3.277 \(\int \sec ^{\frac{14}{3}}(e+f x) \sin ^4(e+f x) \, dx\)

Optimal. Leaf size=53 \[ \frac{3 \sin (e+f x) \sec ^{\frac{11}{3}}(e+f x) \, _2F_1\left (-\frac{11}{6},-\frac{3}{2};-\frac{5}{6};\cos ^2(e+f x)\right )}{11 f \sqrt{\sin ^2(e+f x)}} \]

[Out]

(3*Hypergeometric2F1[-11/6, -3/2, -5/6, Cos[e + f*x]^2]*Sec[e + f*x]^(11/3)*Sin[e + f*x])/(11*f*Sqrt[Sin[e + f
*x]^2])

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Rubi [A]  time = 0.05791, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2632, 2576} \[ \frac{3 \sin (e+f x) \sec ^{\frac{11}{3}}(e+f x) \, _2F_1\left (-\frac{11}{6},-\frac{3}{2};-\frac{5}{6};\cos ^2(e+f x)\right )}{11 f \sqrt{\sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^(14/3)*Sin[e + f*x]^4,x]

[Out]

(3*Hypergeometric2F1[-11/6, -3/2, -5/6, Cos[e + f*x]^2]*Sec[e + f*x]^(11/3)*Sin[e + f*x])/(11*f*Sqrt[Sin[e + f
*x]^2])

Rule 2632

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^2*(a*Sec[e
+ f*x])^(m - 1)*(b*Csc[e + f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/b^2, Int[1/((a*Co
s[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]

Rule 2576

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

Rubi steps

\begin{align*} \int \sec ^{\frac{14}{3}}(e+f x) \sin ^4(e+f x) \, dx &=\left (\cos ^{\frac{2}{3}}(e+f x) \sec ^{\frac{2}{3}}(e+f x)\right ) \int \frac{\sin ^4(e+f x)}{\cos ^{\frac{14}{3}}(e+f x)} \, dx\\ &=\frac{3 \, _2F_1\left (-\frac{11}{6},-\frac{3}{2};-\frac{5}{6};\cos ^2(e+f x)\right ) \sec ^{\frac{11}{3}}(e+f x) \sin (e+f x)}{11 f \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.859298, size = 78, normalized size = 1.47 \[ \frac{3 \sin (e+f x) \left (\frac{9 \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\sin ^2(e+f x)\right )}{\sqrt [6]{\cos ^2(e+f x)}}-(7 \cos (2 (e+f x))+2) \sec ^4(e+f x)\right )}{55 f \sqrt [3]{\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^(14/3)*Sin[e + f*x]^4,x]

[Out]

(3*((9*Hypergeometric2F1[1/2, 5/6, 3/2, Sin[e + f*x]^2])/(Cos[e + f*x]^2)^(1/6) - (2 + 7*Cos[2*(e + f*x)])*Sec
[e + f*x]^4)*Sin[e + f*x])/(55*f*Sec[e + f*x]^(1/3))

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Maple [F]  time = 0.073, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{{\frac{2}{3}}} \left ( \tan \left ( fx+e \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^(2/3)*tan(f*x+e)^4,x)

[Out]

int(sec(f*x+e)^(2/3)*tan(f*x+e)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (f x + e\right )^{\frac{2}{3}} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^(2/3)*tan(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^(2/3)*tan(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sec \left (f x + e\right )^{\frac{2}{3}} \tan \left (f x + e\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^(2/3)*tan(f*x+e)^4,x, algorithm="fricas")

[Out]

integral(sec(f*x + e)^(2/3)*tan(f*x + e)^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tan ^{4}{\left (e + f x \right )} \sec ^{\frac{2}{3}}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**(2/3)*tan(f*x+e)**4,x)

[Out]

Integral(tan(e + f*x)**4*sec(e + f*x)**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (f x + e\right )^{\frac{2}{3}} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^(2/3)*tan(f*x+e)^4,x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^(2/3)*tan(f*x + e)^4, x)