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]

1/5*(cos(f*x+e)^2)^(13/6)*hypergeom([13/6, 5/2],[7/2],sin(f*x+e)^2)*tan(f*x+e)^5/f/(d*sec(f*x+e))^(2/3)

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.14, 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))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} \tan \left (f x + e\right )^{4}}{d \sec \left (f x + e\right )}, x\right ) \]

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)

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

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)

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}\left (f x +e \right )}{\left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}}}\, dx \]

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)

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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