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

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

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ \frac {\cosh ^4(x)}{4}-\cosh ^2(x)+\log (\cosh (x)) \]

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^4(x) \tanh (x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 1.09, size = 257, normalized size = 14.28 \[ \frac {\cosh \relax (x)^{8} + 8 \, \cosh \relax (x) \sinh \relax (x)^{7} + \sinh \relax (x)^{8} + 4 \, {\left (7 \, \cosh \relax (x)^{2} - 3\right )} \sinh \relax (x)^{6} - 12 \, \cosh \relax (x)^{6} + 8 \, {\left (7 \, \cosh \relax (x)^{3} - 9 \, \cosh \relax (x)\right )} \sinh \relax (x)^{5} - 64 \, x \cosh \relax (x)^{4} + 2 \, {\left (35 \, \cosh \relax (x)^{4} - 90 \, \cosh \relax (x)^{2} - 32 \, x\right )} \sinh \relax (x)^{4} + 8 \, {\left (7 \, \cosh \relax (x)^{5} - 30 \, \cosh \relax (x)^{3} - 32 \, x \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 4 \, {\left (7 \, \cosh \relax (x)^{6} - 45 \, \cosh \relax (x)^{4} - 96 \, x \cosh \relax (x)^{2} - 3\right )} \sinh \relax (x)^{2} - 12 \, \cosh \relax (x)^{2} + 64 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} \sinh \relax (x) + 6 \, \cosh \relax (x)^{2} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4}\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 8 \, {\left (\cosh \relax (x)^{7} - 9 \, \cosh \relax (x)^{5} - 32 \, x \cosh \relax (x)^{3} - 3 \, \cosh \relax (x)\right )} \sinh \relax (x) + 1}{64 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} \sinh \relax (x) + 6 \, \cosh \relax (x)^{2} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4}\right )}} \]

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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giac [B]  time = 0.61, size = 43, normalized size = 2.39 \[ \frac {1}{64} \, {\left (48 \, e^{\left (4 \, x\right )} - 12 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-4 \, x\right )} - x + \frac {1}{64} \, e^{\left (4 \, x\right )} - \frac {3}{16} \, e^{\left (2 \, x\right )} + \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

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)

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maple [A]  time = 0.04, size = 17, normalized size = 0.94

method	result	size
default	\(\frac {\left (\sinh ^{4}\relax (x )\right )}{4}-\frac {\left (\sinh ^{2}\relax (x )\right )}{2}+\ln \left (\cosh \relax (x )\right )\)	\(17\)
risch	\(-x +\frac {{\mathrm e}^{4 x}}{64}-\frac {3 \,{\mathrm e}^{2 x}}{16}-\frac {3 \,{\mathrm e}^{-2 x}}{16}+\frac {{\mathrm e}^{-4 x}}{64}+\ln \left (1+{\mathrm e}^{2 x}\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/sech(x)^4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.96, size = 35, normalized size = 1.94 \[ -\frac {1}{64} \, {\left (12 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (4 \, x\right )} + x - \frac {3}{16} \, e^{\left (-2 \, x\right )} + \frac {1}{64} \, e^{\left (-4 \, x\right )} + \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

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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mupad [B]  time = 0.36, size = 35, normalized size = 1.94 \[ \ln \left ({\mathrm {e}}^{2\,x}+1\right )-x-\frac {3\,{\mathrm {e}}^{-2\,x}}{16}-\frac {3\,{\mathrm {e}}^{2\,x}}{16}+\frac {{\mathrm {e}}^{-4\,x}}{64}+\frac {{\mathrm {e}}^{4\,x}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4*tanh(x)^5,x)

[Out]

log(exp(2*x) + 1) - x - (3*exp(-2*x))/16 - (3*exp(2*x))/16 + exp(-4*x)/64 + exp(4*x)/64

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{5}{\relax (x )}}{\operatorname {sech}^{4}{\relax (x )}}\, dx \]

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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