3.96 \(\int \text {sech}(x) \tanh ^5(x) \, dx\)

Optimal. Leaf size=21 \[ -\frac {1}{5} \text {sech}^5(x)+\frac {2 \text {sech}^3(x)}{3}-\text {sech}(x) \]

[Out]

-sech(x)+2/3*sech(x)^3-1/5*sech(x)^5

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2606, 194} \[ -\frac {1}{5} \text {sech}^5(x)+\frac {2 \text {sech}^3(x)}{3}-\text {sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]*Tanh[x]^5,x]

[Out]

-Sech[x] + (2*Sech[x]^3)/3 - Sech[x]^5/5

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 2606

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \[ -\frac {1}{5} \text {sech}^5(x)+\frac {2 \text {sech}^3(x)}{3}-\text {sech}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]*Tanh[x]^5,x]

[Out]

-Sech[x] + (2*Sech[x]^3)/3 - Sech[x]^5/5

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fricas [B]  time = 0.44, size = 185, normalized size = 8.81 \[ -\frac {2 \, {\left (15 \, \cosh \relax (x)^{5} + 75 \, \cosh \relax (x) \sinh \relax (x)^{4} + 15 \, \sinh \relax (x)^{5} + 5 \, {\left (30 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{3} + 35 \, \cosh \relax (x)^{3} + 15 \, {\left (10 \, \cosh \relax (x)^{3} + 7 \, \cosh \relax (x)\right )} \sinh \relax (x)^{2} + {\left (75 \, \cosh \relax (x)^{4} + 15 \, \cosh \relax (x)^{2} + 38\right )} \sinh \relax (x) + 78 \, \cosh \relax (x)\right )}}{15 \, {\left (\cosh \relax (x)^{6} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + 3 \, {\left (5 \, \cosh \relax (x)^{2} + 2\right )} \sinh \relax (x)^{4} + 6 \, \cosh \relax (x)^{4} + 4 \, {\left (5 \, \cosh \relax (x)^{3} + 4 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 3 \, {\left (5 \, \cosh \relax (x)^{4} + 12 \, \cosh \relax (x)^{2} + 5\right )} \sinh \relax (x)^{2} + 15 \, \cosh \relax (x)^{2} + 2 \, {\left (3 \, \cosh \relax (x)^{5} + 8 \, \cosh \relax (x)^{3} + 5 \, \cosh \relax (x)\right )} \sinh \relax (x) + 10\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*cosh(x)^5 + 75*cosh(x)*sinh(x)^4 + 15*sinh(x)^5 + 5*(30*cosh(x)^2 + 1)*sinh(x)^3 + 35*cosh(x)^3 + 15
*(10*cosh(x)^3 + 7*cosh(x))*sinh(x)^2 + (75*cosh(x)^4 + 15*cosh(x)^2 + 38)*sinh(x) + 78*cosh(x))/(cosh(x)^6 +
6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 + 2)*sinh(x)^4 + 6*cosh(x)^4 + 4*(5*cosh(x)^3 + 4*cosh(x))*si
nh(x)^3 + 3*(5*cosh(x)^4 + 12*cosh(x)^2 + 5)*sinh(x)^2 + 15*cosh(x)^2 + 2*(3*cosh(x)^5 + 8*cosh(x)^3 + 5*cosh(
x))*sinh(x) + 10)

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giac [B]  time = 0.13, size = 35, normalized size = 1.67 \[ -\frac {2 \, {\left (15 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 40 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 48\right )}}{15 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*(e^(-x) + e^x)^4 - 40*(e^(-x) + e^x)^2 + 48)/(e^(-x) + e^x)^5

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maple [A]  time = 0.08, size = 28, normalized size = 1.33 \[ -\frac {\sinh ^{4}\relax (x )}{\cosh \relax (x )^{5}}-\frac {4 \left (\sinh ^{2}\relax (x )\right )}{3 \cosh \relax (x )^{5}}-\frac {8}{15 \cosh \relax (x )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)*tanh(x)^5,x)

[Out]

-sinh(x)^4/cosh(x)^5-4/3*sinh(x)^2/cosh(x)^5-8/15/cosh(x)^5

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maxima [B]  time = 0.32, size = 191, normalized size = 9.10 \[ -\frac {2 \, e^{\left (-x\right )}}{5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1} - \frac {8 \, e^{\left (-3 \, x\right )}}{3 \, {\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} - \frac {116 \, e^{\left (-5 \, x\right )}}{15 \, {\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} - \frac {8 \, e^{\left (-7 \, x\right )}}{3 \, {\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} - \frac {2 \, e^{\left (-9 \, x\right )}}{5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^(-x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1) - 8/3*e^(-3*x)/(5*e^(-2*x) + 1
0*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1) - 116/15*e^(-5*x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*
x) + 5*e^(-8*x) + e^(-10*x) + 1) - 8/3*e^(-7*x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*
x) + 1) - 2*e^(-9*x)/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1)

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mupad [B]  time = 1.43, size = 129, normalized size = 6.14 \[ \frac {64\,{\mathrm {e}}^x}{5\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )}-\frac {2\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}+1}-\frac {176\,{\mathrm {e}}^x}{15\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {32\,{\mathrm {e}}^x}{5\,\left (5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}+5\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{10\,x}+1\right )}+\frac {16\,{\mathrm {e}}^x}{3\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/cosh(x),x)

[Out]

(64*exp(x))/(5*(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1)) - (2*exp(x))/(exp(2*x) + 1) - (176*exp(x
))/(15*(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1)) - (32*exp(x))/(5*(5*exp(2*x) + 10*exp(4*x) + 10*exp(6*x) + 5*
exp(8*x) + exp(10*x) + 1)) + (16*exp(x))/(3*(2*exp(2*x) + exp(4*x) + 1))

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sympy [A]  time = 1.47, size = 29, normalized size = 1.38 \[ - \frac {\tanh ^{4}{\relax (x )} \operatorname {sech}{\relax (x )}}{5} - \frac {4 \tanh ^{2}{\relax (x )} \operatorname {sech}{\relax (x )}}{15} - \frac {8 \operatorname {sech}{\relax (x )}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-tanh(x)**4*sech(x)/5 - 4*tanh(x)**2*sech(x)/15 - 8*sech(x)/15

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