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]

1/3*tanh(b*x+a)/b/(sech(b*x+a)^2)^(3/2)+2/3*tanh(b*x+a)/b/(sech(b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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
]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.39, size = 32, normalized size = 0.63 \[ \frac {\sinh \left (b x + a\right )^{3} + 3 \, {\left (\cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right )}{12 \, b} \]

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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giac [A]  time = 0.11, size = 48, normalized size = 0.94 \[ -\frac {{\left (9 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} - 9 \, e^{\left (b x + a\right )}}{24 \, b} \]

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

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maple [B]  time = 0.44, size = 201, normalized size = 3.94 \[ \frac {{\mathrm e}^{4 b x +4 a}}{24 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}+\frac {3 \,{\mathrm e}^{2 b x +2 a}}{8 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {3}{8 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}}-\frac {{\mathrm e}^{-2 b x -2 a}}{24 b \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {\frac {{\mathrm e}^{2 b x +2 a}}{\left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/b/(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(4*b*x+4*a)+3/8/b/(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)-3/8/b/(1+exp(2*b*x+2*a))/(1/(1+exp(2*b*x+2*a))^
2*exp(2*b*x+2*a))^(1/2)-1/24/b/(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
)

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maxima [A]  time = 0.33, size = 54, normalized size = 1.06 \[ \frac {e^{\left (3 \, b x + 3 \, a\right )}}{24 \, b} + \frac {3 \, e^{\left (b x + a\right )}}{8 \, b} - \frac {3 \, e^{\left (-b x - a\right )}}{8 \, b} - \frac {e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (\frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^2}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cosh(a + b*x)^2)^(3/2),x)

[Out]

int(1/(1/cosh(a + b*x)^2)^(3/2), x)

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sympy [A]  time = 18.85, size = 54, normalized size = 1.06 \[ \begin {cases} - \frac {2 \tanh ^{3}{\left (a + b x \right )}}{3 b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {3}{2}}} + \frac {\tanh {\left (a + b x \right )}}{b \left (\operatorname {sech}^{2}{\left (a + b x \right )}\right )^{\frac {3}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\operatorname {sech}^{2}{\relax (a )}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

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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