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

Optimal. Leaf size=64 \[ \frac {2 \cos ^2(e+f x)^{17/12} \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac {5}{4},\frac {17}{12};\frac {9}{4};\sin ^2(e+f x)\right )}{5 d f} \]

[Out]

2/5*(cos(f*x+e)^2)^(17/12)*hypergeom([5/4, 17/12],[9/4],sin(f*x+e)^2)*(b*sec(f*x+e))^(1/3)*(d*tan(f*x+e))^(5/2
)/d/f

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Rubi [A]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2617} \[ \frac {2 \cos ^2(e+f x)^{17/12} \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac {5}{4},\frac {17}{12};\frac {9}{4};\sin ^2(e+f x)\right )}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^(17/12)*Hypergeometric2F1[5/4, 17/12, 9/4, Sin[e + f*x]^2]*(b*Sec[e + f*x])^(1/3)*(d*Tan[e
+ f*x])^(5/2))/(5*d*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]{b \sec (e+f x)} (d \tan (e+f x))^{3/2} \, dx &=\frac {2 \cos ^2(e+f x)^{17/12} \, _2F_1\left (\frac {5}{4},\frac {17}{12};\frac {9}{4};\sin ^2(e+f x)\right ) \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{5/2}}{5 d f}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 62, normalized size = 0.97 \[ \frac {3 d \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, _2F_1\left (-\frac {1}{4},\frac {1}{6};\frac {7}{6};\sec ^2(e+f x)\right )}{f \sqrt [4]{-\tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*Hypergeometric2F1[-1/4, 1/6, 7/6, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(1/3)*Sqrt[d*Tan[e + f*x]])/(f*(-Tan[e
+ f*x]^2)^(1/4))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}} \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}} \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x +e \right )\right )^{\frac {1}{3}} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}} \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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