3.1.42 \(\int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx\) [42]

Optimal. Leaf size=48 \[ -\frac {2 i F\left (\left .\frac {i x}{2}\right |2\right )}{3 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}} \]

[Out]

-2/3*I*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticF(I*sinh(1/2*x),2^(1/2))/cosh(x)^(3/2)/(a*sech(x)^3)^(1/2)+2/
3*tanh(x)/(a*sech(x)^3)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4208, 3854, 3856, 2720} \begin {gather*} \frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}}-\frac {2 i F\left (\left .\frac {i x}{2}\right |2\right )}{3 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Sech[x]^3],x]

[Out]

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

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 3854

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 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]

Rule 4208

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}{\sqrt {a \text {sech}^3(x)}} \, dx &=\frac {\text {sech}^{\frac {3}{2}}(x) \int \frac {1}{\text {sech}^{\frac {3}{2}}(x)} \, dx}{\sqrt {a \text {sech}^3(x)}}\\ &=\frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}}+\frac {\text {sech}^{\frac {3}{2}}(x) \int \sqrt {\text {sech}(x)} \, dx}{3 \sqrt {a \text {sech}^3(x)}}\\ &=\frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}}+\frac {\int \frac {1}{\sqrt {\cosh (x)}} \, dx}{3 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}\\ &=-\frac {2 i F\left (\left .\frac {i x}{2}\right |2\right )}{3 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 38, normalized size = 0.79 \begin {gather*} \frac {-\frac {2 i F\left (\left .\frac {i x}{2}\right |2\right )}{\cosh ^{\frac {3}{2}}(x)}+2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Sech[x]^3],x]

[Out]

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

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Maple [F]
time = 1.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {a \mathrm {sech}\left (x \right )^{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/sqrt(a*sech(x)^3), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 126, normalized size = 2.62 \begin {gather*} \frac {4 \, \sqrt {2} {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + \sqrt {2} {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \sqrt {\frac {a \cosh \left (x\right ) + a \sinh \left (x\right )}{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}}}{6 \, {\left (a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)*sqrt(a)*weierstrassPInverse(-4, 0, cosh(x) + sinh(x
)) + sqrt(2)*(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*s
qrt((a*cosh(x) + a*sinh(x))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)))/(a*cosh(x)^2 + 2*a*cosh(x)*sinh(
x) + a*sinh(x)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a \operatorname {sech}^{3}{\left (x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/sqrt(a*sech(x)^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {\frac {a}{{\mathrm {cosh}\left (x\right )}^3}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a/cosh(x)^3)^(1/2), x)

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