3.44 \(\int \frac {1}{(a \text {sech}^3(x))^{5/2}} \, dx\)

Optimal. Leaf size=121 \[ \frac {26 \tanh (x)}{77 a^2 \sqrt {a \text {sech}^3(x)}}-\frac {26 i F\left (\left .\frac {i x}{2}\right |2\right )}{77 a^2 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {2 \sinh (x) \cosh ^5(x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \sinh (x) \cosh ^3(x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {78 \sinh (x) \cosh (x)}{385 a^2 \sqrt {a \text {sech}^3(x)}} \]

[Out]

-26/77*I*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticF(I*sinh(1/2*x),2^(1/2))/a^2/cosh(x)^(3/2)/(a*sech(x)^3)^(1
/2)+78/385*cosh(x)*sinh(x)/a^2/(a*sech(x)^3)^(1/2)+26/165*cosh(x)^3*sinh(x)/a^2/(a*sech(x)^3)^(1/2)+2/15*cosh(
x)^5*sinh(x)/a^2/(a*sech(x)^3)^(1/2)+26/77*tanh(x)/a^2/(a*sech(x)^3)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4123, 3769, 3771, 2641} \[ \frac {26 \tanh (x)}{77 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {2 \sinh (x) \cosh ^5(x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \sinh (x) \cosh ^3(x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}-\frac {26 i F\left (\left .\frac {i x}{2}\right |2\right )}{77 a^2 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {78 \sinh (x) \cosh (x)}{385 a^2 \sqrt {a \text {sech}^3(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sech[x]^3)^(-5/2),x]

[Out]

(((-26*I)/77)*EllipticF[(I/2)*x, 2])/(a^2*Cosh[x]^(3/2)*Sqrt[a*Sech[x]^3]) + (78*Cosh[x]*Sinh[x])/(385*a^2*Sqr
t[a*Sech[x]^3]) + (26*Cosh[x]^3*Sinh[x])/(165*a^2*Sqrt[a*Sech[x]^3]) + (2*Cosh[x]^5*Sinh[x])/(15*a^2*Sqrt[a*Se
ch[x]^3]) + (26*Tanh[x])/(77*a^2*Sqrt[a*Sech[x]^3])

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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \text {sech}^3(x)\right )^{5/2}} \, dx &=\frac {\text {sech}^{\frac {3}{2}}(x) \int \frac {1}{\text {sech}^{\frac {15}{2}}(x)} \, dx}{a^2 \sqrt {a \text {sech}^3(x)}}\\ &=\frac {2 \cosh ^5(x) \sinh (x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {\left (13 \text {sech}^{\frac {3}{2}}(x)\right ) \int \frac {1}{\text {sech}^{\frac {11}{2}}(x)} \, dx}{15 a^2 \sqrt {a \text {sech}^3(x)}}\\ &=\frac {26 \cosh ^3(x) \sinh (x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^5(x) \sinh (x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {\left (39 \text {sech}^{\frac {3}{2}}(x)\right ) \int \frac {1}{\text {sech}^{\frac {7}{2}}(x)} \, dx}{55 a^2 \sqrt {a \text {sech}^3(x)}}\\ &=\frac {78 \cosh (x) \sinh (x)}{385 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \cosh ^3(x) \sinh (x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^5(x) \sinh (x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {\left (39 \text {sech}^{\frac {3}{2}}(x)\right ) \int \frac {1}{\text {sech}^{\frac {3}{2}}(x)} \, dx}{77 a^2 \sqrt {a \text {sech}^3(x)}}\\ &=\frac {78 \cosh (x) \sinh (x)}{385 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \cosh ^3(x) \sinh (x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^5(x) \sinh (x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \tanh (x)}{77 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {\left (13 \text {sech}^{\frac {3}{2}}(x)\right ) \int \sqrt {\text {sech}(x)} \, dx}{77 a^2 \sqrt {a \text {sech}^3(x)}}\\ &=\frac {78 \cosh (x) \sinh (x)}{385 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \cosh ^3(x) \sinh (x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^5(x) \sinh (x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \tanh (x)}{77 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {13 \int \frac {1}{\sqrt {\cosh (x)}} \, dx}{77 a^2 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}\\ &=-\frac {26 i F\left (\left .\frac {i x}{2}\right |2\right )}{77 a^2 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {78 \cosh (x) \sinh (x)}{385 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \cosh ^3(x) \sinh (x)}{165 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^5(x) \sinh (x)}{15 a^2 \sqrt {a \text {sech}^3(x)}}+\frac {26 \tanh (x)}{77 a^2 \sqrt {a \text {sech}^3(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 63, normalized size = 0.52 \[ \frac {\cosh (x) \sqrt {a \text {sech}^3(x)} \left (19122 \sinh (2 x)+4406 \sinh (4 x)+826 \sinh (6 x)+77 \sinh (8 x)-24960 i \sqrt {\cosh (x)} F\left (\left .\frac {i x}{2}\right |2\right )\right )}{73920 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sech[x]^3)^(-5/2),x]

[Out]

(Cosh[x]*Sqrt[a*Sech[x]^3]*((-24960*I)*Sqrt[Cosh[x]]*EllipticF[(I/2)*x, 2] + 19122*Sinh[2*x] + 4406*Sinh[4*x]
+ 826*Sinh[6*x] + 77*Sinh[8*x]))/(73920*a^3)

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \operatorname {sech}\relax (x)^{3}}}{a^{3} \operatorname {sech}\relax (x)^{9}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)^3)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*sech(x)^3)/(a^3*sech(x)^9), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \operatorname {sech}\relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)^3)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sech(x)^3)^(-5/2), x)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \mathrm {sech}\relax (x )^{3}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sech(x)^3)^(5/2),x)

[Out]

int(1/(a*sech(x)^3)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \operatorname {sech}\relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)^3)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sech(x)^3)^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (\frac {a}{{\mathrm {cosh}\relax (x)}^3}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a/cosh(x)^3)^(5/2),x)

[Out]

int(1/(a/cosh(x)^3)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \operatorname {sech}^{3}{\relax (x )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)**3)**(5/2),x)

[Out]

Integral((a*sech(x)**3)**(-5/2), x)

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