3.70 \(\int \frac{\sinh ^3(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=25 \[ -\frac{\sinh ^5(x)}{5}+\frac{\cosh ^5(x)}{5}-\frac{\cosh ^3(x)}{3} \]

[Out]

-Cosh[x]^3/3 + Cosh[x]^5/5 - Sinh[x]^5/5

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Rubi [A]  time = 0.166771, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {3518, 3108, 3107, 2565, 14, 2564, 30} \[ -\frac{\sinh ^5(x)}{5}+\frac{\cosh ^5(x)}{5}-\frac{\cosh ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(1 + Tanh[x]),x]

[Out]

-Cosh[x]^3/3 + Cosh[x]^5/5 - Sinh[x]^5/5

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3108

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[a^p*b^p, Int[(Cos[c + d*x]^m*Sin[c + d*x]^n)/(b*Cos[c + d*x] + a*Sin
[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a^2 + b^2, 0] && ILtQ[p, 0]

Rule 3107

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*sin[c + d*x]^n*(a*cos[c + d*x] + b*sin[c +
d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0593123, size = 34, normalized size = 1.36 \[ \frac{1}{120} (\cosh (x)-\sinh (x)) (-10 \sinh (2 x)+\sinh (4 x)-20 \cosh (2 x)+4 \cosh (4 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(1 + Tanh[x]),x]

[Out]

((Cosh[x] - Sinh[x])*(-20*Cosh[2*x] + 4*Cosh[4*x] - 10*Sinh[2*x] + Sinh[4*x]))/120

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Maple [B]  time = 0.027, size = 72, normalized size = 2.9 \begin{align*}{\frac{2}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-5}}- \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}+{\frac{2}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\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{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(1+tanh(x)),x)

[Out]

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

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Maxima [A]  time = 1.18165, size = 36, normalized size = 1.44 \begin{align*} -\frac{1}{48} \,{\left (6 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (3 \, x\right )} - \frac{1}{24} \, e^{\left (-3 \, x\right )} + \frac{1}{80} \, e^{\left (-5 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(6*e^(-2*x) - 1)*e^(3*x) - 1/24*e^(-3*x) + 1/80*e^(-5*x)

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Fricas [B]  time = 2.19843, size = 200, normalized size = 8. \begin{align*} \frac{\cosh \left (x\right )^{4} + \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} +{\left (6 \, \cosh \left (x\right )^{2} - 5\right )} \sinh \left (x\right )^{2} - 5 \, \cosh \left (x\right )^{2} +{\left (\cosh \left (x\right )^{3} - 5 \, \cosh \left (x\right )\right )} \sinh \left (x\right )}{30 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(cosh(x)^4 + cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 - 5)*sinh(x)^2 - 5*cosh(x)^2 + (cosh(x)^3 - 5*c
osh(x))*sinh(x))/(cosh(x) + sinh(x))

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Sympy [B]  time = 1.59728, size = 134, normalized size = 5.36 \begin{align*} \frac{3 \sinh ^{3}{\left (x \right )} \tanh{\left (x \right )}}{15 \tanh{\left (x \right )} + 15} - \frac{3 \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{9 \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{6 \sinh{\left (x \right )} \cosh ^{2}{\left (x \right )}}{15 \tanh{\left (x \right )} + 15} - \frac{8 \cosh ^{3}{\left (x \right )} \tanh{\left (x \right )}}{15 \tanh{\left (x \right )} + 15} - \frac{2 \cosh ^{3}{\left (x \right )}}{15 \tanh{\left (x \right )} + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19442, size = 34, normalized size = 1.36 \begin{align*} -\frac{1}{240} \,{\left (10 \, e^{\left (2 \, x\right )} - 3\right )} e^{\left (-5 \, x\right )} + \frac{1}{48} \, e^{\left (3 \, x\right )} - \frac{1}{8} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(10*e^(2*x) - 3)*e^(-5*x) + 1/48*e^(3*x) - 1/8*e^x