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

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

[Out]

((Cos[e + f*x]^2)^(4/3)*Hypergeometric2F1[4/3, 3/2, 5/2, Sin[e + f*x]^2]*Tan[e + f*x]^3)/(3*f*(d*Sec[e + f*x])
^(1/3))

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Rubi [A]  time = 0.0425088, 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} \[ \frac{\cos ^2(e+f x)^{4/3} \tan ^3(e+f x) \, _2F_1\left (\frac{4}{3},\frac{3}{2};\frac{5}{2};\sin ^2(e+f x)\right )}{3 f \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2/(d*Sec[e + f*x])^(1/3),x]

[Out]

((Cos[e + f*x]^2)^(4/3)*Hypergeometric2F1[4/3, 3/2, 5/2, Sin[e + f*x]^2]*Tan[e + f*x]^3)/(3*f*(d*Sec[e + f*x])
^(1/3))

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

Mathematica [A]  time = 0.213616, size = 80, normalized size = 1.4 \[ \frac{3 \left (2 \cos ^2(e+f x)^{2/3} \tan (e+f x)-\sin (2 (e+f x)) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{3}{2};\sin ^2(e+f x)\right )\right )}{4 f \cos ^2(e+f x)^{2/3} \sqrt [3]{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^2/(d*Sec[e + f*x])^(1/3),x]

[Out]

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

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{d\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

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

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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{2}{3}} \tan \left (f x + e\right )^{2}}{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)^2/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(d*sec(f*x+e))^(1/3),x, algorithm="giac")

[Out]

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