3.88 \(\int \text{sech}^3(a+b x) \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac{\text{sech}^5(a+b x)}{5 b}-\frac{\text{sech}^3(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0354046, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2606, 14} \[ \frac{\text{sech}^5(a+b x)}{5 b}-\frac{\text{sech}^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^3*Tanh[a + b*x]^3,x]

[Out]

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

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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0559811, size = 31, normalized size = 1. \[ \frac{\text{sech}^5(a+b x)}{5 b}-\frac{\text{sech}^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^3*Tanh[a + b*x]^3,x]

[Out]

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

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Maple [B]  time = 0.023, size = 68, normalized size = 2.2 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{5\, \left ( \cosh \left ( bx+a \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{2\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{15\,\cosh \left ( bx+a \right ) }}-{\frac{2\,\cosh \left ( bx+a \right ) }{15}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^3*tanh(b*x+a)^3,x)

[Out]

1/b*(-1/5*sinh(b*x+a)^2/cosh(b*x+a)^5+2/15*sinh(b*x+a)^2/cosh(b*x+a)^3+2/15*sinh(b*x+a)^2/cosh(b*x+a)-2/15*cos
h(b*x+a))

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Maxima [B]  time = 1.16571, size = 289, normalized size = 9.32 \begin{align*} -\frac{8 \, e^{\left (-3 \, b x - 3 \, 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 \, e^{\left (-5 \, b x - 5 \, a\right )}}{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 )}} - \frac{8 \, e^{\left (-7 \, b x - 7 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*tanh(b*x+a)^3,x, algorithm="maxima")

[Out]

-8/3*e^(-3*b*x - 3*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*e^(-5*b*x - 5*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)) - 8/3*e^(-7*b*x - 7*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))

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Fricas [B]  time = 1.82287, size = 946, normalized size = 30.52 \begin{align*} -\frac{8 \,{\left (5 \, \cosh \left (b x + a\right )^{4} + 20 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 5 \, \sinh \left (b x + a\right )^{4} + 2 \,{\left (15 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (5 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 5\right )}}{15 \,{\left (b \cosh \left (b x + a\right )^{7} + 7 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + b \sinh \left (b x + a\right )^{7} + 5 \, b \cosh \left (b x + a\right )^{5} +{\left (21 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )^{5} + 5 \,{\left (7 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 11 \, b \cosh \left (b x + a\right )^{3} +{\left (35 \, b \cosh \left (b x + a\right )^{4} + 50 \, b \cosh \left (b x + a\right )^{2} + 9 \, b\right )} \sinh \left (b x + a\right )^{3} +{\left (21 \, b \cosh \left (b x + a\right )^{5} + 50 \, b \cosh \left (b x + a\right )^{3} + 33 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 15 \, b \cosh \left (b x + a\right ) +{\left (7 \, b \cosh \left (b x + a\right )^{6} + 25 \, b \cosh \left (b x + a\right )^{4} + 27 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*tanh(b*x+a)^3,x, algorithm="fricas")

[Out]

-8/15*(5*cosh(b*x + a)^4 + 20*cosh(b*x + a)*sinh(b*x + a)^3 + 5*sinh(b*x + a)^4 + 2*(15*cosh(b*x + a)^2 - 1)*s
inh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(5*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 5)/(b*cosh(b*x + a)
^7 + 7*b*cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 + 5*b*cosh(b*x + a)^5 + (21*b*cosh(b*x + a)^2 + 5*b
)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a)^3 + 5*b*cosh(b*x + a))*sinh(b*x + a)^4 + 11*b*cosh(b*x + a)^3 + (35*b
*cosh(b*x + a)^4 + 50*b*cosh(b*x + a)^2 + 9*b)*sinh(b*x + a)^3 + (21*b*cosh(b*x + a)^5 + 50*b*cosh(b*x + a)^3
+ 33*b*cosh(b*x + a))*sinh(b*x + a)^2 + 15*b*cosh(b*x + a) + (7*b*cosh(b*x + a)^6 + 25*b*cosh(b*x + a)^4 + 27*
b*cosh(b*x + a)^2 + 5*b)*sinh(b*x + a))

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Sympy [A]  time = 4.51699, size = 46, normalized size = 1.48 \begin{align*} \begin{cases} - \frac{\tanh ^{2}{\left (a + b x \right )} \operatorname{sech}^{3}{\left (a + b x \right )}}{5 b} - \frac{2 \operatorname{sech}^{3}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x \tanh ^{3}{\left (a \right )} \operatorname{sech}^{3}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**3*tanh(b*x+a)**3,x)

[Out]

Piecewise((-tanh(a + b*x)**2*sech(a + b*x)**3/(5*b) - 2*sech(a + b*x)**3/(15*b), Ne(b, 0)), (x*tanh(a)**3*sech
(a)**3, True))

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Giac [A]  time = 1.19908, size = 70, normalized size = 2.26 \begin{align*} -\frac{8 \,{\left (5 \, e^{\left (7 \, b x + 7 \, a\right )} - 2 \, e^{\left (5 \, b x + 5 \, a\right )} + 5 \, e^{\left (3 \, b x + 3 \, a\right )}\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)^3*tanh(b*x+a)^3,x, algorithm="giac")

[Out]

-8/15*(5*e^(7*b*x + 7*a) - 2*e^(5*b*x + 5*a) + 5*e^(3*b*x + 3*a))/(b*(e^(2*b*x + 2*a) + 1)^5)