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

Optimal. Leaf size=57 \[ \frac{\cos ^2(e+f x)^{13/6} \tan ^3(e+f x) (d \sec (e+f x))^{4/3} \, _2F_1\left (\frac{3}{2},\frac{13}{6};\frac{5}{2};\sin ^2(e+f x)\right )}{3 f} \]

[Out]

((Cos[e + f*x]^2)^(13/6)*Hypergeometric2F1[3/2, 13/6, 5/2, Sin[e + f*x]^2]*(d*Sec[e + f*x])^(4/3)*Tan[e + f*x]
^3)/(3*f)

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Rubi [A]  time = 0.043448, 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} \[ \frac{\cos ^2(e+f x)^{13/6} \tan ^3(e+f x) (d \sec (e+f x))^{4/3} \, _2F_1\left (\frac{3}{2},\frac{13}{6};\frac{5}{2};\sin ^2(e+f x)\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^(13/6)*Hypergeometric2F1[3/2, 13/6, 5/2, Sin[e + f*x]^2]*(d*Sec[e + f*x])^(4/3)*Tan[e + f*x]
^3)/(3*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))^{4/3} \tan ^2(e+f x) \, dx &=\frac{\cos ^2(e+f x)^{13/6} \, _2F_1\left (\frac{3}{2},\frac{13}{6};\frac{5}{2};\sin ^2(e+f x)\right ) (d \sec (e+f x))^{4/3} \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.280316, size = 80, normalized size = 1.4 \[ \frac{3 d \sqrt [3]{d \sec (e+f x)} \left (2 \sin (e+f x) \sqrt [6]{\cos ^2(e+f x)} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )-3 \sin (e+f x)+\tan (e+f x) \sec (e+f x)\right )}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{4}{3}}} \left ( \tan \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*sec(f*x + e))^(1/3)*d*sec(f*x + e)*tan(f*x + e)^2, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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