3.279 \(\int \sec ^{\frac{11}{3}}(e+f x) \sin ^4(e+f x) \, dx\)

Optimal. Leaf size=53 \[ \frac{3 \sin (e+f x) \sec ^{\frac{8}{3}}(e+f x) \, _2F_1\left (-\frac{3}{2},-\frac{4}{3};-\frac{1}{3};\cos ^2(e+f x)\right )}{8 f \sqrt{\sin ^2(e+f x)}} \]

[Out]

(3*Hypergeometric2F1[-3/2, -4/3, -1/3, Cos[e + f*x]^2]*Sec[e + f*x]^(8/3)*Sin[e + f*x])/(8*f*Sqrt[Sin[e + f*x]
^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0602202, 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} \[ \frac{3 \sin (e+f x) \sec ^{\frac{8}{3}}(e+f x) \, _2F_1\left (-\frac{3}{2},-\frac{4}{3};-\frac{1}{3};\cos ^2(e+f x)\right )}{8 f \sqrt{\sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^(11/3)*Sin[e + f*x]^4,x]

[Out]

(3*Hypergeometric2F1[-3/2, -4/3, -1/3, Cos[e + f*x]^2]*Sec[e + f*x]^(8/3)*Sin[e + f*x])/(8*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{11}{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{11}{3}}(e+f x)} \, dx\\ &=\frac{3 \, _2F_1\left (-\frac{3}{2},-\frac{4}{3};-\frac{1}{3};\cos ^2(e+f x)\right ) \sec ^{\frac{8}{3}}(e+f x) \sin (e+f x)}{8 f \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.194314, size = 78, normalized size = 1.47 \[ \frac{3 \sec ^{\frac{2}{3}}(e+f x) \left (9 \sin (e+f x) \sqrt [3]{\cos ^2(e+f x)} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )-11 \sin (e+f x)+2 \tan (e+f x) \sec (e+f x)\right )}{16 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^(11/3)*Sin[e + f*x]^4,x]

[Out]

(3*Sec[e + f*x]^(2/3)*(-11*Sin[e + f*x] + 9*(Cos[e + f*x]^2)^(1/3)*Hypergeometric2F1[1/3, 1/2, 3/2, Sin[e + f*
x]^2]*Sin[e + f*x] + 2*Sec[e + f*x]*Tan[e + f*x]))/(16*f)

________________________________________________________________________________________

Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( fx+e \right ) \right ) ^{4}{\frac{1}{\sqrt [3]{\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/sec(f*x+e)^(1/3),x, algorithm="maxima")

[Out]

integrate(tan(f*x + e)^4/sec(f*x + e)^(1/3), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\tan \left (f x + e\right )^{4}}{\sec \left (f x + e\right )^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/sec(f*x+e)^(1/3),x, algorithm="fricas")

[Out]

integral(tan(f*x + e)^4/sec(f*x + e)^(1/3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**4/sec(f*x+e)**(1/3),x)

[Out]

Integral(tan(e + f*x)**4/sec(e + f*x)**(1/3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/sec(f*x+e)^(1/3),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^4/sec(f*x + e)^(1/3), x)