3.1.9 \(\int \text {sech}^{\frac {5}{2}}(a+b x) \, dx\) [9]

Optimal. Leaf size=66 \[ -\frac {2 i \sqrt {\cosh (a+b x)} F\left (\left .\frac {1}{2} i (a+b x)\right |2\right ) \sqrt {\text {sech}(a+b x)}}{3 b}+\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \sinh (a+b x)}{3 b} \]

[Out]

2/3*sech(b*x+a)^(3/2)*sinh(b*x+a)/b-2/3*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/2*a+1/2*b*x)*EllipticF(I*sinh(1
/2*a+1/2*b*x),2^(1/2))*cosh(b*x+a)^(1/2)*sech(b*x+a)^(1/2)/b

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Rubi [A]
time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3853, 3856, 2720} \begin {gather*} \frac {2 \sinh (a+b x) \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {2 i \sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)} F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^(5/2),x]

[Out]

(((-2*I)/3)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2]*Sqrt[Sech[a + b*x]])/b + (2*Sech[a + b*x]^(3/2)*
Sinh[a + b*x])/(3*b)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx &=\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \sinh (a+b x)}{3 b}+\frac {1}{3} \int \sqrt {\text {sech}(a+b x)} \, dx\\ &=\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \sinh (a+b x)}{3 b}+\frac {1}{3} \left (\sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)}\right ) \int \frac {1}{\sqrt {\cosh (a+b x)}} \, dx\\ &=-\frac {2 i \sqrt {\cosh (a+b x)} F\left (\left .\frac {1}{2} i (a+b x)\right |2\right ) \sqrt {\text {sech}(a+b x)}}{3 b}+\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \sinh (a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 51, normalized size = 0.77 \begin {gather*} \frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \left (-i \cosh ^{\frac {3}{2}}(a+b x) F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )+\sinh (a+b x)\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^(5/2),x]

[Out]

(2*Sech[a + b*x]^(3/2)*((-I)*Cosh[a + b*x]^(3/2)*EllipticF[(I/2)*(a + b*x), 2] + Sinh[a + b*x]))/(3*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(216\) vs. \(2(82)=164\).
time = 1.46, size = 217, normalized size = 3.29

method	result	size
default	\(\frac {2 \left (2 \sqrt {-\left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \sqrt {-2 \left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cosh \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sqrt {-\left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \sqrt {-2 \left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cosh \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \left (\cosh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}}{3 \sqrt {2 \left (\sinh ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sinh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \left (2 \left (\cosh ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )^{\frac {3}{2}} \sinh \left (\frac {b x}{2}+\frac {a}{2}\right ) b}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(2*(-sinh(1/2*b*x+1/2*a)^2)^(1/2)*(-2*sinh(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cosh(1/2*b*x+1/2*a),2^(1/2)
)*sinh(1/2*b*x+1/2*a)^2+2*cosh(1/2*b*x+1/2*a)*sinh(1/2*b*x+1/2*a)^2+(-sinh(1/2*b*x+1/2*a)^2)^(1/2)*(-2*sinh(1/
2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cosh(1/2*b*x+1/2*a),2^(1/2)))*((2*cosh(1/2*b*x+1/2*a)^2-1)*sinh(1/2*b*x+1/2*
a)^2)^(1/2)/(2*sinh(1/2*b*x+1/2*a)^4+sinh(1/2*b*x+1/2*a)^2)^(1/2)/(2*cosh(1/2*b*x+1/2*a)^2-1)^(3/2)/sinh(1/2*b
*x+1/2*a)/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sech(b*x + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.15, size = 190, normalized size = 2.88 \begin {gather*} \frac {2 \, {\left (\sqrt {2} {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \sqrt {\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}} + {\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2} + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(sqrt(2)*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*sqrt((cosh(b*x + a) + sin
h(b*x + a))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) + (sqrt(2)*cosh(b*x + a)^
2 + 2*sqrt(2)*cosh(b*x + a)*sinh(b*x + a) + sqrt(2)*sinh(b*x + a)^2 + sqrt(2))*weierstrassPInverse(-4, 0, cosh
(b*x + a) + sinh(b*x + a)))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 + b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {sech}^{\frac {5}{2}}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(a + b*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sech(b*x + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (\frac {1}{\mathrm {cosh}\left (a+b\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cosh(a + b*x))^(5/2),x)

[Out]

int((1/cosh(a + b*x))^(5/2), x)

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