\(\int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [50]

Optimal result

Integrand size = 39, antiderivative size = 146 \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b^2 (4 A+C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 b C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d} \] Output:

-3/8*b^2*(4*A+C)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b* 
sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)+3*b*B*hypergeom([-1/6, 1/2],[5/6],c 
os(d*x+c)^2)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)+3/4*b* 
C*(b*sec(d*x+c))^(1/3)*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.79 \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \cot (c+d x) \sqrt [3]{b \sec (c+d x)} \left (C \tan ^2(c+d x)+(4 A+C) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}+B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right ) \sec (c+d x) \sqrt {-\tan ^2(c+d x)}\right )}{4 d} \] Input:

Integrate[Cos[c + d*x]*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(3*b*Cot[c + d*x]*(b*Sec[c + d*x])^(1/3)*(C*Tan[c + d*x]^2 + (4*A + C)*Hyp 
ergeometric2F1[1/6, 1/2, 7/6, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2] + B*Hy 
pergeometric2F1[1/2, 2/3, 5/3, Sec[c + d*x]^2]*Sec[c + d*x]*Sqrt[-Tan[c + 
d*x]^2]))/(4*d)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 2030, 4535, 3042, 4259, 3042, 3122, 4534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b \left (-\frac {3 b (4 A+C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d}\right )\)

Input:

Int[Cos[c + d*x]*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d* 
x]^2),x]
 

Output:

b*((-3*b*(4*A + C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c 
+ d*x])/(8*d*(b*Sec[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeom 
etric2F1[-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*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/( 
4*d))
 

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 4534

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

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

Input:

int(cos(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

int(cos(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Fricas [F]

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

integrate(cos(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2 
),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)*(b*sec(d*x+c))^(4/3)*(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))^(4/3)*c 
os(d*x + c), x)
 

Giac [F]

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

integrate(cos(d*x+c)*(b*sec(d*x+c))^(4/3)*(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))^(4/3)*c 
os(d*x + c), x)
 

Mupad [F(-1)]

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

int(cos(c + d*x)*(b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d* 
x)^2),x)
 

Output:

int(cos(c + d*x)*(b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d* 
x)^2), x)
 

Reduce [F]

\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=b^{\frac {4}{3}} \left (\left (\int \sec \left (d x +c \right )^{\frac {10}{3}} \cos \left (d x +c \right )d x \right ) c +\left (\int \sec \left (d x +c \right )^{\frac {7}{3}} \cos \left (d x +c \right )d x \right ) b +\left (\int \sec \left (d x +c \right )^{\frac {4}{3}} \cos \left (d x +c \right )d x \right ) a \right ) \] Input:

int(cos(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
                                                                                    
                                                                                    
 

Output:

b**(1/3)*b*(int(sec(c + d*x)**(1/3)*cos(c + d*x)*sec(c + d*x)**3,x)*c + in 
t(sec(c + d*x)**(1/3)*cos(c + d*x)*sec(c + d*x)**2,x)*b + int(sec(c + d*x) 
**(1/3)*cos(c + d*x)*sec(c + d*x),x)*a)