3.581 \(\int \sinh ^4(x) \tanh (x) \, dx\)

Optimal. Leaf size=18 \[ \frac{\cosh ^4(x)}{4}-\cosh ^2(x)+\log (\cosh (x)) \]

[Out]

-Cosh[x]^2 + Cosh[x]^4/4 + Log[Cosh[x]]

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Rubi [A]  time = 0.0225872, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2590, 266, 43} \[ \frac{\cosh ^4(x)}{4}-\cosh ^2(x)+\log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4*Tanh[x],x]

[Out]

-Cosh[x]^2 + Cosh[x]^4/4 + Log[Cosh[x]]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0060303, size = 18, normalized size = 1. \[ \frac{\cosh ^4(x)}{4}-\cosh ^2(x)+\log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4*Tanh[x],x]

[Out]

-Cosh[x]^2 + Cosh[x]^4/4 + Log[Cosh[x]]

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Maple [A]  time = 0.013, size = 17, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{4}}{4}}-{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{2}}+\ln \left ( \cosh \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/sech(x)^4,x)

[Out]

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

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Maxima [B]  time = 1.40772, size = 47, normalized size = 2.61 \begin{align*} -\frac{1}{64} \,{\left (12 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (4 \, x\right )} + x - \frac{3}{16} \, e^{\left (-2 \, x\right )} + \frac{1}{64} \, e^{\left (-4 \, x\right )} + \log \left (e^{\left (-2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/sech(x)^4,x, algorithm="maxima")

[Out]

-1/64*(12*e^(-2*x) - 1)*e^(4*x) + x - 3/16*e^(-2*x) + 1/64*e^(-4*x) + log(e^(-2*x) + 1)

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Fricas [B]  time = 2.1511, size = 863, normalized size = 47.94 \begin{align*} \frac{\cosh \left (x\right )^{8} + 8 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + \sinh \left (x\right )^{8} + 4 \,{\left (7 \, \cosh \left (x\right )^{2} - 3\right )} \sinh \left (x\right )^{6} - 12 \, \cosh \left (x\right )^{6} + 8 \,{\left (7 \, \cosh \left (x\right )^{3} - 9 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} - 64 \, x \cosh \left (x\right )^{4} + 2 \,{\left (35 \, \cosh \left (x\right )^{4} - 90 \, \cosh \left (x\right )^{2} - 32 \, x\right )} \sinh \left (x\right )^{4} + 8 \,{\left (7 \, \cosh \left (x\right )^{5} - 30 \, \cosh \left (x\right )^{3} - 32 \, x \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 4 \,{\left (7 \, \cosh \left (x\right )^{6} - 45 \, \cosh \left (x\right )^{4} - 96 \, x \cosh \left (x\right )^{2} - 3\right )} \sinh \left (x\right )^{2} - 12 \, \cosh \left (x\right )^{2} + 64 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4}\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 8 \,{\left (\cosh \left (x\right )^{7} - 9 \, \cosh \left (x\right )^{5} - 32 \, x \cosh \left (x\right )^{3} - 3 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}{64 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/sech(x)^4,x, algorithm="fricas")

[Out]

1/64*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 - 3)*sinh(x)^6 - 12*cosh(x)^6 + 8*(7*cosh(x
)^3 - 9*cosh(x))*sinh(x)^5 - 64*x*cosh(x)^4 + 2*(35*cosh(x)^4 - 90*cosh(x)^2 - 32*x)*sinh(x)^4 + 8*(7*cosh(x)^
5 - 30*cosh(x)^3 - 32*x*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 45*cosh(x)^4 - 96*x*cosh(x)^2 - 3)*sinh(x)^2 - 1
2*cosh(x)^2 + 64*(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4)*l
og(2*cosh(x)/(cosh(x) - sinh(x))) + 8*(cosh(x)^7 - 9*cosh(x)^5 - 32*x*cosh(x)^3 - 3*cosh(x))*sinh(x) + 1)/(cos
h(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{5}{\left (x \right )}}{\operatorname{sech}^{4}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**5/sech(x)**4,x)

[Out]

Integral(tanh(x)**5/sech(x)**4, x)

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Giac [B]  time = 1.1268, size = 58, normalized size = 3.22 \begin{align*} \frac{1}{64} \,{\left (48 \, e^{\left (4 \, x\right )} - 12 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-4 \, x\right )} - x + \frac{1}{64} \, e^{\left (4 \, x\right )} - \frac{3}{16} \, e^{\left (2 \, x\right )} + \log \left (e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/sech(x)^4,x, algorithm="giac")

[Out]

1/64*(48*e^(4*x) - 12*e^(2*x) + 1)*e^(-4*x) - x + 1/64*e^(4*x) - 3/16*e^(2*x) + log(e^(2*x) + 1)