3.322 \(\int \frac{\sec ^2(e+f x)}{(b \sin (e+f x))^{5/3}} \, dx\)

Optimal. Leaf size=58 \[ -\frac{3 \sqrt{\cos ^2(e+f x)} \sec (e+f x) \, _2F_1\left (-\frac{1}{3},\frac{3}{2};\frac{2}{3};\sin ^2(e+f x)\right )}{2 b f (b \sin (e+f x))^{2/3}} \]

[Out]

(-3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-1/3, 3/2, 2/3, Sin[e + f*x]^2]*Sec[e + f*x])/(2*b*f*(b*Sin[e + f*x
])^(2/3))

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Rubi [A]  time = 0.0455026, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2577} \[ -\frac{3 \sqrt{\cos ^2(e+f x)} \sec (e+f x) \, _2F_1\left (-\frac{1}{3},\frac{3}{2};\frac{2}{3};\sin ^2(e+f x)\right )}{2 b f (b \sin (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(b*Sin[e + f*x])^(5/3),x]

[Out]

(-3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-1/3, 3/2, 2/3, Sin[e + f*x]^2]*Sec[e + f*x])/(2*b*f*(b*Sin[e + f*x
])^(2/3))

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

Mathematica [A]  time = 0.0424613, size = 55, normalized size = 0.95 \[ -\frac{3 \sqrt{\cos ^2(e+f x)} \tan (e+f x) \, _2F_1\left (-\frac{1}{3},\frac{3}{2};\frac{2}{3};\sin ^2(e+f x)\right )}{2 f (b \sin (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(b*Sin[e + f*x])^(5/3),x]

[Out]

(-3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[-1/3, 3/2, 2/3, Sin[e + f*x]^2]*Tan[e + f*x])/(2*f*(b*Sin[e + f*x])
^(5/3))

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( fx+e \right ) \right ) ^{2} \left ( b\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)

[Out]

int(sec(f*x+e)^2/(b*sin(f*x+e))^(5/3),x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (b \sin \left (f x + e\right )\right )^{\frac{1}{3}} \sec \left (f x + e\right )^{2}}{b^{2} \cos \left (f x + e\right )^{2} - b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

integral(-(b*sin(f*x + e))^(1/3)*sec(f*x + e)^2/(b^2*cos(f*x + e)^2 - b^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2/(b*sin(f*x+e))**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{2}}{\left (b \sin \left (f x + e\right )\right )^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(b*sin(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^2/(b*sin(f*x + e))^(5/3), x)