\(\int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [45]

Optimal result

Integrand size = 41, antiderivative size = 150 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b (A+4 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^2 \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 \cot (c+d x) \left (A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right )+4 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )-2 C \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{2/3} \sqrt {-\tan ^2(c+d x)}}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 2030, 4535, 3042, 4259, 3042, 3122, 4533, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b^2 \left (\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}\right )\)

\(\Big \downarrow \) 4533

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

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {(A+4 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{4 b^2}+\frac {3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}\right )\)

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {(A+4 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{2/3}}dx}{4 b^2}+\frac {3 A \tan (c+d x)}{4 d (b \sec (c+d x))^{4/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}\right )\)

\(\Big \downarrow \) 3122

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

Input:

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

Output:

b^2*((-3*B*Hypergeometric2F1[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]) - (3*(A + 4*C)*Hypergeom 
etric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(4*b*d*(b*Sec[c + d* 
x])^(1/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*Tan[c + d*x])/(4*d*(b*Sec[c + d*x]) 
^(4/3)))
 

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 4533

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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