\(\int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) [902]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}} \] Output:

3/2*A*hypergeom([-1/3, 1/2],[2/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d*x+c) 
)^(2/3)/(sin(d*x+c)^2)^(1/2)-3*B*(b*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2] 
,[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 \csc (c+d x) \left (A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )-2 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{2 d (b \cos (c+d x))^{2/3}} \] Input:

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

Output:

(3*Csc[c + d*x]*(A*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2] - 2*B 
*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2])*Sqrt[Sin[c 
 + d*x]^2])/(2*d*(b*Cos[c + d*x])^(2/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 2030, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b \left (\frac {3 A \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 b d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{b^2 d \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

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

Output:

b*((3*A*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2 
*b*d*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^ 
(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b^2* 
d*Sqrt[Sin[c + d*x]^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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {\left (\int \frac {\sec \left (d x +c \right )}{\cos \left (d x +c \right )^{\frac {2}{3}}}d x \right ) a +\left (\int \cos \left (d x +c \right )^{\frac {1}{3}} \sec \left (d x +c \right )d x \right ) b}{b^{\frac {2}{3}}} \] Input:

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

Output:

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