3.26 \(\int \text{sech}^2(a+b x)^{3/2} \, dx\)

Optimal. Leaf size=40 \[ \frac{\sin ^{-1}(\tanh (a+b x))}{2 b}+\frac{\tanh (a+b x) \sqrt{\text{sech}^2(a+b x)}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0168868, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4122, 195, 216} \[ \frac{\sin ^{-1}(\tanh (a+b x))}{2 b}+\frac{\tanh (a+b x) \sqrt{\text{sech}^2(a+b x)}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(3/2),x]

[Out]

ArcSin[Tanh[a + b*x]]/(2*b) + (Sqrt[Sech[a + b*x]^2]*Tanh[a + b*x])/(2*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[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 \text{sech}^2(a+b x)^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac{\sqrt{\text{sech}^2(a+b x)} \tanh (a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=\frac{\sin ^{-1}(\tanh (a+b x))}{2 b}+\frac{\sqrt{\text{sech}^2(a+b x)} \tanh (a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0407185, size = 46, normalized size = 1.15 \[ \frac{\text{sech}(a+b x) \left (\tan ^{-1}(\sinh (a+b x))+\tanh (a+b x) \text{sech}(a+b x)\right )}{2 b \sqrt{\text{sech}^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(3/2),x]

[Out]

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

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Maple [C]  time = 0.12, size = 183, normalized size = 4.6 \begin{align*}{\frac{{{\rm e}^{2\,bx+2\,a}}-1}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) b}\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}+{\frac{{\frac{i}{2}} \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{{\frac{i}{2}} \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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.54095, size = 88, normalized size = 2.2 \begin{align*} -\frac{\arctan \left (e^{\left (-b x - a\right )}\right )}{b} + \frac{e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

-arctan(e^(-b*x - a))/b + (e^(-b*x - a) - e^(-3*b*x - 3*a))/(b*(2*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a) + 1))

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Fricas [B]  time = 2.15229, size = 757, normalized size = 18.92 \begin{align*} \frac{\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} +{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) +{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh
(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x +
a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (3*cosh(b*x + a)^2 - 1)*sinh(
b*x + a) - cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 + 2*b*cos
h(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 + b*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 \left (\operatorname{sech}^{2}{\left (a + b x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)**2)**(3/2),x)

[Out]

Integral((sech(a + b*x)**2)**(3/2), x)

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Giac [B]  time = 1.12229, size = 107, normalized size = 2.68 \begin{align*} \frac{\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

1/4*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b + (e^(b*x + a) - e^(-b*x - a))/(((e^(b*x + a) -
e^(-b*x - a))^2 + 4)*b)