3.69 \(\int \frac{\sinh ^4(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=60 \[ \frac{x}{16}-\frac{1}{8 (1-\tanh (x))}-\frac{3}{16 (\tanh (x)+1)}+\frac{1}{32 (1-\tanh (x))^2}+\frac{5}{32 (\tanh (x)+1)^2}-\frac{1}{24 (\tanh (x)+1)^3} \]

[Out]

x/16 + 1/(32*(1 - Tanh[x])^2) - 1/(8*(1 - Tanh[x])) - 1/(24*(1 + Tanh[x])^3) + 5/(32*(1 + Tanh[x])^2) - 3/(16*
(1 + Tanh[x]))

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Rubi [A]  time = 0.0724269, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3516, 848, 88, 207} \[ \frac{x}{16}-\frac{1}{8 (1-\tanh (x))}-\frac{3}{16 (\tanh (x)+1)}+\frac{1}{32 (1-\tanh (x))^2}+\frac{5}{32 (\tanh (x)+1)^2}-\frac{1}{24 (\tanh (x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(1 + Tanh[x]),x]

[Out]

x/16 + 1/(32*(1 - Tanh[x])^2) - 1/(8*(1 - Tanh[x])) - 1/(24*(1 + Tanh[x])^3) + 5/(32*(1 + Tanh[x])^2) - 3/(16*
(1 + Tanh[x]))

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

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

Rubi steps

\begin{align*} \int \frac{\sinh ^4(x)}{1+\tanh (x)} \, dx &=-\operatorname{Subst}\left (\int \frac{x^4}{(1+x) \left (-1+x^2\right )^3} \, dx,x,\tanh (x)\right )\\ &=-\operatorname{Subst}\left (\int \frac{x^4}{(-1+x)^3 (1+x)^4} \, dx,x,\tanh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{1}{16 (-1+x)^3}+\frac{1}{8 (-1+x)^2}-\frac{1}{8 (1+x)^4}+\frac{5}{16 (1+x)^3}-\frac{3}{16 (1+x)^2}+\frac{1}{16 \left (-1+x^2\right )}\right ) \, dx,x,\tanh (x)\right )\\ &=\frac{1}{32 (1-\tanh (x))^2}-\frac{1}{8 (1-\tanh (x))}-\frac{1}{24 (1+\tanh (x))^3}+\frac{5}{32 (1+\tanh (x))^2}-\frac{3}{16 (1+\tanh (x))}-\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{x}{16}+\frac{1}{32 (1-\tanh (x))^2}-\frac{1}{8 (1-\tanh (x))}-\frac{1}{24 (1+\tanh (x))^3}+\frac{5}{32 (1+\tanh (x))^2}-\frac{3}{16 (1+\tanh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0763446, size = 42, normalized size = 0.7 \[ \frac{1}{192} (12 x-3 \sinh (2 x)-3 \sinh (4 x)+\sinh (6 x)-15 \cosh (2 x)+6 \cosh (4 x)-\cosh (6 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(1 + Tanh[x]),x]

[Out]

(12*x - 15*Cosh[2*x] + 6*Cosh[4*x] - Cosh[6*x] - 3*Sinh[2*x] - 3*Sinh[4*x] + Sinh[6*x])/192

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Maple [B]  time = 0.031, size = 98, normalized size = 1.6 \begin{align*} -{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-6}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-5}-{\frac{7}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}}+{\frac{1}{12} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{16}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-4}}+{\frac{1}{4} \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}{16}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(1+tanh(x)),x)

[Out]

-1/3/(tanh(1/2*x)+1)^6+1/(tanh(1/2*x)+1)^5-7/8/(tanh(1/2*x)+1)^4+1/12/(tanh(1/2*x)+1)^3+1/8/(tanh(1/2*x)+1)^2+
1/16*ln(tanh(1/2*x)+1)+1/8/(tanh(1/2*x)-1)^4+1/4/(tanh(1/2*x)-1)^3-1/8/(tanh(1/2*x)-1)-1/16*ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.12109, size = 49, normalized size = 0.82 \begin{align*} -\frac{1}{128} \,{\left (6 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (4 \, x\right )} + \frac{1}{16} \, x - \frac{1}{32} \, e^{\left (-2 \, x\right )} + \frac{3}{128} \, e^{\left (-4 \, x\right )} - \frac{1}{192} \, e^{\left (-6 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(1+tanh(x)),x, algorithm="maxima")

[Out]

-1/128*(6*e^(-2*x) - 1)*e^(4*x) + 1/16*x - 1/32*e^(-2*x) + 3/128*e^(-4*x) - 1/192*e^(-6*x)

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Fricas [B]  time = 2.24885, size = 319, normalized size = 5.32 \begin{align*} \frac{\cosh \left (x\right )^{5} + 5 \, \cosh \left (x\right ) \sinh \left (x\right )^{4} + 5 \, \sinh \left (x\right )^{5} +{\left (50 \, \cosh \left (x\right )^{2} - 27\right )} \sinh \left (x\right )^{3} - 9 \, \cosh \left (x\right )^{3} +{\left (10 \, \cosh \left (x\right )^{3} - 27 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 12 \,{\left (2 \, x - 1\right )} \cosh \left (x\right ) +{\left (25 \, \cosh \left (x\right )^{4} - 81 \, \cosh \left (x\right )^{2} + 24 \, x + 12\right )} \sinh \left (x\right )}{384 \,{\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)^4/(1+tanh(x)),x, algorithm="fricas")

[Out]

1/384*(cosh(x)^5 + 5*cosh(x)*sinh(x)^4 + 5*sinh(x)^5 + (50*cosh(x)^2 - 27)*sinh(x)^3 - 9*cosh(x)^3 + (10*cosh(
x)^3 - 27*cosh(x))*sinh(x)^2 + 12*(2*x - 1)*cosh(x) + (25*cosh(x)^4 - 81*cosh(x)^2 + 24*x + 12)*sinh(x))/(cosh
(x) + sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{4}{\left (x \right )}}{\tanh{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(1+tanh(x)),x)

[Out]

Integral(sinh(x)**4/(tanh(x) + 1), x)

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Giac [A]  time = 1.21964, size = 57, normalized size = 0.95 \begin{align*} -\frac{1}{384} \,{\left (22 \, e^{\left (6 \, x\right )} + 12 \, e^{\left (4 \, x\right )} - 9 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-6 \, x\right )} + \frac{1}{16} \, x + \frac{1}{128} \, e^{\left (4 \, x\right )} - \frac{3}{64} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(1+tanh(x)),x, algorithm="giac")

[Out]

-1/384*(22*e^(6*x) + 12*e^(4*x) - 9*e^(2*x) + 2)*e^(-6*x) + 1/16*x + 1/128*e^(4*x) - 3/64*e^(2*x)