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

Optimal. Leaf size=77 \[ \frac {14 \sinh (x)}{45 a \sqrt {a \text {sech}^3(x)}}-\frac {14 i E\left (\left .\frac {i x}{2}\right |2\right )}{15 a \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {2 \sinh (x) \cosh ^2(x)}{9 a \sqrt {a \text {sech}^3(x)}} \]

[Out]

-14/15*I*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticE(I*sinh(1/2*x),2^(1/2))/a/cosh(x)^(3/2)/(a*sech(x)^3)^(1/2
)+14/45*sinh(x)/a/(a*sech(x)^3)^(1/2)+2/9*cosh(x)^2*sinh(x)/a/(a*sech(x)^3)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4123, 3769, 3771, 2639} \[ \frac {14 \sinh (x)}{45 a \sqrt {a \text {sech}^3(x)}}+\frac {2 \sinh (x) \cosh ^2(x)}{9 a \sqrt {a \text {sech}^3(x)}}-\frac {14 i E\left (\left .\frac {i x}{2}\right |2\right )}{15 a \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-14*I)/15)*EllipticE[(I/2)*x, 2])/(a*Cosh[x]^(3/2)*Sqrt[a*Sech[x]^3]) + (14*Sinh[x])/(45*a*Sqrt[a*Sech[x]^3
]) + (2*Cosh[x]^2*Sinh[x])/(9*a*Sqrt[a*Sech[x]^3])

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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 )^{3/2}} \, dx &=\frac {\text {sech}^{\frac {3}{2}}(x) \int \frac {1}{\text {sech}^{\frac {9}{2}}(x)} \, dx}{a \sqrt {a \text {sech}^3(x)}}\\ &=\frac {2 \cosh ^2(x) \sinh (x)}{9 a \sqrt {a \text {sech}^3(x)}}+\frac {\left (7 \text {sech}^{\frac {3}{2}}(x)\right ) \int \frac {1}{\text {sech}^{\frac {5}{2}}(x)} \, dx}{9 a \sqrt {a \text {sech}^3(x)}}\\ &=\frac {14 \sinh (x)}{45 a \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^2(x) \sinh (x)}{9 a \sqrt {a \text {sech}^3(x)}}+\frac {\left (7 \text {sech}^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sqrt {\text {sech}(x)}} \, dx}{15 a \sqrt {a \text {sech}^3(x)}}\\ &=\frac {14 \sinh (x)}{45 a \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^2(x) \sinh (x)}{9 a \sqrt {a \text {sech}^3(x)}}+\frac {7 \int \sqrt {\cosh (x)} \, dx}{15 a \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}\\ &=-\frac {14 i E\left (\left .\frac {i x}{2}\right |2\right )}{15 a \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {14 \sinh (x)}{45 a \sqrt {a \text {sech}^3(x)}}+\frac {2 \cosh ^2(x) \sinh (x)}{9 a \sqrt {a \text {sech}^3(x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 47, normalized size = 0.61 \[ \frac {33 \sinh (x)+5 \sinh (3 x)-\frac {84 i E\left (\left .\frac {i x}{2}\right |2\right )}{\cosh ^{\frac {3}{2}}(x)}}{90 a \sqrt {a \text {sech}^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-84*I)*EllipticE[(I/2)*x, 2])/Cosh[x]^(3/2) + 33*Sinh[x] + 5*Sinh[3*x])/(90*a*Sqrt[a*Sech[x]^3])

________________________________________________________________________________________

fricas [F]  time = 1.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \operatorname {sech}\relax (x)^{3}}}{a^{2} \operatorname {sech}\relax (x)^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sech(x)^3)/(a^2*sech(x)^6), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sech(x)^3)^(-3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sech(x)^3)^(-3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a/cosh(x)^3)^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sech(x)**3)**(-3/2), x)

________________________________________________________________________________________