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

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 48 \[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {i x}{2},2\right )}{3 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}}+\frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4208, 3854, 3856, 2720} \[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\frac {2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}}-\frac {2 i \operatorname {EllipticF}\left (\frac {i x}{2},2\right )}{3 \cosh ^{\frac {3}{2}}(x) \sqrt {a \text {sech}^3(x)}} \]

[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*} \text {integral}& = \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 \operatorname {EllipticF}\left (\frac {i x}{2},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*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\frac {-\frac {2 i \operatorname {EllipticF}\left (\frac {i x}{2},2\right )}{\cosh ^{\frac {3}{2}}(x)}+2 \tanh (x)}{3 \sqrt {a \text {sech}^3(x)}} \]

[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])

Maple [F]

\[\int \frac {1}{\sqrt {a \operatorname {sech}\left (x \right )^{3}}}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.62 \[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\frac {4 \, \sqrt {2} {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + \sqrt {2} {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \sqrt {\frac {a \cosh \left (x\right ) + a \sinh \left (x\right )}{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}}}{6 \, {\left (a \cosh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) \sinh \left (x\right ) + a \sinh \left (x\right )^{2}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\int \frac {1}{\sqrt {a \operatorname {sech}^{3}{\left (x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a*sech(x)**3), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\int { \frac {1}{\sqrt {a \operatorname {sech}\left (x\right )^{3}}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(a*sech(x)^3), x)

Giac [F]

\[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\int { \frac {1}{\sqrt {a \operatorname {sech}\left (x\right )^{3}}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(a*sech(x)^3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a \text {sech}^3(x)}} \, dx=\int \frac {1}{\sqrt {\frac {a}{{\mathrm {cosh}\left (x\right )}^3}}} \,d x \]

[In]

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

[Out]

int(1/(a/cosh(x)^3)^(1/2), x)