\(\int (b \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [149]

Optimal result

Integrand size = 31, antiderivative size = 89 \[ \int (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}-\frac {3 (5 A+2 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d \sqrt {\sin ^2(c+d x)}} \] Output:

3/5*C*(b*cos(d*x+c))^(2/3)*sin(d*x+c)/d-3/10*(5*A+2*C)*(b*cos(d*x+c))^(2/3 
)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/ 
2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {3 b \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )+C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{8 d \sqrt [3]{b \cos (c+d x)}} \] Input:

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

Output:

(-3*b*Cot[c + d*x]*(4*A*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2] + 
 C*Cos[c + d*x]^2*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2])*Sqrt[S 
in[c + d*x]^2])/(8*d*(b*Cos[c + d*x])^(1/3))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 2030, 3493, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b \left (\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}-\frac {3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{10 b d \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

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

Output:

b*((3*C*(b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*b*d) - (3*(5*A + 2*C)*(b*C 
os[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c 
+ d*x])/(10*b*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 3493

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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