3.16 \(\int (b \text {sech}(c+d x))^{5/2} \, dx\)

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

[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 = 0.04, 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 = {3768, 3771, 2641} \[ \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}-\frac {2 i b^2 \sqrt {\cosh (c+d x)} F\left (\left .\frac {1}{2} i (c+d x)\right |2\right ) \sqrt {b \text {sech}(c+d x)}}{3 d} \]

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 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]

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]

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 = 0.07, size = 56, normalized size = 0.76 \[ \frac {2 b^2 \sqrt {b \text {sech}(c+d x)} \left (\tanh (c+d x)-i \sqrt {\cosh (c+d x)} F\left (\left .\frac {1}{2} i (c+d x)\right |2\right )\right )}{3 d} \]

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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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \operatorname {sech}\left (d x + c\right )} b^{2} \operatorname {sech}\left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sech(d*x + c))*b^2*sech(d*x + c)^2, x)

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

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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maple [F]  time = 0.28, 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 \[ \int \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

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

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

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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