\(\int (a \text {sech}^2(x))^{5/2} \, dx\) [32]

Optimal result

Integrand size = 10, antiderivative size = 65 \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\frac {3}{8} a^{5/2} \arctan \left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a \text {sech}^2(x)}}\right )+\frac {3}{8} a^2 \sqrt {a \text {sech}^2(x)} \tanh (x)+\frac {1}{4} a \left (a \text {sech}^2(x)\right )^{3/2} \tanh (x) \]

[Out]

3/8*a^(5/2)*arctan(a^(1/2)*tanh(x)/(a*sech(x)^2)^(1/2))+1/4*a*(a*sech(x)^2)^(3/2)*tanh(x)+3/8*a^2*(a*sech(x)^2
)^(1/2)*tanh(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4207, 201, 223, 209} \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\frac {3}{8} a^{5/2} \arctan \left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a \text {sech}^2(x)}}\right )+\frac {3}{8} a^2 \tanh (x) \sqrt {a \text {sech}^2(x)}+\frac {1}{4} a \tanh (x) \left (a \text {sech}^2(x)\right )^{3/2} \]

[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 201

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 209

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

Rule 223

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 4207

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*} \text {integral}& = a \text {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 ) \text {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 ) \text {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 ) \text {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} \arctan \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] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.58 \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\frac {1}{8} \cosh (x) \left (a \text {sech}^2(x)\right )^{5/2} \left (3 \arctan (\sinh (x)) \cosh ^4(x)+2 \sinh (x)+3 \cosh ^2(x) \sinh (x)\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.91 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.95

method	result	size
risch	\(\frac {a^{2} \sqrt {\frac {{\mathrm e}^{2 x} a}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 x}+11 \,{\mathrm e}^{4 x}-11 \,{\mathrm e}^{2 x}-3\right )}{4 \left (1+{\mathrm e}^{2 x}\right )^{3}}+\frac {3 i a^{2} {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {{\mathrm e}^{2 x} a}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \ln \left ({\mathrm e}^{x}+i\right )}{8}-\frac {3 i a^{2} {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {{\mathrm e}^{2 x} a}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \ln \left ({\mathrm e}^{x}-i\right )}{8}\)	\(127\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1082 vs. \(2 (49) = 98\).

Time = 0.28 (sec) , antiderivative size = 1082, normalized size of antiderivative = 16.65 \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\text {Too large to display} \]

[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)

Sympy [F]

\[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\int \left (a \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

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

[Out]

Integral((a*sech(x)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\frac {3}{4} \, a^{\frac {5}{2}} \arctan \left (e^{x}\right ) + \frac {3 \, a^{\frac {5}{2}} e^{\left (7 \, x\right )} + 11 \, a^{\frac {5}{2}} e^{\left (5 \, x\right )} - 11 \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} - 3 \, a^{\frac {5}{2}} e^{x}}{4 \, {\left (e^{\left (8 \, x\right )} + 4 \, e^{\left (6 \, x\right )} + 6 \, e^{\left (4 \, x\right )} + 4 \, e^{\left (2 \, x\right )} + 1\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\frac {1}{16} \, {\left (3 \, \pi - \frac {4 \, {\left (3 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 20 \, e^{\left (-x\right )} - 20 \, e^{x}\right )}}{{\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2}} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} a^{\frac {5}{2}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (a \text {sech}^2(x)\right )^{5/2} \, dx=\int {\left (\frac {a}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2} \,d x \]

[In]

int((a/cosh(x)^2)^(5/2),x)

[Out]

int((a/cosh(x)^2)^(5/2), x)