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]

3/11*hypergeom([-11/6, -3/2],[-5/6],cos(f*x+e)^2)*sec(f*x+e)^(11/3)*sin(f*x+e)/f/(sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.86, 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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sec \left (f x + e\right )^{\frac {2}{3}} \tan \left (f x + e\right )^{4}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (f x + e\right )^{\frac {2}{3}} \tan \left (f x + e\right )^{4}\,{d x} \]

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)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{\frac {2}{3}}\left (f x +e \right )\right ) \left (\tan ^{4}\left (f x +e \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (f x + e\right )^{\frac {2}{3}} \tan \left (f x + e\right )^{4}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^4*(1/cos(e + f*x))^(2/3),x)

[Out]

int(tan(e + f*x)^4*(1/cos(e + f*x))^(2/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tan ^{4}{\left (e + f x \right )} \sec ^{\frac {2}{3}}{\left (e + f x \right )}\, dx \]

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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