\(\int \cos ^2(c+d x) (b \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \, dx\) [146]

Optimal result

Integrand size = 33, antiderivative size = 95 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{11/3} \sin (c+d x)}{14 b^3 d}-\frac {3 (14 A+11 C) (b \cos (c+d x))^{11/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{154 b^3 d \sqrt {\sin ^2(c+d x)}} \] Output:

3/14*C*(b*cos(d*x+c))^(11/3)*sin(d*x+c)/b^3/d-3/154*(14*A+11*C)*(b*cos(d*x 
+c))^(11/3)*hypergeom([1/2, 11/6],[17/6],cos(d*x+c)^2)*sin(d*x+c)/b^3/d/(s 
in(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (b \cos (c+d x))^{2/3} \cot (c+d x) \left (17 A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\cos ^2(c+d x)\right )+11 C \cos ^4(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {17}{6},\frac {23}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{187 d} \] Input:

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

Output:

(-3*(b*Cos[c + d*x])^(2/3)*Cot[c + d*x]*(17*A*Cos[c + d*x]^2*Hypergeometri 
c2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2] + 11*C*Cos[c + d*x]^4*Hypergeometric 
2F1[1/2, 17/6, 23/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(187*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2030, 3042, 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.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{8/3} \left (C \cos ^2(c+d x)+A\right )dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{8/3} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{b^2}\)

\(\Big \downarrow \) 3493

\(\displaystyle \frac {\frac {1}{14} (14 A+11 C) \int (b \cos (c+d x))^{8/3}dx+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{11/3}}{14 b d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{14} (14 A+11 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{8/3}dx+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{11/3}}{14 b d}}{b^2}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {3 C \sin (c+d x) (b \cos (c+d x))^{11/3}}{14 b d}-\frac {3 (14 A+11 C) \sin (c+d x) (b \cos (c+d x))^{11/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\cos ^2(c+d x)\right )}{154 b d \sqrt {\sin ^2(c+d x)}}}{b^2}\)

Input:

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

Output:

((3*C*(b*Cos[c + d*x])^(11/3)*Sin[c + d*x])/(14*b*d) - (3*(14*A + 11*C)*(b 
*Cos[c + d*x])^(11/3)*Hypergeometric2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2]*S 
in[c + d*x])/(154*b*d*Sqrt[Sin[c + d*x]^2]))/b^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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos ^2(c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=b^{\frac {2}{3}} \left (\left (\int \cos \left (d x +c \right )^{\frac {14}{3}}d x \right ) c +\left (\int \cos \left (d x +c \right )^{\frac {8}{3}}d x \right ) a \right ) \] Input:

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

Output:

b**(2/3)*(int(cos(c + d*x)**(2/3)*cos(c + d*x)**4,x)*c + int(cos(c + d*x)* 
*(2/3)*cos(c + d*x)**2,x)*a)