(x)**3/(15*tanh(x) + 15)

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 31, normalized size = 1.07 \begin {gather*} -\frac {1}{240} \, {\left (90 \, e^{\left (4 \, x\right )} + 20 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-5 \, x\right )} + \frac {1}{48} \, e^{\left (3 \, x\right )} + \frac {1}{4} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(1+tanh(x)),x, algorithm="giac")

[Out]

-1/240*(90*e^(4*x) + 20*e^(2*x) + 3)*e^(-5*x) + 1/48*e^(3*x) + 1/4*e^x

________________________________________________________________________________________

Mupad [B]
time = 1.19, size = 29, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^{3\,x}}{48}-\frac {{\mathrm {e}}^{-3\,x}}{12}-\frac {3\,{\mathrm {e}}^{-x}}{8}-\frac {{\mathrm {e}}^{-5\,x}}{80}+\frac {{\mathrm {e}}^x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(tanh(x) + 1),x)

[Out]

exp(3*x)/48 - exp(-3*x)/12 - (3*exp(-x))/8 - exp(-5*x)/80 + exp(x)/4

________________________________________________________________________________________