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

Optimal. Leaf size=11 \[ \frac{\sin ^{-1}(\tanh (a+b x))}{b} \]

[Out]

ArcSin[Tanh[a + b*x]]/b

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Rubi [A]  time = 0.0109753, antiderivative size = 11, 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, 216} \[ \frac{\sin ^{-1}(\tanh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sech[a + b*x]^2],x]

[Out]

ArcSin[Tanh[a + b*x]]/b

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [B]  time = 0.01561, size = 29, normalized size = 2.64 \[ \frac{\cosh (a+b x) \sqrt{\text{sech}^2(a+b x)} \tan ^{-1}(\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sech[a + b*x]^2],x]

[Out]

(ArcTan[Sinh[a + b*x]]*Cosh[a + b*x]*Sqrt[Sech[a + b*x]^2])/b

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Maple [C]  time = 0.132, size = 130, normalized size = 11.8 \begin{align*}{\frac{i \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) \ln \left ({{\rm e}^{bx}}+i{{\rm e}^{-a}} \right ){{\rm e}^{-bx-a}}}{b}\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}-{\frac{i \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) \ln \left ({{\rm e}^{bx}}-i{{\rm e}^{-a}} \right ){{\rm e}^{-bx-a}}}{b}\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((sech(b*x+a)^2)^(1/2),x)

[Out]

I*(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*ln(exp(b*x)+I*exp(-a))*exp(-b*x-a)-I*(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*ln(exp(b*x)-I*exp(-a))*exp(-b*x-a)

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Maxima [A]  time = 0.995687, size = 15, normalized size = 1.36 \begin{align*} \frac{\arctan \left (\sinh \left (b x + a\right )\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sinh(b*x + a))/b

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Fricas [A]  time = 2.00202, size = 58, normalized size = 5.27 \begin{align*} \frac{2 \, \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(cosh(b*x + a) + sinh(b*x + a))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\operatorname{sech}^{2}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sech(a + b*x)**2), x)

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Giac [A]  time = 1.14072, size = 16, normalized size = 1.45 \begin{align*} \frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(e^(b*x + a))/b