\(\int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx\) [636]

Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\frac {33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac {3 \left (8 a^2-3 b^2\right ) d \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{16 f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \] Output:

33/40*a*b*(d*sec(f*x+e))^(5/3)/f+3/16*(8*a^2-3*b^2)*d*hypergeom([-1/3, 1/2 
],[2/3],cos(f*x+e)^2)*(d*sec(f*x+e))^(2/3)*sin(f*x+e)/f/(sin(f*x+e)^2)^(1/ 
2)+3/8*b*(d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))/f
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\frac {3 (d \sec (e+f x))^{5/3} \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{5 f \sqrt {-\tan ^2(e+f x)}} \] Input:

Integrate[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])^2,x]
 

Output:

(3*(d*Sec[e + f*x])^(5/3)*(b^2*Hypergeometric2F1[-1/2, 5/6, 11/6, Sec[e + 
f*x]^2]*Tan[e + f*x] + a*(-(a*Hypergeometric2F1[1/2, 5/6, 11/6, Sec[e + f* 
x]^2]*Tan[e + f*x]) + 2*b*Sqrt[-Tan[e + f*x]^2])))/(5*f*Sqrt[-Tan[e + f*x] 
^2])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3993, 27, 3042, 3967, 3042, 4259, 3042, 3122}

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.

\(\displaystyle \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2dx\)

\(\Big \downarrow \) 3993

\(\displaystyle \frac {3}{8} \int \frac {1}{3} (d \sec (e+f x))^{5/3} \left (8 a^2+11 b \tan (e+f x) a-3 b^2\right )dx+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} \int (d \sec (e+f x))^{5/3} \left (8 a^2+11 b \tan (e+f x) a-3 b^2\right )dx+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \int (d \sec (e+f x))^{5/3} \left (8 a^2+11 b \tan (e+f x) a-3 b^2\right )dx+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 3967

\(\displaystyle \frac {1}{8} \left (\left (8 a^2-3 b^2\right ) \int (d \sec (e+f x))^{5/3}dx+\frac {33 a b (d \sec (e+f x))^{5/3}}{5 f}\right )+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \left (\left (8 a^2-3 b^2\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/3}dx+\frac {33 a b (d \sec (e+f x))^{5/3}}{5 f}\right )+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {1}{8} \left (\left (8 a^2-3 b^2\right ) \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3} \int \frac {1}{\left (\frac {\cos (e+f x)}{d}\right )^{5/3}}dx+\frac {33 a b (d \sec (e+f x))^{5/3}}{5 f}\right )+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} \left (\left (8 a^2-3 b^2\right ) \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}\right )^{5/3}}dx+\frac {33 a b (d \sec (e+f x))^{5/3}}{5 f}\right )+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {1}{8} \left (\frac {3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(e+f x)\right )}{2 f \sqrt {\sin ^2(e+f x)}}+\frac {33 a b (d \sec (e+f x))^{5/3}}{5 f}\right )+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\)

Input:

Int[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])^2,x]
 

Output:

((33*a*b*(d*Sec[e + f*x])^(5/3))/(5*f) + (3*(8*a^2 - 3*b^2)*d*Hypergeometr 
ic2F1[-1/3, 1/2, 2/3, Cos[e + f*x]^2]*(d*Sec[e + f*x])^(2/3)*Sin[e + f*x]) 
/(2*f*Sqrt[Sin[e + f*x]^2]))/8 + (3*b*(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e 
+ f*x]))/(8*f)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3967

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a   Int[(d 
*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] 
|| NeQ[a^2 + b^2, 0])
 

rule 3993

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^2, x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m 
 + 1))), x] + Simp[1/(m + 1)   Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^2 + 
a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^ 
2 + b^2, 0] &&  !IntegerQ[m]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]

Input:

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))^2,x)
 

Output:

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))^2,x)
 

Fricas [F]

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

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(5/3)*(a+b*tan(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

integrate((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a)^2, x)
 

Giac [F]

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

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \] Input:

int((d/cos(e + f*x))^(5/3)*(a + b*tan(e + f*x))^2,x)
 

Output:

int((d/cos(e + f*x))^(5/3)*(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx=\frac {d^{\frac {5}{3}} \left (6 \sec \left (f x +e \right )^{\frac {5}{3}} a b +5 \left (\int \sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )^{2}d x \right ) b^{2} f +5 \left (\int \sec \left (f x +e \right )^{\frac {5}{3}}d x \right ) a^{2} f \right )}{5 f} \] Input:

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e))^2,x)
                                                                                    
                                                                                    
 

Output:

(d**(2/3)*d*(6*sec(e + f*x)**(2/3)*sec(e + f*x)*a*b + 5*int(sec(e + f*x)** 
(2/3)*sec(e + f*x)*tan(e + f*x)**2,x)*b**2*f + 5*int(sec(e + f*x)**(2/3)*s 
ec(e + f*x),x)*a**2*f))/(5*f)