3.29 \(\int \frac{1}{\text{sech}^2(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=51 \[ \frac{2 \tanh (a+b x)}{3 b \sqrt{\text{sech}^2(a+b x)}}+\frac{\tanh (a+b x)}{3 b \text{sech}^2(a+b x)^{3/2}} \]

[Out]

Tanh[a + b*x]/(3*b*(Sech[a + b*x]^2)^(3/2)) + (2*Tanh[a + b*x])/(3*b*Sqrt[Sech[a + b*x]^2])

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Rubi [A]  time = 0.0197899, antiderivative size = 51, 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, 192, 191} \[ \frac{2 \tanh (a+b x)}{3 b \sqrt{\text{sech}^2(a+b x)}}+\frac{\tanh (a+b x)}{3 b \text{sech}^2(a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(-3/2),x]

[Out]

Tanh[a + b*x]/(3*b*(Sech[a + b*x]^2)^(3/2)) + (2*Tanh[a + b*x])/(3*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 192

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

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}{\text{sech}^2(a+b x)^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{5/2}} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac{\tanh (a+b x)}{3 b \text{sech}^2(a+b x)^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{3/2}} \, dx,x,\tanh (a+b x)\right )}{3 b}\\ &=\frac{\tanh (a+b x)}{3 b \text{sech}^2(a+b x)^{3/2}}+\frac{2 \tanh (a+b x)}{3 b \sqrt{\text{sech}^2(a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0657508, size = 44, normalized size = 0.86 \[ \frac{\tanh ^3(a+b x)+3 \tanh (a+b x) \text{sech}^2(a+b x)}{3 b \text{sech}^2(a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(-3/2),x]

[Out]

(3*Sech[a + b*x]^2*Tanh[a + b*x] + Tanh[a + b*x]^3)/(3*b*(Sech[a + b*x]^2)^(3/2))

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Maple [B]  time = 0.104, size = 201, normalized size = 3.9 \begin{align*}{\frac{{{\rm e}^{4\,bx+4\,a}}}{ \left ( 24+24\,{{\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{3\,{{\rm e}^{2\,bx+2\,a}}}{ \left ( 8+8\,{{\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{3}{ \left ( 8+8\,{{\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{{{\rm e}^{-2\,bx-2\,a}}}{ \left ( 24+24\,{{\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)^(3/2),x)

[Out]

1/24/(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(4*b*x+4*a)+3/8/(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)-3/8/(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-1/24/(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
)

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Maxima [A]  time = 1.01207, size = 73, normalized size = 1.43 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )}}{24 \, b} + \frac{3 \, e^{\left (b x + a\right )}}{8 \, b} - \frac{3 \, e^{\left (-b x - a\right )}}{8 \, b} - \frac{e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

1/24*e^(3*b*x + 3*a)/b + 3/8*e^(b*x + a)/b - 3/8*e^(-b*x - a)/b - 1/24*e^(-3*b*x - 3*a)/b

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Fricas [A]  time = 2.04689, size = 89, normalized size = 1.75 \begin{align*} \frac{\sinh \left (b x + a\right )^{3} + 3 \,{\left (\cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

1/12*(sinh(b*x + a)^3 + 3*(cosh(b*x + a)^2 + 3)*sinh(b*x + a))/b

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Sympy [A]  time = 20.3354, size = 54, normalized size = 1.06 \begin{align*} \begin{cases} - \frac{2 \tanh ^{3}{\left (a + b x \right )}}{3 b \left (\operatorname{sech}^{2}{\left (a + b x \right )}\right )^{\frac{3}{2}}} + \frac{\tanh{\left (a + b x \right )}}{b \left (\operatorname{sech}^{2}{\left (a + b x \right )}\right )^{\frac{3}{2}}} & \text{for}\: b \neq 0 \\\frac{x}{\left (\operatorname{sech}^{2}{\left (a \right )}\right )^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)**2)**(3/2),x)

[Out]

Piecewise((-2*tanh(a + b*x)**3/(3*b*(sech(a + b*x)**2)**(3/2)) + tanh(a + b*x)/(b*(sech(a + b*x)**2)**(3/2)),
Ne(b, 0)), (x/(sech(a)**2)**(3/2), True))

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Giac [A]  time = 1.13752, size = 65, normalized size = 1.27 \begin{align*} -\frac{{\left (9 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} - 9 \, e^{\left (b x + a\right )}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

-1/24*((9*e^(2*b*x + 2*a) + 1)*e^(-3*b*x - 3*a) - e^(3*b*x + 3*a) - 9*e^(b*x + a))/b