3.339 \(\int (b \sec (e+f x))^{4/3} (d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=64 \[ \frac{2 \cos ^2(e+f x)^{23/12} (b \sec (e+f x))^{4/3} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{5}{4},\frac{23}{12};\frac{9}{4};\sin ^2(e+f x)\right )}{5 d f} \]

[Out]

(2*(Cos[e + f*x]^2)^(23/12)*Hypergeometric2F1[5/4, 23/12, 9/4, Sin[e + f*x]^2]*(b*Sec[e + f*x])^(4/3)*(d*Tan[e
+ f*x])^(5/2))/(5*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.0580486, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2617} \[ \frac{2 \cos ^2(e+f x)^{23/12} (b \sec (e+f x))^{4/3} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{5}{4},\frac{23}{12};\frac{9}{4};\sin ^2(e+f x)\right )}{5 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^(23/12)*Hypergeometric2F1[5/4, 23/12, 9/4, Sin[e + f*x]^2]*(b*Sec[e + f*x])^(4/3)*(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 (b \sec (e+f x))^{4/3} (d \tan (e+f x))^{3/2} \, dx &=\frac{2 \cos ^2(e+f x)^{23/12} \, _2F_1\left (\frac{5}{4},\frac{23}{12};\frac{9}{4};\sin ^2(e+f x)\right ) (b \sec (e+f x))^{4/3} (d \tan (e+f x))^{5/2}}{5 d f}\\ \end{align*}

Mathematica [A]  time = 0.144881, size = 64, normalized size = 1. \[ \frac{3 d (b \sec (e+f x))^{4/3} \sqrt{d \tan (e+f x)} \, _2F_1\left (-\frac{1}{4},\frac{2}{3};\frac{5}{3};\sec ^2(e+f x)\right )}{4 f \sqrt [4]{-\tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.146, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( fx+e \right ) \right ) ^{{\frac{4}{3}}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(4/3)*(d*tan(f*x+e))^(3/2),x)

[Out]

int((b*sec(f*x+e))^(4/3)*(d*tan(f*x+e))^(3/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{\frac{4}{3}} \left (d \tan \left (f x + e\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(4/3)*(d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(4/3)*(d*tan(f*x + e))^(3/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \sec \left (f x + e\right )\right )^{\frac{1}{3}} \sqrt{d \tan \left (f x + e\right )} b d \sec \left (f x + e\right ) \tan \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(4/3)*(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))*b*d*sec(f*x + e)*tan(f*x + e), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**(4/3)*(d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{\frac{4}{3}} \left (d \tan \left (f x + e\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(4/3)*(d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(4/3)*(d*tan(f*x + e))^(3/2), x)