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

Optimal. Leaf size=55 \[ \frac{8 \tanh (x)}{15 a^2 \sqrt{a \text{sech}^2(x)}}+\frac{4 \tanh (x)}{15 a \left (a \text{sech}^2(x)\right )^{3/2}}+\frac{\tanh (x)}{5 \left (a \text{sech}^2(x)\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0291431, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4122, 192, 191} \[ \frac{8 \tanh (x)}{15 a^2 \sqrt{a \text{sech}^2(x)}}+\frac{4 \tanh (x)}{15 a \left (a \text{sech}^2(x)\right )^{3/2}}+\frac{\tanh (x)}{5 \left (a \text{sech}^2(x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sech[x]^2)^(-5/2),x]

[Out]

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

Mathematica [A]  time = 0.0388786, size = 36, normalized size = 0.65 \[ \frac{(150 \sinh (x)+25 \sinh (3 x)+3 \sinh (5 x)) \cosh (x) \sqrt{a \text{sech}^2(x)}}{240 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sech[x]^2)^(-5/2),x]

[Out]

(Cosh[x]*Sqrt[a*Sech[x]^2]*(150*Sinh[x] + 25*Sinh[3*x] + 3*Sinh[5*x]))/(240*a^3)

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Maple [B]  time = 0.054, size = 196, normalized size = 3.6 \begin{align*}{\frac{{{\rm e}^{6\,x}}}{160\,{a}^{2} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+{\frac{5\,{{\rm e}^{4\,x}}}{96\,{a}^{2} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}+{\frac{5\,{{\rm e}^{2\,x}}}{16\,{a}^{2} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{5}{16\,{a}^{2} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{5\,{{\rm e}^{-2\,x}}}{96\,{a}^{2} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}}-{\frac{{{\rm e}^{-4\,x}}}{160\,{a}^{2} \left ({{\rm e}^{2\,x}}+1 \right ) }{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sech(x)^2)^(5/2),x)

[Out]

1/160/a^2*exp(6*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)+5/96/a^2*exp(4*x)/(exp(2*x)+1)/(a*exp(2*x)/(
exp(2*x)+1)^2)^(1/2)+5/16/a^2*exp(2*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)-5/16/a^2/(exp(2*x)+1)/(a
*exp(2*x)/(exp(2*x)+1)^2)^(1/2)-5/96/a^2*exp(-2*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)-1/160/a^2*ex
p(-4*x)/(exp(2*x)+1)/(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)

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Maxima [A]  time = 1.82914, size = 72, normalized size = 1.31 \begin{align*} \frac{e^{\left (5 \, x\right )}}{160 \, a^{\frac{5}{2}}} + \frac{5 \, e^{\left (3 \, x\right )}}{96 \, a^{\frac{5}{2}}} - \frac{5 \, e^{\left (-x\right )}}{16 \, a^{\frac{5}{2}}} - \frac{5 \, e^{\left (-3 \, x\right )}}{96 \, a^{\frac{5}{2}}} - \frac{e^{\left (-5 \, x\right )}}{160 \, a^{\frac{5}{2}}} + \frac{5 \, e^{x}}{16 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*e^(5*x)/a^(5/2) + 5/96*e^(3*x)/a^(5/2) - 5/16*e^(-x)/a^(5/2) - 5/96*e^(-3*x)/a^(5/2) - 1/160*e^(-5*x)/a^
(5/2) + 5/16*e^x/a^(5/2)

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Fricas [B]  time = 2.14912, size = 1885, normalized size = 34.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*(e^(2*x) + 1)*sinh(x)^10 + 3*cosh(x)^10 + 30*(cosh(x)*e^(2*x) + cosh(x))*sinh(x)^9 + 5*(27*cosh(x)^2
+ (27*cosh(x)^2 + 5)*e^(2*x) + 5)*sinh(x)^8 + 25*cosh(x)^8 + 40*(9*cosh(x)^3 + (9*cosh(x)^3 + 5*cosh(x))*e^(2*
x) + 5*cosh(x))*sinh(x)^7 + 10*(63*cosh(x)^4 + 70*cosh(x)^2 + (63*cosh(x)^4 + 70*cosh(x)^2 + 15)*e^(2*x) + 15)
*sinh(x)^6 + 150*cosh(x)^6 + 4*(189*cosh(x)^5 + 350*cosh(x)^3 + (189*cosh(x)^5 + 350*cosh(x)^3 + 225*cosh(x))*
e^(2*x) + 225*cosh(x))*sinh(x)^5 + 10*(63*cosh(x)^6 + 175*cosh(x)^4 + 225*cosh(x)^2 + (63*cosh(x)^6 + 175*cosh
(x)^4 + 225*cosh(x)^2 - 15)*e^(2*x) - 15)*sinh(x)^4 - 150*cosh(x)^4 + 40*(9*cosh(x)^7 + 35*cosh(x)^5 + 75*cosh
(x)^3 + (9*cosh(x)^7 + 35*cosh(x)^5 + 75*cosh(x)^3 - 15*cosh(x))*e^(2*x) - 15*cosh(x))*sinh(x)^3 + 5*(27*cosh(
x)^8 + 140*cosh(x)^6 + 450*cosh(x)^4 - 180*cosh(x)^2 + (27*cosh(x)^8 + 140*cosh(x)^6 + 450*cosh(x)^4 - 180*cos
h(x)^2 - 5)*e^(2*x) - 5)*sinh(x)^2 - 25*cosh(x)^2 + (3*cosh(x)^10 + 25*cosh(x)^8 + 150*cosh(x)^6 - 150*cosh(x)
^4 - 25*cosh(x)^2 - 3)*e^(2*x) + 10*(3*cosh(x)^9 + 20*cosh(x)^7 + 90*cosh(x)^5 - 60*cosh(x)^3 + (3*cosh(x)^9 +
20*cosh(x)^7 + 90*cosh(x)^5 - 60*cosh(x)^3 - 5*cosh(x))*e^(2*x) - 5*cosh(x))*sinh(x) - 3)*sqrt(a/(e^(4*x) + 2
*e^(2*x) + 1))*e^x/(a^3*cosh(x)^5*e^x + 5*a^3*cosh(x)^4*e^x*sinh(x) + 10*a^3*cosh(x)^3*e^x*sinh(x)^2 + 10*a^3*
cosh(x)^2*e^x*sinh(x)^3 + 5*a^3*cosh(x)*e^x*sinh(x)^4 + a^3*e^x*sinh(x)^5)

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Sympy [A]  time = 15.0781, size = 60, normalized size = 1.09 \begin{align*} \frac{8 \tanh ^{5}{\left (x \right )}}{15 a^{\frac{5}{2}} \left (\operatorname{sech}^{2}{\left (x \right )}\right )^{\frac{5}{2}}} - \frac{4 \tanh ^{3}{\left (x \right )}}{3 a^{\frac{5}{2}} \left (\operatorname{sech}^{2}{\left (x \right )}\right )^{\frac{5}{2}}} + \frac{\tanh{\left (x \right )}}{a^{\frac{5}{2}} \left (\operatorname{sech}^{2}{\left (x \right )}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)**2)**(5/2),x)

[Out]

8*tanh(x)**5/(15*a**(5/2)*(sech(x)**2)**(5/2)) - 4*tanh(x)**3/(3*a**(5/2)*(sech(x)**2)**(5/2)) + tanh(x)/(a**(
5/2)*(sech(x)**2)**(5/2))

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Giac [A]  time = 1.13454, size = 55, normalized size = 1. \begin{align*} -\frac{{\left (150 \, e^{\left (4 \, x\right )} + 25 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-5 \, x\right )} - 3 \, e^{\left (5 \, x\right )} - 25 \, e^{\left (3 \, x\right )} - 150 \, e^{x}}{480 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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