3.1.58 \(\int \frac {1}{\sqrt {a \sec ^3(x)}} \, dx\) [58]

Optimal. Leaf size=44 \[ \frac {2 F\left (\left .\frac {x}{2}\right |2\right )}{3 \cos ^{\frac {3}{2}}(x) \sqrt {a \sec ^3(x)}}+\frac {2 \tan (x)}{3 \sqrt {a \sec ^3(x)}} \]

[Out]

2/3*(cos(1/2*x)^2)^(1/2)/cos(1/2*x)*EllipticF(sin(1/2*x),2^(1/2))/cos(x)^(3/2)/(a*sec(x)^3)^(1/2)+2/3*tan(x)/(
a*sec(x)^3)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 44, 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 \tan (x)}{3 \sqrt {a \sec ^3(x)}}+\frac {2 F\left (\left .\frac {x}{2}\right |2\right )}{3 \cos ^{\frac {3}{2}}(x) \sqrt {a \sec ^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*EllipticF[x/2, 2])/(3*Cos[x]^(3/2)*Sqrt[a*Sec[x]^3]) + (2*Tan[x])/(3*Sqrt[a*Sec[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 \sec ^3(x)}} \, dx &=\frac {\sec ^{\frac {3}{2}}(x) \int \frac {1}{\sec ^{\frac {3}{2}}(x)} \, dx}{\sqrt {a \sec ^3(x)}}\\ &=\frac {2 \tan (x)}{3 \sqrt {a \sec ^3(x)}}+\frac {\sec ^{\frac {3}{2}}(x) \int \sqrt {\sec (x)} \, dx}{3 \sqrt {a \sec ^3(x)}}\\ &=\frac {2 \tan (x)}{3 \sqrt {a \sec ^3(x)}}+\frac {\int \frac {1}{\sqrt {\cos (x)}} \, dx}{3 \cos ^{\frac {3}{2}}(x) \sqrt {a \sec ^3(x)}}\\ &=\frac {2 F\left (\left .\frac {x}{2}\right |2\right )}{3 \cos ^{\frac {3}{2}}(x) \sqrt {a \sec ^3(x)}}+\frac {2 \tan (x)}{3 \sqrt {a \sec ^3(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 31, normalized size = 0.70 \begin {gather*} \frac {2 \left (\frac {F\left (\left .\frac {x}{2}\right |2\right )}{\cos ^{\frac {3}{2}}(x)}+\tan (x)\right )}{3 \sqrt {a \sec ^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(EllipticF[x/2, 2]/Cos[x]^(3/2) + Tan[x]))/(3*Sqrt[a*Sec[x]^3])

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Maple [C] Result contains complex when optimal does not.
time = 0.50, size = 76, normalized size = 1.73

method	result	size
default	\(\frac {2 \left (-1+\cos \left (x \right )\right ) \left (-i \sqrt {\frac {1}{\cos \left (x \right )+1}}\, \sqrt {\frac {\cos \left (x \right )}{\cos \left (x \right )+1}}\, \sin \left (x \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}, i\right )+\cos ^{2}\left (x \right )-\cos \left (x \right )\right ) \left (\cos \left (x \right )+1\right )^{2}}{3 \cos \left (x \right )^{2} \sin \left (x \right )^{3} \sqrt {\frac {a}{\cos \left (x \right )^{3}}}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(-1+cos(x))*(-I*(1/(cos(x)+1))^(1/2)*(cos(x)/(cos(x)+1))^(1/2)*sin(x)*EllipticF(I*(-1+cos(x))/sin(x),I)+co
s(x)^2-cos(x))*(cos(x)+1)^2/cos(x)^2/sin(x)^3/(a/cos(x)^3)^(1/2)

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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*sec(x)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(a*sec(x)^3), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.79, size = 58, normalized size = 1.32 \begin {gather*} \frac {2 \, \sqrt {\frac {a}{\cos \left (x\right )^{3}}} \cos \left (x\right )^{2} \sin \left (x\right ) + i \, \sqrt {2} \sqrt {a} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) - i \, \sqrt {2} \sqrt {a} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (x\right ) - i \, \sin \left (x\right )\right )}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*sqrt(a/cos(x)^3)*cos(x)^2*sin(x) + I*sqrt(2)*sqrt(a)*weierstrassPInverse(-4, 0, cos(x) + I*sin(x)) - I*
sqrt(2)*sqrt(a)*weierstrassPInverse(-4, 0, cos(x) - I*sin(x)))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/sqrt(a*sec(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}{{\cos \left (x\right )}^3}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a/cos(x)^3)^(1/2), x)

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