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

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

[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)

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Rubi [A]  time = 0.031515, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3768, 3771, 2641} \[ \frac{2 \sinh (a+b x) \text{sech}^{\frac{3}{2}}(a+b x)}{3 b}-\frac{2 i \sqrt{\cosh (a+b x)} \sqrt{\text{sech}(a+b x)} F\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{3 b} \]

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 3768

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 3771

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]

Rule 2641

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

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*}

Mathematica [A]  time = 0.0745525, size = 51, normalized size = 0.77 \[ \frac{2 \text{sech}^{\frac{3}{2}}(a+b x) \left (\sinh (a+b x)-i \cosh ^{\frac{3}{2}}(a+b x) \text{EllipticF}\left (\frac{1}{2} i (a+b x),2\right )\right )}{3 b} \]

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]  time = 0.276, size = 217, normalized size = 3.3 \begin{align*}{\frac{2}{3\,b} \left ( 2\,\sqrt{- \left ( \sinh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \sinh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cosh \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) \left ( \sinh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+\sqrt{- \left ( \sinh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}\sqrt{-2\, \left ( \sinh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cosh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) ,\sqrt{2} \right ) +2\,\cosh \left ( 1/2\,bx+a/2 \right ) \left ( \sinh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2} \right ) \sqrt{ \left ( 2\, \left ( \cosh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sinh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}{\frac{1}{\sqrt{2\, \left ( \sinh \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+ \left ( \sinh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}}} \left ( 2\, \left ( \cosh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) ^{-{\frac{3}{2}}} \left ( \sinh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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+(-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)*sinh(1/2*b*x+1/2*a)^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., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}\left (b x + a\right )^{\frac{5}{2}}\,{d x} \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\operatorname{sech}\left (b x + a\right )^{\frac{5}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{sech}\left (b x + a\right )^{\frac{5}{2}}\,{d x} \end{align*}

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)