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

Optimal. Leaf size=64 \[ \frac {3 \cos ^2(e+f x)^{11/12} \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, _2F_1\left (\frac {2}{3},\frac {11}{12};\frac {5}{3};\sin ^2(e+f x)\right )}{4 d f} \]

[Out]

3/4*(cos(f*x+e)^2)^(11/12)*hypergeom([2/3, 11/12],[5/3],sin(f*x+e)^2)*(b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3
)/d/f

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Rubi [A]  time = 0.04, 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} \[ \frac {3 \cos ^2(e+f x)^{11/12} \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, _2F_1\left (\frac {2}{3},\frac {11}{12};\frac {5}{3};\sin ^2(e+f x)\right )}{4 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[e + f*x]^2)^(11/12)*Hypergeometric2F1[2/3, 11/12, 5/3, Sin[e + f*x]^2]*Sqrt[b*Sec[e + f*x]]*(d*Tan[e +
f*x])^(4/3))/(4*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 {b \sec (e+f x)} \sqrt [3]{d \tan (e+f x)} \, dx &=\frac {3 \cos ^2(e+f x)^{11/12} \, _2F_1\left (\frac {2}{3},\frac {11}{12};\frac {5}{3};\sin ^2(e+f x)\right ) \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3}}{4 d f}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 62, normalized size = 0.97 \[ \frac {2 d \sqrt [3]{-\tan ^2(e+f x)} \sqrt {b \sec (e+f x)} \, _2F_1\left (\frac {1}{4},\frac {1}{3};\frac {5}{4};\sec ^2(e+f x)\right )}{f (d \tan (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (f x +e \right )}\, \left (d \tan \left (f x +e \right )\right )^{\frac {1}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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