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

Optimal. Leaf size=65 \[ \frac{3 \sin ^{-1}(\tanh (a+b x))}{8 b}+\frac{\tanh (a+b x) \text{sech}^2(a+b x)^{3/2}}{4 b}+\frac{3 \tanh (a+b x) \sqrt{\text{sech}^2(a+b x)}}{8 b} \]

[Out]

(3*ArcSin[Tanh[a + b*x]])/(8*b) + (3*Sqrt[Sech[a + b*x]^2]*Tanh[a + b*x])/(8*b) + ((Sech[a + b*x]^2)^(3/2)*Tan
h[a + b*x])/(4*b)

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Rubi [A]  time = 0.0220126, antiderivative size = 65, 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, 195, 216} \[ \frac{3 \sin ^{-1}(\tanh (a+b x))}{8 b}+\frac{\tanh (a+b x) \text{sech}^2(a+b x)^{3/2}}{4 b}+\frac{3 \tanh (a+b x) \sqrt{\text{sech}^2(a+b x)}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[a + b*x]^2)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.114824, size = 55, normalized size = 0.85 \[ \frac{\text{sech}^2(a+b x)^{3/2} \left (3 \sinh (2 (a+b x))+4 \tanh (a+b x)+6 \cosh ^3(a+b x) \tan ^{-1}(\sinh (a+b x))\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[a + b*x]^2)^(5/2),x]

[Out]

((Sech[a + b*x]^2)^(3/2)*(6*ArcTan[Sinh[a + b*x]]*Cosh[a + b*x]^3 + 3*Sinh[2*(a + b*x)] + 4*Tanh[a + b*x]))/(1
6*b)

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Maple [C]  time = 0.129, size = 208, normalized size = 3.2 \begin{align*}{\frac{3\,{{\rm e}^{6\,bx+6\,a}}+11\,{{\rm e}^{4\,bx+4\,a}}-11\,{{\rm e}^{2\,bx+2\,a}}-3}{4\, \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{3}b}\sqrt{{\frac{{{\rm e}^{2\,bx+2\,a}}}{ \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}}}+{\frac{{\frac{3\,i}{8}} \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{3\,i}{8}} \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)^(5/2),x)

[Out]

1/4/(1+exp(2*b*x+2*a))^3*(1/(1+exp(2*b*x+2*a))^2*exp(2*b*x+2*a))^(1/2)*(3*exp(6*b*x+6*a)+11*exp(4*b*x+4*a)-11*
exp(2*b*x+2*a)-3)/b+3/8*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)-3/8*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 [B]  time = 1.5103, size = 151, normalized size = 2.32 \begin{align*} -\frac{3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac{3 \, e^{\left (-b x - a\right )} + 11 \, e^{\left (-3 \, b x - 3 \, a\right )} - 11 \, e^{\left (-5 \, b x - 5 \, a\right )} - 3 \, e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b{\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*arctan(e^(-b*x - a))/b + 1/4*(3*e^(-b*x - a) + 11*e^(-3*b*x - 3*a) - 11*e^(-5*b*x - 5*a) - 3*e^(-7*b*x -
7*a))/(b*(4*e^(-2*b*x - 2*a) + 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 1))

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Fricas [B]  time = 2.20845, size = 2263, normalized size = 34.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*cosh(b*x + a)^7 + 21*cosh(b*x + a)*sinh(b*x + a)^6 + 3*sinh(b*x + a)^7 + (63*cosh(b*x + a)^2 + 11)*sinh
(b*x + a)^5 + 11*cosh(b*x + a)^5 + 5*(21*cosh(b*x + a)^3 + 11*cosh(b*x + a))*sinh(b*x + a)^4 + (105*cosh(b*x +
a)^4 + 110*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^3 - 11*cosh(b*x + a)^3 + (63*cosh(b*x + a)^5 + 110*cosh(b*x +
a)^3 - 33*cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a
)^8 + 4*(7*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 + 3*cosh(b*x + a))*
sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 + 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*c
osh(b*x + a)^5 + 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 + 15*cosh(b*x +
a)^4 + 9*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 + 3*cosh(b*x + a)^5 + 3
*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (21*cosh(b*x + a)
^6 + 55*cosh(b*x + a)^4 - 33*cosh(b*x + a)^2 - 3)*sinh(b*x + a) - 3*cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*co
sh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 + 4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 + b)*sinh(b*x +
a)^6 + 8*(7*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x +
a)^4 + 30*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 + 10*b*cosh(b*x + a)^3 + 3*b*cosh
(b*x + a))*sinh(b*x + a)^3 + 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*x + a)^6 + 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*
x + a)^2 + b)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 + 3*b*cosh(b*x + a)^5 + 3*b*cosh(b*x + a)^3 + b*cosh(b*x
+ a))*sinh(b*x + a) + b)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(b*x+a)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14046, size = 140, normalized size = 2.15 \begin{align*} \frac{3 \,{\left (\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 )\right )}}{16 \, b} + \frac{3 \,{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 20 \, e^{\left (b x + a\right )} - 20 \, e^{\left (-b x - a\right )}}{4 \,{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b + 1/4*(3*(e^(b*x + a) - e^(-b*x - a))^3 + 20*e^
(b*x + a) - 20*e^(-b*x - a))/(((e^(b*x + a) - e^(-b*x - a))^2 + 4)^2*b)