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

Optimal. Leaf size=57 \[ \frac {\cos ^2(e+f x)^{17/6} \tan ^5(e+f x) (d \sec (e+f x))^{2/3} \, _2F_1\left (\frac {5}{2},\frac {17}{6};\frac {7}{2};\sin ^2(e+f x)\right )}{5 f} \]

[Out]

1/5*(cos(f*x+e)^2)^(17/6)*hypergeom([5/2, 17/6],[7/2],sin(f*x+e)^2)*(d*sec(f*x+e))^(2/3)*tan(f*x+e)^5/f

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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} \[ \frac {\cos ^2(e+f x)^{17/6} \tan ^5(e+f x) (d \sec (e+f x))^{2/3} \, _2F_1\left (\frac {5}{2},\frac {17}{6};\frac {7}{2};\sin ^2(e+f x)\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^(17/6)*Hypergeometric2F1[5/2, 17/6, 7/2, Sin[e + f*x]^2]*(d*Sec[e + f*x])^(2/3)*Tan[e + f*x]
^5)/(5*f)

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 (d \sec (e+f x))^{2/3} \tan ^4(e+f x) \, dx &=\frac {\cos ^2(e+f x)^{17/6} \, _2F_1\left (\frac {5}{2},\frac {17}{6};\frac {7}{2};\sin ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \tan ^5(e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 69, normalized size = 1.21 \[ \frac {3 \tan (e+f x) (d \sec (e+f x))^{2/3} \left (9 \cos ^2(e+f x)^{5/6} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\sin ^2(e+f x)\right )+5 \sec ^2(e+f x)-14\right )}{55 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(d*Sec[e + f*x])^(2/3)*(-14 + 9*(Cos[e + f*x]^2)^(5/6)*Hypergeometric2F1[1/2, 5/6, 3/2, Sin[e + f*x]^2] + 5
*Sec[e + f*x]^2)*Tan[e + f*x])/(55*f)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} \tan \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*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 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} \tan \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (\tan ^{4}\left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*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 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} \tan \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*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 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2/3} \,d x \]

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 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{\frac {2}{3}} \tan ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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