3.4.48 \(\int (b \sec (e+f x))^{3/2} \sqrt [3]{d \tan (e+f x)} \, dx\) [348]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {2697} \begin {gather*} \frac {3 \cos ^2(e+f x)^{17/12} (b \sec (e+f x))^{3/2} (d \tan (e+f x))^{4/3} \, _2F_1\left (\frac {2}{3},\frac {17}{12};\frac {5}{3};\sin ^2(e+f x)\right )}{4 d f} \end {gather*}

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)^(17/12)*Hypergeometric2F1[2/3, 17/12, 5/3, Sin[e + f*x]^2]*(b*Sec[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(4/3))/(4*d*f)

Rule 2697

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)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 64, normalized size = 1.00 \begin {gather*} \frac {2 d \, _2F_1\left (\frac {1}{3},\frac {3}{4};\frac {7}{4};\sec ^2(e+f x)\right ) (b \sec (e+f x))^{3/2} \sqrt [3]{-\tan ^2(e+f x)}}{3 f (d \tan (e+f x))^{2/3}} \end {gather*}

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

________________________________________________________________________________________

Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \left (b \sec \left (f x +e \right )\right )^{\frac {3}{2}} \left (d \tan \left (f x +e \right )\right )^{\frac {1}{3}}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(1/3)*b*sec(f*x + e), x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________