\(\int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx\) [341]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\frac {2 \cos ^2(e+f x)^{13/12} \operatorname {Hypergeometric2F1}\left (\frac {13}{12},\frac {5}{4},\frac {9}{4},\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{5 d f \sqrt [3]{b \sec (e+f x)}} \]

[Out]

2/5*(cos(f*x+e)^2)^(13/12)*hypergeom([13/12, 5/4],[9/4],sin(f*x+e)^2)*(d*tan(f*x+e))^(5/2)/d/f/(b*sec(f*x+e))^
(1/3)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2697} \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\frac {2 \cos ^2(e+f x)^{13/12} (d \tan (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {13}{12},\frac {5}{4},\frac {9}{4},\sin ^2(e+f x)\right )}{5 d f \sqrt [3]{b \sec (e+f x)}} \]

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^(13/12)*Hypergeometric2F1[13/12, 5/4, 9/4, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(5/2))/(5*d*f*
(b*Sec[e + f*x])^(1/3))

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*} \text {integral}& = \frac {2 \cos ^2(e+f x)^{13/12} \operatorname {Hypergeometric2F1}\left (\frac {13}{12},\frac {5}{4},\frac {9}{4},\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{5 d f \sqrt [3]{b \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08 \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\frac {3 \cot ^3(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-\frac {1}{6},\frac {5}{6},\sec ^2(e+f x)\right ) (d \tan (e+f x))^{3/2} \left (-\tan ^2(e+f x)\right )^{3/4}}{f \sqrt [3]{b \sec (e+f x)}} \]

[In]

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

[Out]

(3*Cot[e + f*x]^3*Hypergeometric2F1[-1/4, -1/6, 5/6, Sec[e + f*x]^2]*(d*Tan[e + f*x])^(3/2)*(-Tan[e + f*x]^2)^
(3/4))/(f*(b*Sec[e + f*x])^(1/3))

Maple [F]

\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sec (e+f x)}} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \]

[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)