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]

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

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 30

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

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}^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*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \[ -\frac {\text {sech}^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Sech[a + b*x]^2/b

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fricas [B]  time = 0.45, size = 84, normalized size = 5.60 \[ -\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) + {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )} \]

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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giac [B]  time = 0.12, size = 27, normalized size = 1.80 \[ -\frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} \]

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)

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maple [A]  time = 0.06, size = 14, normalized size = 0.93 \[ -\frac {\mathrm {sech}\left (b x +a \right )^{2}}{2 b} \]

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 = 0.30, size = 13, normalized size = 0.87 \[ \frac {\tanh \left (b x + a\right )^{2}}{2 \, b} \]

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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mupad [B]  time = 0.07, size = 13, normalized size = 0.87 \[ -\frac {1}{2\,b\,{\mathrm {cosh}\left (a+b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + b*x)/cosh(a + b*x)^2,x)

[Out]

-1/(2*b*cosh(a + b*x)^2)

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sympy [A]  time = 0.44, size = 22, normalized size = 1.47 \[ \begin {cases} - \frac {\operatorname {sech}^{2}{\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\x \tanh {\relax (a )} \operatorname {sech}^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

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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