### 3.352 $$\int \text{sech}^2(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=15 $-\frac{\text{sech}^2(a+b x)}{2 b}$

[Out]

-Sech[a + b*x]^2/(2*b)

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Rubi [A]  time = 0.0208823, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {2606, 30} $-\frac{\text{sech}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]^2*Tanh[a + b*x],x]

[Out]

-Sech[a + b*x]^2/(2*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0087869, size = 15, normalized size = 1. $-\frac{\text{sech}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[a + b*x]^2*Tanh[a + b*x],x]

[Out]

-Sech[a + b*x]^2/(2*b)

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Maple [A]  time = 0.007, size = 14, normalized size = 0.9 \begin{align*} -{\frac{ \left ({\rm sech} \left (bx+a\right ) \right ) ^{2}}{2\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01246, size = 31, normalized size = 2.07 \begin{align*} -\frac{2}{b{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*(e^(b*x + a) + e^(-b*x - a))^2)

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Fricas [B]  time = 1.7522, size = 235, normalized size = 15.67 \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 )^{3} + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) +{\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (a + b x \right )} \operatorname{sech}^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x)*sech(a + b*x)**3, x)

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Giac [A]  time = 1.16917, size = 36, normalized size = 2.4 \begin{align*} -\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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