\(\int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx\) [8]

Optimal result

Integrand size = 29, antiderivative size = 116 \[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{5 b d \sqrt {\sin ^2(c+d x)}} \] Output:

3/2*A*hypergeom([-1/3, 1/2],[2/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(2/3)*sin(d 
*x+c)/d/(sin(d*x+c)^2)^(1/2)+3/5*B*hypergeom([-5/6, 1/2],[1/6],cos(d*x+c)^ 
2)*(b*sec(d*x+c))^(5/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\frac {3 \csc (c+d x) \left (8 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(c+d x)\right )+5 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{5/3} \sqrt {-\tan ^2(c+d x)}}{40 b d} \] Input:

Integrate[Sec[c + d*x]*(b*Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x]),x]
 

Output:

(3*Csc[c + d*x]*(8*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Sec[c 
+ d*x]^2] + 5*B*Hypergeometric2F1[1/2, 4/3, 7/3, Sec[c + d*x]^2])*(b*Sec[c 
 + d*x])^(5/3)*Sqrt[-Tan[c + d*x]^2])/(40*b*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2030, 3042, 4274, 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.

\(\displaystyle \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \sec (c+d x))^{5/3} (A+B \sec (c+d x))dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {A \int (b \sec (c+d x))^{5/3}dx+\frac {B \int (b \sec (c+d x))^{8/3}dx}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{8/3}dx}{b}}{b}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {A \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{5/3}}dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{8/3}}dx}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{5/3}}dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{8/3}}dx}{b}}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {3 A b \sin (c+d x) (b \sec (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)}}}{b}\)

Input:

Int[Sec[c + d*x]*(b*Sec[c + d*x])^(2/3)*(A + B*Sec[c + d*x]),x]
 

Output:

((3*A*b*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x]) 
^(2/3)*Sin[c + d*x])/(2*d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[- 
5/6, 1/2, 1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(5/3)*Sin[c + d*x])/(5*d*S 
qrt[Sin[c + d*x]^2]))/b
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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 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]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

Maple [F]

\[\int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]

Input:

int(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x)
 

Output:

int(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x)
 

Fricas [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right ) \,d x } \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x, algorithm="f 
ricas")
 

Output:

integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*(b*sec(d*x + c))^(2/3), x)
 

Sympy [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))**(2/3)*(A+B*sec(d*x+c)),x)
 

Output:

Integral((b*sec(c + d*x))**(2/3)*(A + B*sec(c + d*x))*sec(c + d*x), x)
 

Maxima [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right ) \,d x } \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x, algorithm="m 
axima")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)*sec(d*x + c), x)
 

Giac [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right ) \,d x } \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x, algorithm="g 
iac")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)*sec(d*x + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{\cos \left (c+d\,x\right )} \,d x \] Input:

int(((A + B/cos(c + d*x))*(b/cos(c + d*x))^(2/3))/cos(c + d*x),x)
 

Output:

int(((A + B/cos(c + d*x))*(b/cos(c + d*x))^(2/3))/cos(c + d*x), x)
 

Reduce [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=b^{\frac {2}{3}} \left (\left (\int \sec \left (d x +c \right )^{\frac {8}{3}}d x \right ) b +\left (\int \sec \left (d x +c \right )^{\frac {5}{3}}d x \right ) a \right ) \] Input:

int(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x)
 

Output:

b**(2/3)*(int(sec(c + d*x)**(2/3)*sec(c + d*x)**2,x)*b + int(sec(c + d*x)* 
*(2/3)*sec(c + d*x),x)*a)