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

Optimal. Leaf size=65 \[ \frac{3}{8} a^2 \tanh (x) \sqrt{a \text{sech}^2(x)}+\frac{3}{8} a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a \text{sech}^2(x)}}\right )+\frac{1}{4} a \tanh (x) \left (a \text{sech}^2(x)\right )^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.0336254, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4122, 195, 217, 203} \[ \frac{3}{8} a^2 \tanh (x) \sqrt{a \text{sech}^2(x)}+\frac{3}{8} a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a \text{sech}^2(x)}}\right )+\frac{1}{4} a \tanh (x) \left (a \text{sech}^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a \text{sech}^2(x)\right )^{5/2} \, dx &=a \operatorname{Subst}\left (\int \left (a-a x^2\right )^{3/2} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{4} a \left (a \text{sech}^2(x)\right )^{3/2} \tanh (x)+\frac{1}{4} \left (3 a^2\right ) \operatorname{Subst}\left (\int \sqrt{a-a x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{3}{8} a^2 \sqrt{a \text{sech}^2(x)} \tanh (x)+\frac{1}{4} a \left (a \text{sech}^2(x)\right )^{3/2} \tanh (x)+\frac{1}{8} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x^2}} \, dx,x,\tanh (x)\right )\\ &=\frac{3}{8} a^2 \sqrt{a \text{sech}^2(x)} \tanh (x)+\frac{1}{4} a \left (a \text{sech}^2(x)\right )^{3/2} \tanh (x)+\frac{1}{8} \left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{a \text{sech}^2(x)}}\right )\\ &=\frac{3}{8} a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a \text{sech}^2(x)}}\right )+\frac{3}{8} a^2 \sqrt{a \text{sech}^2(x)} \tanh (x)+\frac{1}{4} a \left (a \text{sech}^2(x)\right )^{3/2} \tanh (x)\\ \end{align*}

Mathematica [A]  time = 0.0341739, size = 42, normalized size = 0.65 \[ \frac{1}{8} \cosh (x) \left (a \text{sech}^2(x)\right )^{5/2} \left (2 \sinh (x)+3 \sinh (x) \cosh ^2(x)+6 \cosh ^4(x) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[x]*(a*Sech[x]^2)^(5/2)*(6*ArcTan[Tanh[x/2]]*Cosh[x]^4 + 2*Sinh[x] + 3*Cosh[x]^2*Sinh[x]))/8

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Maple [C]  time = 0.08, size = 127, normalized size = 2. \begin{align*}{\frac{{a}^{2} \left ( 3\,{{\rm e}^{6\,x}}+11\,{{\rm e}^{4\,x}}-11\,{{\rm e}^{2\,x}}-3 \right ) }{4\, \left ({{\rm e}^{2\,x}}+1 \right ) ^{3}}\sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}+{\frac{3\,i}{8}}{a}^{2}{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}+1 \right ) \sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}\ln \left ({{\rm e}^{x}}+i \right ) -{\frac{3\,i}{8}}{a}^{2}{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}+1 \right ) \sqrt{{\frac{a{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}\ln \left ({{\rm e}^{x}}-i \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2/(exp(2*x)+1)^3*(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)*(3*exp(6*x)+11*exp(4*x)-11*exp(2*x)-3)+3/8*I*a^2*exp(
-x)*(exp(2*x)+1)*(a*exp(2*x)/(exp(2*x)+1)^2)^(1/2)*ln(exp(x)+I)-3/8*I*a^2*exp(-x)*(exp(2*x)+1)*(a*exp(2*x)/(ex
p(2*x)+1)^2)^(1/2)*ln(exp(x)-I)

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Maxima [A]  time = 1.68599, size = 97, normalized size = 1.49 \begin{align*} \frac{3}{4} \, a^{\frac{5}{2}} \arctan \left (e^{x}\right ) + \frac{3 \, a^{\frac{5}{2}} e^{\left (7 \, x\right )} + 11 \, a^{\frac{5}{2}} e^{\left (5 \, x\right )} - 11 \, a^{\frac{5}{2}} e^{\left (3 \, x\right )} - 3 \, a^{\frac{5}{2}} e^{x}}{4 \,{\left (e^{\left (8 \, x\right )} + 4 \, e^{\left (6 \, x\right )} + 6 \, e^{\left (4 \, x\right )} + 4 \, e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.23797, size = 3067, normalized size = 47.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*a^2*cosh(x)^7 + 3*(a^2*e^(2*x) + a^2)*sinh(x)^7 + 11*a^2*cosh(x)^5 + 21*(a^2*cosh(x)*e^(2*x) + a^2*cosh
(x))*sinh(x)^6 + (63*a^2*cosh(x)^2 + 11*a^2 + (63*a^2*cosh(x)^2 + 11*a^2)*e^(2*x))*sinh(x)^5 - 11*a^2*cosh(x)^
3 + 5*(21*a^2*cosh(x)^3 + 11*a^2*cosh(x) + (21*a^2*cosh(x)^3 + 11*a^2*cosh(x))*e^(2*x))*sinh(x)^4 + (105*a^2*c
osh(x)^4 + 110*a^2*cosh(x)^2 - 11*a^2 + (105*a^2*cosh(x)^4 + 110*a^2*cosh(x)^2 - 11*a^2)*e^(2*x))*sinh(x)^3 -
3*a^2*cosh(x) + (63*a^2*cosh(x)^5 + 110*a^2*cosh(x)^3 - 33*a^2*cosh(x) + (63*a^2*cosh(x)^5 + 110*a^2*cosh(x)^3
- 33*a^2*cosh(x))*e^(2*x))*sinh(x)^2 + 3*(a^2*cosh(x)^8 + (a^2*e^(2*x) + a^2)*sinh(x)^8 + 4*a^2*cosh(x)^6 + 8
*(a^2*cosh(x)*e^(2*x) + a^2*cosh(x))*sinh(x)^7 + 4*(7*a^2*cosh(x)^2 + a^2 + (7*a^2*cosh(x)^2 + a^2)*e^(2*x))*s
inh(x)^6 + 6*a^2*cosh(x)^4 + 8*(7*a^2*cosh(x)^3 + 3*a^2*cosh(x) + (7*a^2*cosh(x)^3 + 3*a^2*cosh(x))*e^(2*x))*s
inh(x)^5 + 2*(35*a^2*cosh(x)^4 + 30*a^2*cosh(x)^2 + 3*a^2 + (35*a^2*cosh(x)^4 + 30*a^2*cosh(x)^2 + 3*a^2)*e^(2
*x))*sinh(x)^4 + 4*a^2*cosh(x)^2 + 8*(7*a^2*cosh(x)^5 + 10*a^2*cosh(x)^3 + 3*a^2*cosh(x) + (7*a^2*cosh(x)^5 +
10*a^2*cosh(x)^3 + 3*a^2*cosh(x))*e^(2*x))*sinh(x)^3 + 4*(7*a^2*cosh(x)^6 + 15*a^2*cosh(x)^4 + 9*a^2*cosh(x)^2
+ a^2 + (7*a^2*cosh(x)^6 + 15*a^2*cosh(x)^4 + 9*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^2 + a^2 + (a^2*cosh(x)^
8 + 4*a^2*cosh(x)^6 + 6*a^2*cosh(x)^4 + 4*a^2*cosh(x)^2 + a^2)*e^(2*x) + 8*(a^2*cosh(x)^7 + 3*a^2*cosh(x)^5 +
3*a^2*cosh(x)^3 + a^2*cosh(x) + (a^2*cosh(x)^7 + 3*a^2*cosh(x)^5 + 3*a^2*cosh(x)^3 + a^2*cosh(x))*e^(2*x))*sin
h(x))*arctan(cosh(x) + sinh(x)) + (3*a^2*cosh(x)^7 + 11*a^2*cosh(x)^5 - 11*a^2*cosh(x)^3 - 3*a^2*cosh(x))*e^(2
*x) + (21*a^2*cosh(x)^6 + 55*a^2*cosh(x)^4 - 33*a^2*cosh(x)^2 - 3*a^2 + (21*a^2*cosh(x)^6 + 55*a^2*cosh(x)^4 -
33*a^2*cosh(x)^2 - 3*a^2)*e^(2*x))*sinh(x))*sqrt(a/(e^(4*x) + 2*e^(2*x) + 1))*e^x/(8*cosh(x)*e^x*sinh(x)^7 +
e^x*sinh(x)^8 + 4*(7*cosh(x)^2 + 1)*e^x*sinh(x)^6 + 8*(7*cosh(x)^3 + 3*cosh(x))*e^x*sinh(x)^5 + 2*(35*cosh(x)^
4 + 30*cosh(x)^2 + 3)*e^x*sinh(x)^4 + 8*(7*cosh(x)^5 + 10*cosh(x)^3 + 3*cosh(x))*e^x*sinh(x)^3 + 4*(7*cosh(x)^
6 + 15*cosh(x)^4 + 9*cosh(x)^2 + 1)*e^x*sinh(x)^2 + 8*(cosh(x)^7 + 3*cosh(x)^5 + 3*cosh(x)^3 + cosh(x))*e^x*si
nh(x) + (cosh(x)^8 + 4*cosh(x)^6 + 6*cosh(x)^4 + 4*cosh(x)^2 + 1)*e^x)

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

[Out]

Timed out

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Giac [A]  time = 1.1358, size = 88, normalized size = 1.35 \begin{align*} \frac{1}{16} \,{\left (3 \, \pi - \frac{4 \,{\left (3 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 20 \, e^{\left (-x\right )} - 20 \, e^{x}\right )}}{{\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2}} + 6 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} a^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*pi - 4*(3*(e^(-x) - e^x)^3 + 20*e^(-x) - 20*e^x)/((e^(-x) - e^x)^2 + 4)^2 + 6*arctan(1/2*(e^(2*x) - 1)
*e^(-x)))*a^(5/2)