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

Optimal. Leaf size=64 \[ \frac {3 \cos ^2(e+f x)^{13/12} (b \sec (e+f x))^{3/2} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {13}{12};\frac {4}{3};\sin ^2(e+f x)\right )}{2 d f} \]

[Out]

3/2*(cos(f*x+e)^2)^(13/12)*hypergeom([1/3, 13/12],[4/3],sin(f*x+e)^2)*(b*sec(f*x+e))^(3/2)*(d*tan(f*x+e))^(2/3
)/d/f

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Rubi [A]  time = 0.06, 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} \[ \frac {3 \cos ^2(e+f x)^{13/12} (b \sec (e+f x))^{3/2} (d \tan (e+f x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {13}{12};\frac {4}{3};\sin ^2(e+f x)\right )}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[e + f*x]^2)^(13/12)*Hypergeometric2F1[1/3, 13/12, 4/3, Sin[e + f*x]^2]*(b*Sec[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(2/3))/(2*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 \frac {(b \sec (e+f x))^{3/2}}{\sqrt [3]{d \tan (e+f x)}} \, dx &=\frac {3 \cos ^2(e+f x)^{13/12} \, _2F_1\left (\frac {1}{3},\frac {13}{12};\frac {4}{3};\sin ^2(e+f x)\right ) (b \sec (e+f x))^{3/2} (d \tan (e+f x))^{2/3}}{2 d f}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 64, normalized size = 1.00 \[ \frac {2 d \left (-\tan ^2(e+f x)\right )^{2/3} (b \sec (e+f x))^{3/2} \, _2F_1\left (\frac {2}{3},\frac {3}{4};\frac {7}{4};\sec ^2(e+f x)\right )}{3 f (d \tan (e+f x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac {2}{3}} b \sec \left (f x + e\right )}{d \tan \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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