3.6 \(\int \text{sech}^6(a+b x) \, dx\)

Optimal. Leaf size=41 \[ \frac{\tanh ^5(a+b x)}{5 b}-\frac{2 \tanh ^3(a+b x)}{3 b}+\frac{\tanh (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0147582, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3767} \[ \frac{\tanh ^5(a+b x)}{5 b}-\frac{2 \tanh ^3(a+b x)}{3 b}+\frac{\tanh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^6,x]

[Out]

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

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

Mathematica [A]  time = 0.0105993, size = 41, normalized size = 1. \[ \frac{\tanh ^5(a+b x)}{5 b}-\frac{2 \tanh ^3(a+b x)}{3 b}+\frac{\tanh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^6,x]

[Out]

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

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Maple [A]  time = 0.013, size = 33, normalized size = 0.8 \begin{align*}{\frac{\tanh \left ( bx+a \right ) }{b} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (bx+a\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (bx+a\right ) \right ) ^{2}}{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^6,x)

[Out]

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

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Maxima [B]  time = 1.01809, size = 277, normalized size = 6.76 \begin{align*} \frac{16 \, e^{\left (-2 \, b x - 2 \, a\right )}}{3 \, b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac{32 \, e^{\left (-4 \, b x - 4 \, a\right )}}{3 \, b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac{16}{15 \, b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^6,x, algorithm="maxima")

[Out]

16/3*e^(-2*b*x - 2*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a)
+ e^(-10*b*x - 10*a) + 1)) + 32/3*e^(-4*b*x - 4*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x
- 6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1)) + 16/15/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) +
10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1))

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Fricas [B]  time = 2.09174, size = 954, normalized size = 23.27 \begin{align*} -\frac{16 \,{\left (11 \, \cosh \left (b x + a\right )^{2} + 18 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 11 \, \sinh \left (b x + a\right )^{2} + 5\right )}}{15 \,{\left (b \cosh \left (b x + a\right )^{8} + 8 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + b \sinh \left (b x + a\right )^{8} + 5 \, b \cosh \left (b x + a\right )^{6} +{\left (28 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )^{6} + 2 \,{\left (28 \, b \cosh \left (b x + a\right )^{3} + 15 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + 10 \, b \cosh \left (b x + a\right )^{4} + 5 \,{\left (14 \, b \cosh \left (b x + a\right )^{4} + 15 \, b \cosh \left (b x + a\right )^{2} + 2 \, b\right )} \sinh \left (b x + a\right )^{4} + 4 \,{\left (14 \, b \cosh \left (b x + a\right )^{5} + 25 \, b \cosh \left (b x + a\right )^{3} + 10 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 11 \, b \cosh \left (b x + a\right )^{2} +{\left (28 \, b \cosh \left (b x + a\right )^{6} + 75 \, b \cosh \left (b x + a\right )^{4} + 60 \, b \cosh \left (b x + a\right )^{2} + 11 \, b\right )} \sinh \left (b x + a\right )^{2} + 2 \,{\left (4 \, b \cosh \left (b x + a\right )^{7} + 15 \, b \cosh \left (b x + a\right )^{5} + 20 \, b \cosh \left (b x + a\right )^{3} + 9 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 5 \, b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^6,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}^{6}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**6,x)

[Out]

Integral(sech(a + b*x)**6, x)

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Giac [A]  time = 1.12395, size = 57, normalized size = 1.39 \begin{align*} -\frac{16 \,{\left (10 \, e^{\left (4 \, b x + 4 \, a\right )} + 5 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{15 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^6,x, algorithm="giac")

[Out]

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