### 3.81 $$\int \text{sech}(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=11 $-\frac{\text{sech}(a+b x)}{b}$

[Out]

-(Sech[a + b*x]/b)

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Rubi [A]  time = 0.0130824, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {2606, 8} $-\frac{\text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]*Tanh[a + b*x],x]

[Out]

-(Sech[a + b*x]/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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0069194, size = 11, normalized size = 1. $-\frac{\text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sech[a + b*x]/b)

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Maple [A]  time = 0.007, size = 12, normalized size = 1.1 \begin{align*} -{\frac{{\rm sech} \left (bx+a\right )}{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sech(b*x+a)/b

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Maxima [B]  time = 1.134, size = 31, normalized size = 2.82 \begin{align*} -\frac{2}{b{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*(e^(b*x + a) + e^(-b*x - a)))

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Fricas [B]  time = 1.69006, size = 154, normalized size = 14. \begin{align*} -\frac{2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.320807, size = 17, normalized size = 1.55 \begin{align*} \begin{cases} - \frac{\operatorname{sech}{\left (a + b x \right )}}{b} & \text{for}\: b \neq 0 \\x \tanh{\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),x)

[Out]

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

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Giac [B]  time = 1.16443, size = 32, normalized size = 2.91 \begin{align*} -\frac{2 \, e^{\left (b x + a\right )}}{b{\left (e^{\left (2 \, b x + 2 \, 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),x, algorithm="giac")

[Out]

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