3.290 \(\int \frac{\tan ^4(e+f x)}{(d \sec (e+f x))^{2/3}} \, dx\)

Optimal. Leaf size=57 \[ \frac{\cos ^2(e+f x)^{13/6} \tan ^5(e+f x) \, _2F_1\left (\frac{13}{6},\frac{5}{2};\frac{7}{2};\sin ^2(e+f x)\right )}{5 f (d \sec (e+f x))^{2/3}} \]

[Out]

((Cos[e + f*x]^2)^(13/6)*Hypergeometric2F1[13/6, 5/2, 7/2, Sin[e + f*x]^2]*Tan[e + f*x]^5)/(5*f*(d*Sec[e + f*x
])^(2/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0446449, antiderivative size = 57, 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 = {2617} \[ \frac{\cos ^2(e+f x)^{13/6} \tan ^5(e+f x) \, _2F_1\left (\frac{13}{6},\frac{5}{2};\frac{7}{2};\sin ^2(e+f x)\right )}{5 f (d \sec (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^4/(d*Sec[e + f*x])^(2/3),x]

[Out]

((Cos[e + f*x]^2)^(13/6)*Hypergeometric2F1[13/6, 5/2, 7/2, Sin[e + f*x]^2]*Tan[e + f*x]^5)/(5*f*(d*Sec[e + f*x
])^(2/3))

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin{align*} \int \frac{\tan ^4(e+f x)}{(d \sec (e+f x))^{2/3}} \, dx &=\frac{\cos ^2(e+f x)^{13/6} \, _2F_1\left (\frac{13}{6},\frac{5}{2};\frac{7}{2};\sin ^2(e+f x)\right ) \tan ^5(e+f x)}{5 f (d \sec (e+f x))^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.14309, size = 67, normalized size = 1.18 \[ \frac{3 \tan (e+f x) \left (9 \sqrt [6]{\cos ^2(e+f x)} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )+\sec ^2(e+f x)-10\right )}{7 f (d \sec (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4/(d*Sec[e + f*x])^(2/3),x]

[Out]

(3*(-10 + 9*(Cos[e + f*x]^2)^(1/6)*Hypergeometric2F1[1/6, 1/2, 3/2, Sin[e + f*x]^2] + Sec[e + f*x]^2)*Tan[e +
f*x])/(7*f*(d*Sec[e + f*x])^(2/3))

________________________________________________________________________________________

Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( fx+e \right ) \right ) ^{4} \left ( d\sec \left ( fx+e \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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