3.28 \(\int \frac{1}{\sqrt{\text{sech}^2(a+b x)}} \, dx\)

Optimal. Leaf size=22 \[ \frac{\tanh (a+b x)}{b \sqrt{\text{sech}^2(a+b x)}} \]

[Out]

Tanh[a + b*x]/(b*Sqrt[Sech[a + b*x]^2])

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Rubi [A]  time = 0.0164002, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4122, 191} \[ \frac{\tanh (a+b x)}{b \sqrt{\text{sech}^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[Sech[a + b*x]^2],x]

[Out]

Tanh[a + b*x]/(b*Sqrt[Sech[a + b*x]^2])

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0255771, size = 22, normalized size = 1. \[ \frac{\tanh (a+b x)}{b \sqrt{\text{sech}^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[Sech[a + b*x]^2],x]

[Out]

Tanh[a + b*x]/(b*Sqrt[Sech[a + b*x]^2])

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Maple [B]  time = 0.106, size = 97, normalized size = 4.4 \begin{align*}{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 2+2\,{{\rm e}^{2\,bx+2\,a}} \right ) b}{\frac{1}{\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}}}-{\frac{1}{ \left ( 2+2\,{{\rm e}^{2\,bx+2\,a}} \right ) b}{\frac{1}{\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.976915, size = 35, normalized size = 1.59 \begin{align*} \frac{e^{\left (b x + a\right )}}{2 \, b} - \frac{e^{\left (-b x - a\right )}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.97329, size = 23, normalized size = 1.05 \begin{align*} \frac{\sinh \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

sinh(b*x + a)/b

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Sympy [A]  time = 16.0804, size = 29, normalized size = 1.32 \begin{align*} \begin{cases} \frac{\tanh{\left (a + b x \right )}}{b \sqrt{\operatorname{sech}^{2}{\left (a + b x \right )}}} & \text{for}\: b \neq 0 \\\frac{x}{\sqrt{\operatorname{sech}^{2}{\left (a \right )}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.15026, size = 31, normalized size = 1.41 \begin{align*} \frac{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(e^(b*x + a) - e^(-b*x - a))/b