### 3.87 $$\int \text{sech}(a+b x) \tanh ^3(a+b x) \, dx$$

Optimal. Leaf size=27 $\frac{\text{sech}^3(a+b x)}{3 b}-\frac{\text{sech}(a+b x)}{b}$

[Out]

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

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Rubi [A]  time = 0.0244432, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.067, Rules used = {2606} $\frac{\text{sech}^3(a+b x)}{3 b}-\frac{\text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]*Tanh[a + b*x]^3,x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0311728, size = 27, normalized size = 1. $\frac{\text{sech}^3(a+b x)}{3 b}-\frac{\text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 50, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{2\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,\cosh \left ( bx+a \right ) }}-{\frac{2\,\cosh \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/3*sinh(b*x+a)^2/cosh(b*x+a)^3+2/3*sinh(b*x+a)^2/cosh(b*x+a)-2/3*cosh(b*x+a))

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Maxima [B]  time = 1.02041, size = 200, normalized size = 7.41 \begin{align*} -\frac{2 \, e^{\left (-b x - a\right )}}{b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )}} - \frac{4 \, e^{\left (-3 \, b x - 3 \, a\right )}}{3 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )}} - \frac{2 \, e^{\left (-5 \, b x - 5 \, a\right )}}{b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^(-b*x - a)/(b*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a) + 1)) - 4/3*e^(-3*b*x - 3*a)/(b
*(3*e^(-2*b*x - 2*a) + 3*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a) + 1)) - 2*e^(-5*b*x - 5*a)/(b*(3*e^(-2*b*x - 2*a)
+ 3*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a) + 1))

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Fricas [B]  time = 1.74834, size = 467, normalized size = 17.3 \begin{align*} -\frac{2 \,{\left (3 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 3 \, \sinh \left (b x + a\right )^{3} +{\left (9 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) + 5 \, \cosh \left (b x + a\right )\right )}}{3 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} + 2 \, b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 3 \, b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*cosh(b*x + a)^3 + 9*cosh(b*x + a)*sinh(b*x + a)^2 + 3*sinh(b*x + a)^3 + (9*cosh(b*x + a)^2 - 1)*sinh(b
*x + a) + 5*cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 + 4*b*co
sh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + 2*b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b
*x + a) + 3*b)

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Sympy [A]  time = 1.28128, size = 41, normalized size = 1.52 \begin{align*} \begin{cases} - \frac{\tanh ^{2}{\left (a + b x \right )} \operatorname{sech}{\left (a + b x \right )}}{3 b} - \frac{2 \operatorname{sech}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x \tanh ^{3}{\left (a \right )} \operatorname{sech}{\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)*tanh(b*x+a)**3,x)

[Out]

Piecewise((-tanh(a + b*x)**2*sech(a + b*x)/(3*b) - 2*sech(a + b*x)/(3*b), Ne(b, 0)), (x*tanh(a)**3*sech(a), Tr
ue))

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Giac [A]  time = 1.24873, size = 66, normalized size = 2.44 \begin{align*} -\frac{2 \,{\left (3 \, e^{\left (5 \, b x + 5 \, a\right )} + 2 \, e^{\left (3 \, b x + 3 \, a\right )} + 3 \, e^{\left (b x + a\right )}\right )}}{3 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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