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

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 = 0.0203543, 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 = 0.0108386, 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 = 0.005, 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)^2*tanh(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.00907, size = 18, normalized size = 1.2 \begin{align*} \frac{\tanh \left (b x + a\right )^{2}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^2*tanh(b*x+a),x, algorithm="maxima")

[Out]

1/2*tanh(b*x + a)^2/b

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Fricas [B]  time = 1.75892, 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)^2*tanh(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 [A]  time = 0.678421, size = 22, normalized size = 1.47 \begin{align*} \begin{cases} - \frac{\operatorname{sech}^{2}{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\x \tanh{\left (a \right )} \operatorname{sech}^{2}{\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)**2*tanh(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.19404, 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)^2*tanh(b*x+a),x, algorithm="giac")

[Out]

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