\(\int \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{3/2} \, dx\) [340]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\frac {2 \cos ^2(e+f x)^{17/12} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {17}{12},\frac {9}{4},\sin ^2(e+f x)\right ) \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{5/2}}{5 d f} \] Output:

2/5*(cos(f*x+e)^2)^(17/12)*hypergeom([5/4, 17/12],[9/4],sin(f*x+e)^2)*(b*s 
ec(f*x+e))^(1/3)*(d*tan(f*x+e))^(5/2)/d/f
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97 \[ \int \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\frac {3 d \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{6},\frac {7}{6},\sec ^2(e+f x)\right ) \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)}}{f \sqrt [4]{-\tan ^2(e+f x)}} \] Input:

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

Output:

(3*d*Hypergeometric2F1[-1/4, 1/6, 7/6, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(1 
/3)*Sqrt[d*Tan[e + f*x]])/(f*(-Tan[e + f*x]^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3042, 3097}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3097

\(\displaystyle \frac {2 \cos ^2(e+f x)^{17/12} \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {17}{12},\frac {9}{4},\sin ^2(e+f x)\right )}{5 d f}\)

Input:

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

Output:

(2*(Cos[e + f*x]^2)^(17/12)*Hypergeometric2F1[5/4, 17/12, 9/4, Sin[e + f*x 
]^2]*(b*Sec[e + f*x])^(1/3)*(d*Tan[e + f*x])^(5/2))/(5*d*f)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3097

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] &&  !IntegerQ[m/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\frac {3 \sqrt {d}\, b^{\frac {1}{3}} d \left (2 \sqrt {\tan \left (f x +e \right )}\, \sec \left (f x +e \right )^{\frac {1}{3}}-\left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sec \left (f x +e \right )^{\frac {1}{3}}}{\tan \left (f x +e \right )}d x \right ) f \right )}{5 f} \] Input:

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

Output:

(3*sqrt(d)*b**(1/3)*d*(2*sqrt(tan(e + f*x))*sec(e + f*x)**(1/3) - int((sqr 
t(tan(e + f*x))*sec(e + f*x)**(1/3))/tan(e + f*x),x)*f))/(5*f)