3.35 \(\int \frac{1}{(c \sec (a+b x))^{2/3}} \, dx\)

Optimal. Leaf size=56 \[ -\frac{3 c \sin (a+b x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{11}{6},\cos ^2(a+b x)\right )}{5 b \sqrt{\sin ^2(a+b x)} (c \sec (a+b x))^{5/3}} \]

[Out]

(-3*c*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[a + b*x]^2]*Sin[a + b*x])/(5*b*(c*Sec[a + b*x])^(5/3)*Sqrt[Sin[a +
b*x]^2])

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Rubi [A]  time = 0.0282795, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3772, 2643} \[ -\frac{3 c \sin (a+b x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(a+b x)\right )}{5 b \sqrt{\sin ^2(a+b x)} (c \sec (a+b x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(-2/3),x]

[Out]

(-3*c*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[a + b*x]^2]*Sin[a + b*x])/(5*b*(c*Sec[a + b*x])^(5/3)*Sqrt[Sin[a +
b*x]^2])

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{1}{(c \sec (a+b x))^{2/3}} \, dx &=\sqrt [3]{\frac{\cos (a+b x)}{c}} \sqrt [3]{c \sec (a+b x)} \int \left (\frac{\cos (a+b x)}{c}\right )^{2/3} \, dx\\ &=-\frac{3 \cos ^2(a+b x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(a+b x)\right ) \sqrt [3]{c \sec (a+b x)} \sin (a+b x)}{5 b c \sqrt{\sin ^2(a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0526139, size = 57, normalized size = 1.02 \[ -\frac{3 \sqrt{-\tan ^2(a+b x)} \cot (a+b x) \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\sec ^2(a+b x)\right )}{2 b (c \sec (a+b x))^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*Sec[a + b*x])^(-2/3),x]

[Out]

(-3*Cot[a + b*x]*Hypergeometric2F1[-1/3, 1/2, 2/3, Sec[a + b*x]^2]*Sqrt[-Tan[a + b*x]^2])/(2*b*(c*Sec[a + b*x]
)^(2/3))

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Maple [F]  time = 0.097, size = 0, normalized size = 0. \begin{align*} \int \left ( c\sec \left ( bx+a \right ) \right ) ^{-{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sec(b*x+a))^(2/3),x)

[Out]

int(1/(c*sec(b*x+a))^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))^(2/3),x, algorithm="maxima")

[Out]

integrate((c*sec(b*x + a))^(-2/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (c \sec \left (b x + a\right )\right )^{\frac{1}{3}}}{c \sec \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*sec(b*x + a))^(1/3)/(c*sec(b*x + a)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec{\left (a + b x \right )}\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sec(a + b*x))**(-2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))^(2/3),x, algorithm="giac")

[Out]

integrate((c*sec(b*x + a))^(-2/3), x)