\(\int \frac {(d \tan (e+f x))^{3/2}}{(b \sec (e+f x))^{4/3}} \, dx\) [342]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(b \sec (e+f x))^{4/3}} \, dx=\frac {2 \cos ^2(e+f x)^{7/12} \operatorname {Hypergeometric2F1}\left (\frac {7}{12},\frac {5}{4},\frac {9}{4},\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{5 d f (b \sec (e+f x))^{4/3}} \] Output:

2/5*(cos(f*x+e)^2)^(7/12)*hypergeom([7/12, 5/4],[9/4],sin(f*x+e)^2)*(d*tan 
(f*x+e))^(5/2)/d/f/(b*sec(f*x+e))^(4/3)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(b \sec (e+f x))^{4/3}} \, dx=\frac {3 \cot ^3(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-\frac {1}{4},\frac {1}{3},\sec ^2(e+f x)\right ) (d \tan (e+f x))^{3/2} \left (-\tan ^2(e+f x)\right )^{3/4}}{4 f (b \sec (e+f x))^{4/3}} \] Input:

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

Output:

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

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3097

\(\displaystyle \frac {2 \cos ^2(e+f x)^{7/12} (d \tan (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{12},\frac {5}{4},\frac {9}{4},\sin ^2(e+f x)\right )}{5 d f (b \sec (e+f x))^{4/3}}\)

Input:

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

Output:

(2*(Cos[e + f*x]^2)^(7/12)*Hypergeometric2F1[7/12, 5/4, 9/4, Sin[e + f*x]^ 
2]*(d*Tan[e + f*x])^(5/2))/(5*d*f*(b*Sec[e + f*x])^(4/3))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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