\(\int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx\) [11]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 A b^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \sec (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}} \] Output:

-3/7*A*b^3*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d 
*x+c))^(7/3)/(sin(d*x+c)^2)^(1/2)-3/4*b^2*B*hypergeom([1/2, 2/3],[5/3],cos 
(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx=-\frac {3 b \cot (c+d x) \left (A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right )+4 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{4 d \sqrt [3]{b \sec (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 118, 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 = {3042, 2030, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b^2 \int \frac {A+B \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{4/3}}dx\)

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b^2 \left (-\frac {3 A b \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}\right )\)

Input:

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

Output:

b^2*((-3*A*b*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x 
])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeometric 
2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*(b*Sec[c + d*x])^(4/ 
3)*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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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