\(\int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx\) [350]

Optimal result

Integrand size = 25, antiderivative size = 62 \[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=-\frac {3 \cos ^2(e+f x)^{7/12} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {7}{12},\frac {5}{6},\sin ^2(e+f x)\right ) (b \sec (e+f x))^{3/2}}{d f \sqrt [3]{d \tan (e+f x)}} \] Output:

-3*(cos(f*x+e)^2)^(7/12)*hypergeom([-1/6, 7/12],[5/6],sin(f*x+e)^2)*(b*sec 
(f*x+e))^(3/2)/d/f/(d*tan(f*x+e))^(1/3)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\frac {2 d \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{6},\frac {7}{4},\sec ^2(e+f x)\right ) (b \sec (e+f x))^{3/2} \left (-\tan ^2(e+f x)\right )^{7/6}}{3 f (d \tan (e+f x))^{7/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 62, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3097

\(\displaystyle -\frac {3 \cos ^2(e+f x)^{7/12} (b \sec (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {7}{12},\frac {5}{6},\sin ^2(e+f x)\right )}{d f \sqrt [3]{d \tan (e+f x)}}\)

Input:

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

Output:

(-3*(Cos[e + f*x]^2)^(7/12)*Hypergeometric2F1[-1/6, 7/12, 5/6, Sin[e + f*x 
]^2]*(b*Sec[e + f*x])^(3/2))/(d*f*(d*Tan[e + f*x])^(1/3))
 

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]

\[\int \frac {\left (b \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x +e \right )\right )^{\frac {4}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\int { \frac {\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x + e\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\int \frac {\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {4}{3}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\int { \frac {\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x + e\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\int { \frac {\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \tan \left (f x + e\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{4/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(b \sec (e+f x))^{3/2}}{(d \tan (e+f x))^{4/3}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )}{\tan \left (f x +e \right )^{\frac {4}{3}}}d x \right ) b}{d^{\frac {4}{3}}} \] Input:

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

Output:

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