3.317 \(\int \frac{\sec ^2(e+f x)}{\sqrt [3]{b \sin (e+f x)}} \, dx\)

Optimal. Leaf size=58 \[ \frac{3 \sqrt{\cos ^2(e+f x)} \sec (e+f x) (b \sin (e+f x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{3}{2};\frac{4}{3};\sin ^2(e+f x)\right )}{2 b f} \]

[Out]

(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[1/3, 3/2, 4/3, Sin[e + f*x]^2]*Sec[e + f*x]*(b*Sin[e + f*x])^(2/3))/
(2*b*f)

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Rubi [A]  time = 0.0397975, 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} \[ \frac{3 \sqrt{\cos ^2(e+f x)} \sec (e+f x) (b \sin (e+f x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{3}{2};\frac{4}{3};\sin ^2(e+f x)\right )}{2 b f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(b*Sin[e + f*x])^(1/3),x]

[Out]

(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[1/3, 3/2, 4/3, Sin[e + f*x]^2]*Sec[e + f*x]*(b*Sin[e + f*x])^(2/3))/
(2*b*f)

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)}{\sqrt [3]{b \sin (e+f x)}} \, dx &=\frac{3 \sqrt{\cos ^2(e+f x)} \, _2F_1\left (\frac{1}{3},\frac{3}{2};\frac{4}{3};\sin ^2(e+f x)\right ) \sec (e+f x) (b \sin (e+f x))^{2/3}}{2 b f}\\ \end{align*}

Mathematica [A]  time = 0.0417054, size = 55, normalized size = 0.95 \[ \frac{3 \sqrt{\cos ^2(e+f x)} \tan (e+f x) \, _2F_1\left (\frac{1}{3},\frac{3}{2};\frac{4}{3};\sin ^2(e+f x)\right )}{2 f \sqrt [3]{b \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(b*Sin[e + f*x])^(1/3),x]

[Out]

(3*Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[1/3, 3/2, 4/3, Sin[e + f*x]^2]*Tan[e + f*x])/(2*f*(b*Sin[e + f*x])^(
1/3))

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( fx+e \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{b\sin \left ( fx+e \right ) }}}}\, 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))^(1/3),x)

[Out]

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

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Maxima [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{1}{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))^(1/3),x, algorithm="maxima")

[Out]

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

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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{2}{3}} \sec \left (f x + e\right )^{2}}{b \sin \left (f x + e\right )}, 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))^(1/3),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2/(b*sin(f*x+e))**(1/3),x)

[Out]

Integral(sec(e + f*x)**2/(b*sin(e + f*x))**(1/3), x)

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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{1}{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))^(1/3),x, algorithm="giac")

[Out]

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