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

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

[Out]

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

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Rubi [A]  time = 0.0364821, 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} \[ \frac{\cos ^2(e+f x)^{5/3} \tan ^3(e+f x) \sqrt [3]{d \sec (e+f x)} \, _2F_1\left (\frac{3}{2},\frac{5}{3};\frac{5}{2};\sin ^2(e+f x)\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^(5/3)*Hypergeometric2F1[3/2, 5/3, 5/2, Sin[e + f*x]^2]*(d*Sec[e + f*x])^(1/3)*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 \sqrt [3]{d \sec (e+f x)} \tan ^2(e+f x) \, dx &=\frac{\cos ^2(e+f x)^{5/3} \, _2F_1\left (\frac{3}{2},\frac{5}{3};\frac{5}{2};\sin ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.280346, size = 80, normalized size = 1.4 \[ \frac{3 \sqrt [3]{d \sec (e+f x)} \left (2 \sqrt [3]{\cos ^2(e+f x)} \tan (e+f x)-\sin (2 (e+f x)) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{3}{2};\sin ^2(e+f x)\right )\right )}{8 f \sqrt [3]{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(d*Sec[e + f*x])^(1/3)*(-(Hypergeometric2F1[1/2, 2/3, 3/2, Sin[e + f*x]^2]*Sin[2*(e + f*x)]) + 2*(Cos[e + f
*x]^2)^(1/3)*Tan[e + f*x]))/(8*f*(Cos[e + f*x]^2)^(1/3))

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{1}{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))^(1/3)*tan(f*x+e)^2,x, algorithm="giac")

[Out]

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