\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx\) [17]

Optimal result

Integrand size = 31, antiderivative size = 117 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}} \] Output:

3*A*hypergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(1/3)*sin(d*x 
+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/4*B*hypergeom([-2/3, 1/2],[1/3],cos(d*x+c)^ 
2)*(b*sec(d*x+c))^(4/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx=-\frac {3 \csc ^3(c+d x) \left (7 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )+4 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right )\right ) \left (-\tan ^2(c+d x)\right )^{3/2}}{28 d (b \sec (c+d x))^{2/3}} \] Input:

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

Output:

(-3*Csc[c + d*x]^3*(7*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Sec[ 
c + d*x]^2] + 4*B*Hypergeometric2F1[1/2, 7/6, 13/6, Sec[c + d*x]^2])*(-Tan 
[c + d*x]^2)^(3/2))/(28*d*(b*Sec[c + d*x])^(2/3))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, 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.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \sec (c+d x))^{4/3} (A+B \sec (c+d x))dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {A \int (b \sec (c+d x))^{4/3}dx+\frac {B \int (b \sec (c+d x))^{7/3}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx}{b}}{b^2}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{4/3}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{4/3}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{7/3}}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {3 A b \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}}{b^2}\)

Input:

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

Output:

((3*A*b*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x]) 
^(1/3)*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[-2/ 
3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(4/3)*Sin[c + d*x])/(4*d*Sqr 
t[Sin[c + d*x]^2]))/b^2
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*(b*sec(d*x + c))^(1/3)/b, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{2/3}} \, dx=\frac {\left (\int \sec \left (d x +c \right )^{\frac {7}{3}}d x \right ) b +\left (\int \sec \left (d x +c \right )^{\frac {4}{3}}d x \right ) a}{b^{\frac {2}{3}}} \] Input:

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

Output:

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