3.30 \(\int \frac{1}{\text{sech}^2(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac{8 \tanh (a+b x)}{15 b \sqrt{\text{sech}^2(a+b x)}}+\frac{4 \tanh (a+b x)}{15 b \text{sech}^2(a+b x)^{3/2}}+\frac{\tanh (a+b x)}{5 b \text{sech}^2(a+b x)^{5/2}} \]

[Out]

Tanh[a + b*x]/(5*b*(Sech[a + b*x]^2)^(5/2)) + (4*Tanh[a + b*x])/(15*b*(Sech[a + b*x]^2)^(3/2)) + (8*Tanh[a + b
*x])/(15*b*Sqrt[Sech[a + b*x]^2])

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Rubi [A]  time = 0.0266368, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4122, 192, 191} \[ \frac{8 \tanh (a+b x)}{15 b \sqrt{\text{sech}^2(a+b x)}}+\frac{4 \tanh (a+b x)}{15 b \text{sech}^2(a+b x)^{3/2}}+\frac{\tanh (a+b x)}{5 b \text{sech}^2(a+b x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(-5/2),x]

[Out]

Tanh[a + b*x]/(5*b*(Sech[a + b*x]^2)^(5/2)) + (4*Tanh[a + b*x])/(15*b*(Sech[a + b*x]^2)^(3/2)) + (8*Tanh[a + b
*x])/(15*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)^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{7/2}} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac{\tanh (a+b x)}{5 b \text{sech}^2(a+b x)^{5/2}}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{5/2}} \, dx,x,\tanh (a+b x)\right )}{5 b}\\ &=\frac{\tanh (a+b x)}{5 b \text{sech}^2(a+b x)^{5/2}}+\frac{4 \tanh (a+b x)}{15 b \text{sech}^2(a+b x)^{3/2}}+\frac{8 \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{3/2}} \, dx,x,\tanh (a+b x)\right )}{15 b}\\ &=\frac{\tanh (a+b x)}{5 b \text{sech}^2(a+b x)^{5/2}}+\frac{4 \tanh (a+b x)}{15 b \text{sech}^2(a+b x)^{3/2}}+\frac{8 \tanh (a+b x)}{15 b \sqrt{\text{sech}^2(a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0817697, size = 47, normalized size = 0.62 \[ \frac{\left (3 \sinh ^4(a+b x)+10 \sinh ^2(a+b x)+15\right ) \tanh (a+b x)}{15 b \sqrt{\text{sech}^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(-5/2),x]

[Out]

((15 + 10*Sinh[a + b*x]^2 + 3*Sinh[a + b*x]^4)*Tanh[a + b*x])/(15*b*Sqrt[Sech[a + b*x]^2])

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Maple [B]  time = 0.098, size = 305, normalized size = 4. \begin{align*}{\frac{{{\rm e}^{6\,bx+6\,a}}}{ \left ( 160+160\,{{\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{5\,{{\rm e}^{4\,bx+4\,a}}}{ \left ( 96+96\,{{\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{5\,{{\rm e}^{2\,bx+2\,a}}}{ \left ( 16+16\,{{\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{5}{ \left ( 16+16\,{{\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{5\,{{\rm e}^{-2\,bx-2\,a}}}{ \left ( 96+96\,{{\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}^{-4\,bx-4\,a}}}{ \left ( 160+160\,{{\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)^(5/2),x)

[Out]

1/160/(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(6*b*x+6*a)+5/96/(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)+5/16/(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)-5/16/(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-5/96/(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)-1/160/(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)

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Maxima [A]  time = 1.03074, size = 111, normalized size = 1.46 \begin{align*} \frac{e^{\left (5 \, b x + 5 \, a\right )}}{160 \, b} + \frac{5 \, e^{\left (3 \, b x + 3 \, a\right )}}{96 \, b} + \frac{5 \, e^{\left (b x + a\right )}}{16 \, b} - \frac{5 \, e^{\left (-b x - a\right )}}{16 \, b} - \frac{5 \, e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} - \frac{e^{\left (-5 \, b x - 5 \, a\right )}}{160 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="maxima")

[Out]

1/160*e^(5*b*x + 5*a)/b + 5/96*e^(3*b*x + 3*a)/b + 5/16*e^(b*x + a)/b - 5/16*e^(-b*x - a)/b - 5/96*e^(-3*b*x -
3*a)/b - 1/160*e^(-5*b*x - 5*a)/b

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Fricas [A]  time = 2.12512, size = 182, normalized size = 2.39 \begin{align*} \frac{3 \, \sinh \left (b x + a\right )^{5} + 5 \,{\left (6 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 15 \,{\left (\cosh \left (b x + a\right )^{4} + 5 \, \cosh \left (b x + a\right )^{2} + 10\right )} \sinh \left (b x + a\right )}{240 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="fricas")

[Out]

1/240*(3*sinh(b*x + a)^5 + 5*(6*cosh(b*x + a)^2 + 5)*sinh(b*x + a)^3 + 15*(cosh(b*x + a)^4 + 5*cosh(b*x + a)^2
+ 10)*sinh(b*x + a))/b

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Sympy [A]  time = 54.5431, size = 80, normalized size = 1.05 \begin{align*} \begin{cases} \frac{8 \tanh ^{5}{\left (a + b x \right )}}{15 b \left (\operatorname{sech}^{2}{\left (a + b x \right )}\right )^{\frac{5}{2}}} - \frac{4 \tanh ^{3}{\left (a + b x \right )}}{3 b \left (\operatorname{sech}^{2}{\left (a + b x \right )}\right )^{\frac{5}{2}}} + \frac{\tanh{\left (a + b x \right )}}{b \left (\operatorname{sech}^{2}{\left (a + b x \right )}\right )^{\frac{5}{2}}} & \text{for}\: b \neq 0 \\\frac{x}{\left (\operatorname{sech}^{2}{\left (a \right )}\right )^{\frac{5}{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)**(5/2),x)

[Out]

Piecewise((8*tanh(a + b*x)**5/(15*b*(sech(a + b*x)**2)**(5/2)) - 4*tanh(a + b*x)**3/(3*b*(sech(a + b*x)**2)**(
5/2)) + tanh(a + b*x)/(b*(sech(a + b*x)**2)**(5/2)), Ne(b, 0)), (x/(sech(a)**2)**(5/2), True))

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Giac [A]  time = 1.12807, size = 95, normalized size = 1.25 \begin{align*} -\frac{{\left (150 \, e^{\left (4 \, b x + 4 \, a\right )} + 25 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )} e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (5 \, b x + 5 \, a\right )} - 25 \, e^{\left (3 \, b x + 3 \, a\right )} - 150 \, e^{\left (b x + a\right )}}{480 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(b*x+a)^2)^(5/2),x, algorithm="giac")

[Out]

-1/480*((150*e^(4*b*x + 4*a) + 25*e^(2*b*x + 2*a) + 3)*e^(-5*b*x - 5*a) - 3*e^(5*b*x + 5*a) - 25*e^(3*b*x + 3*
a) - 150*e^(b*x + a))/b