3.1.16 \(\int (b \text {sech}(c+d x))^{5/2} \, dx\) [16]

Optimal. Leaf size=74 \[ -\frac {2 i b^2 \sqrt {\cosh (c+d x)} F\left (\left .\frac {1}{2} i (c+d x)\right |2\right ) \sqrt {b \text {sech}(c+d x)}}{3 d}+\frac {2 b (b \text {sech}(c+d x))^{3/2} \sinh (c+d x)}{3 d} \]

[Out]

2/3*b*(b*sech(d*x+c))^(3/2)*sinh(d*x+c)/d-2/3*I*b^2*(cosh(1/2*d*x+1/2*c)^2)^(1/2)/cosh(1/2*d*x+1/2*c)*Elliptic
F(I*sinh(1/2*d*x+1/2*c),2^(1/2))*cosh(d*x+c)^(1/2)*(b*sech(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3853, 3856, 2720} \begin {gather*} \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}-\frac {2 i b^2 \sqrt {\cosh (c+d x)} F\left (\left .\frac {1}{2} i (c+d x)\right |2\right ) \sqrt {b \text {sech}(c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sech[c + d*x])^(5/2),x]

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 56, normalized size = 0.76 \begin {gather*} \frac {2 b^2 \sqrt {b \text {sech}(c+d x)} \left (-i \sqrt {\cosh (c+d x)} F\left (\left .\frac {1}{2} i (c+d x)\right |2\right )+\tanh (c+d x)\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sech[c + d*x])^(5/2),x]

[Out]

(2*b^2*Sqrt[b*Sech[c + d*x]]*((-I)*Sqrt[Cosh[c + d*x]]*EllipticF[(I/2)*(c + d*x), 2] + Tanh[c + d*x]))/(3*d)

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Maple [F]
time = 1.33, size = 0, normalized size = 0.00 \[\int \left (b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {5}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sech(d*x+c))^(5/2),x)

[Out]

int((b*sech(d*x+c))^(5/2),x)

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

[Out]

integrate((b*sech(d*x + c))^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 215, normalized size = 2.91 \begin {gather*} \frac {2 \, {\left (\sqrt {2} {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + \sqrt {2} {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1}}\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(sqrt(2)*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + b^2)*sqrt(b)*wei
erstrassPInverse(-4, 0, cosh(d*x + c) + sinh(d*x + c)) + sqrt(2)*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*si
nh(d*x + c) + b^2*sinh(d*x + c)^2 - b^2)*sqrt((b*cosh(d*x + c) + b*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*
x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*
x + c)^2 + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sech(c + d*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((b*sech(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sech(d*x + c))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cosh(c + d*x))^(5/2),x)

[Out]

int((b/cosh(c + d*x))^(5/2), x)

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