3.201 \(\int \frac{1}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=56 \[ -\frac{3 b \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{6},\frac{13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}} \]

[Out]

(-3*b*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin[c +
d*x]^2])

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Rubi [A]  time = 0.0259363, 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} \[ -\frac{3 b \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(-4/3),x]

[Out]

(-3*b*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin[c +
d*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}{(b \sec (c+d x))^{4/3}} \, dx &=\left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac{\cos (c+d x)}{b}\right )^{4/3} \, dx\\ &=-\frac{3 \cos ^3(c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{7 b^2 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0035631, size = 57, normalized size = 1.02 \[ -\frac{3 \sqrt{-\tan ^2(c+d x)} \cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{1}{2},\frac{1}{3},\sec ^2(c+d x)\right )}{4 d (b \sec (c+d x))^{4/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Sec[c + d*x])^(-4/3),x]

[Out]

(-3*Cot[c + d*x]*Hypergeometric2F1[-2/3, 1/2, 1/3, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2])/(4*d*(b*Sec[c + d*x]
)^(4/3))

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sec(d*x+c))^(4/3),x)

[Out]

int(1/(b*sec(d*x+c))^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c))^(-4/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^(2/3)/(b^2*sec(d*x + c)^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec{\left (c + d x \right )}\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(d*x+c))**(4/3),x)

[Out]

Integral((b*sec(c + d*x))**(-4/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (d x + c\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(-4/3), x)