3.8 \(\int \text {sech}^6(\pi x) \, dx\)

Optimal. Leaf size=35 \[ \frac {\tanh ^5(\pi x)}{5 \pi }-\frac {2 \tanh ^3(\pi x)}{3 \pi }+\frac {\tanh (\pi x)}{\pi } \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3767} \[ \frac {\tanh ^5(\pi x)}{5 \pi }-\frac {2 \tanh ^3(\pi x)}{3 \pi }+\frac {\tanh (\pi x)}{\pi } \]

Antiderivative was successfully verified.

[In]

Int[Sech[Pi*x]^6,x]

[Out]

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \text {sech}^6(\pi x) \, dx &=\frac {i \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \tanh (\pi x)\right )}{\pi }\\ &=\frac {\tanh (\pi x)}{\pi }-\frac {2 \tanh ^3(\pi x)}{3 \pi }+\frac {\tanh ^5(\pi x)}{5 \pi }\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 35, normalized size = 1.00 \[ \frac {\tanh ^5(\pi x)}{5 \pi }-\frac {2 \tanh ^3(\pi x)}{3 \pi }+\frac {\tanh (\pi x)}{\pi } \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[Pi*x]^6,x]

[Out]

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

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fricas [B]  time = 0.47, size = 280, normalized size = 8.00 \[ -\frac {16 \, {\left (11 \, \cosh \left (\pi x\right )^{2} + 18 \, \cosh \left (\pi x\right ) \sinh \left (\pi x\right ) + 11 \, \sinh \left (\pi x\right )^{2} + 5\right )}}{15 \, {\left (5 \, \pi + \pi \cosh \left (\pi x\right )^{8} + 8 \, \pi \cosh \left (\pi x\right ) \sinh \left (\pi x\right )^{7} + \pi \sinh \left (\pi x\right )^{8} + 5 \, \pi \cosh \left (\pi x\right )^{6} + {\left (5 \, \pi + 28 \, \pi \cosh \left (\pi x\right )^{2}\right )} \sinh \left (\pi x\right )^{6} + 2 \, {\left (28 \, \pi \cosh \left (\pi x\right )^{3} + 15 \, \pi \cosh \left (\pi x\right )\right )} \sinh \left (\pi x\right )^{5} + 10 \, \pi \cosh \left (\pi x\right )^{4} + 5 \, {\left (2 \, \pi + 14 \, \pi \cosh \left (\pi x\right )^{4} + 15 \, \pi \cosh \left (\pi x\right )^{2}\right )} \sinh \left (\pi x\right )^{4} + 4 \, {\left (14 \, \pi \cosh \left (\pi x\right )^{5} + 25 \, \pi \cosh \left (\pi x\right )^{3} + 10 \, \pi \cosh \left (\pi x\right )\right )} \sinh \left (\pi x\right )^{3} + 11 \, \pi \cosh \left (\pi x\right )^{2} + {\left (11 \, \pi + 28 \, \pi \cosh \left (\pi x\right )^{6} + 75 \, \pi \cosh \left (\pi x\right )^{4} + 60 \, \pi \cosh \left (\pi x\right )^{2}\right )} \sinh \left (\pi x\right )^{2} + 2 \, {\left (4 \, \pi \cosh \left (\pi x\right )^{7} + 15 \, \pi \cosh \left (\pi x\right )^{5} + 20 \, \pi \cosh \left (\pi x\right )^{3} + 9 \, \pi \cosh \left (\pi x\right )\right )} \sinh \left (\pi x\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(pi*x)^6,x, algorithm="fricas")

[Out]

-16/15*(11*cosh(pi*x)^2 + 18*cosh(pi*x)*sinh(pi*x) + 11*sinh(pi*x)^2 + 5)/(5*pi + pi*cosh(pi*x)^8 + 8*pi*cosh(
pi*x)*sinh(pi*x)^7 + pi*sinh(pi*x)^8 + 5*pi*cosh(pi*x)^6 + (5*pi + 28*pi*cosh(pi*x)^2)*sinh(pi*x)^6 + 2*(28*pi
*cosh(pi*x)^3 + 15*pi*cosh(pi*x))*sinh(pi*x)^5 + 10*pi*cosh(pi*x)^4 + 5*(2*pi + 14*pi*cosh(pi*x)^4 + 15*pi*cos
h(pi*x)^2)*sinh(pi*x)^4 + 4*(14*pi*cosh(pi*x)^5 + 25*pi*cosh(pi*x)^3 + 10*pi*cosh(pi*x))*sinh(pi*x)^3 + 11*pi*
cosh(pi*x)^2 + (11*pi + 28*pi*cosh(pi*x)^6 + 75*pi*cosh(pi*x)^4 + 60*pi*cosh(pi*x)^2)*sinh(pi*x)^2 + 2*(4*pi*c
osh(pi*x)^7 + 15*pi*cosh(pi*x)^5 + 20*pi*cosh(pi*x)^3 + 9*pi*cosh(pi*x))*sinh(pi*x))

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giac [A]  time = 0.14, size = 30, normalized size = 0.86 \[ -\frac {16 \, {\left (10 \, e^{\left (4 \, \pi x\right )} + 5 \, e^{\left (2 \, \pi x\right )} + 1\right )}}{15 \, \pi {\left (e^{\left (2 \, \pi x\right )} + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(pi*x)^6,x, algorithm="giac")

[Out]

-16/15*(10*e^(4*pi*x) + 5*e^(2*pi*x) + 1)/(pi*(e^(2*pi*x) + 1)^5)

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maple [A]  time = 0.28, size = 27, normalized size = 0.77 \[ \frac {\left (\frac {8}{15}+\frac {\mathrm {sech}\left (\pi x \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (\pi x \right )^{2}}{15}\right ) \tanh \left (\pi x \right )}{\pi } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(Pi*x)^6,x)

[Out]

1/Pi*(8/15+1/5*sech(Pi*x)^4+4/15*sech(Pi*x)^2)*tanh(Pi*x)

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maxima [B]  time = 2.08, size = 137, normalized size = 3.91 \[ \frac {16 \, e^{\left (-2 \, \pi x\right )}}{3 \, \pi {\left (5 \, e^{\left (-2 \, \pi x\right )} + 10 \, e^{\left (-4 \, \pi x\right )} + 10 \, e^{\left (-6 \, \pi x\right )} + 5 \, e^{\left (-8 \, \pi x\right )} + e^{\left (-10 \, \pi x\right )} + 1\right )}} + \frac {32 \, e^{\left (-4 \, \pi x\right )}}{3 \, \pi {\left (5 \, e^{\left (-2 \, \pi x\right )} + 10 \, e^{\left (-4 \, \pi x\right )} + 10 \, e^{\left (-6 \, \pi x\right )} + 5 \, e^{\left (-8 \, \pi x\right )} + e^{\left (-10 \, \pi x\right )} + 1\right )}} + \frac {16}{15 \, \pi {\left (5 \, e^{\left (-2 \, \pi x\right )} + 10 \, e^{\left (-4 \, \pi x\right )} + 10 \, e^{\left (-6 \, \pi x\right )} + 5 \, e^{\left (-8 \, \pi x\right )} + e^{\left (-10 \, \pi x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(pi*x)^6,x, algorithm="maxima")

[Out]

16/3*e^(-2*pi*x)/(pi*(5*e^(-2*pi*x) + 10*e^(-4*pi*x) + 10*e^(-6*pi*x) + 5*e^(-8*pi*x) + e^(-10*pi*x) + 1)) + 3
2/3*e^(-4*pi*x)/(pi*(5*e^(-2*pi*x) + 10*e^(-4*pi*x) + 10*e^(-6*pi*x) + 5*e^(-8*pi*x) + e^(-10*pi*x) + 1)) + 16
/15/(pi*(5*e^(-2*pi*x) + 10*e^(-4*pi*x) + 10*e^(-6*pi*x) + 5*e^(-8*pi*x) + e^(-10*pi*x) + 1))

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mupad [B]  time = 1.52, size = 30, normalized size = 0.86 \[ -\frac {16\,\left (5\,{\mathrm {e}}^{2\,\Pi \,x}+10\,{\mathrm {e}}^{4\,\Pi \,x}+1\right )}{15\,\Pi \,{\left ({\mathrm {e}}^{2\,\Pi \,x}+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cosh(Pi*x)^6,x)

[Out]

-(16*(5*exp(2*Pi*x) + 10*exp(4*Pi*x) + 1))/(15*Pi*(exp(2*Pi*x) + 1)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}^{6}{\left (\pi x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(pi*x)**6,x)

[Out]

Integral(sech(pi*x)**6, x)

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