\(\int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx\) [928]

Optimal result

Integrand size = 31, antiderivative size = 167 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=-\frac {3 A \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{2+m}(c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (8+3 m),\frac {1}{6} (14+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (8+3 m) \sqrt {\sin ^2(c+d x)}} \] Output:

-3*A*cos(d*x+c)^(1+m)*(b*cos(d*x+c))^(2/3)*hypergeom([1/2, 5/6+1/2*m],[11/ 
6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(5+3*m)/(sin(d*x+c)^2)^(1/2)-3*B*cos(d 
*x+c)^(2+m)*(b*cos(d*x+c))^(2/3)*hypergeom([1/2, 4/3+1/2*m],[7/3+1/2*m],co 
s(d*x+c)^2)*sin(d*x+c)/d/(8+3*m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=-\frac {3 \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \csc (c+d x) \left (A (8+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right )+B (5+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (8+3 m),\frac {7}{3}+\frac {m}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (5+3 m) (8+3 m)} \] Input:

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

Output:

(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(A*(8 + 3*m)* 
Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2] + B*(5 + 
 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (8 + 3*m)/6, 7/3 + m/2, Cos[c + 
d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(5 + 3*m)*(8 + 3*m))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2034, 3042, 3227, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {(b \cos (c+d x))^{2/3} \left (A \int \cos ^{m+\frac {2}{3}}(c+d x)dx+B \int \cos ^{m+\frac {5}{3}}(c+d x)dx\right )}{\cos ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b \cos (c+d x))^{2/3} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {5}{3}}dx\right )}{\cos ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {(b \cos (c+d x))^{2/3} \left (-\frac {3 A \sin (c+d x) \cos ^{m+\frac {5}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+5),\frac {1}{6} (3 m+11),\cos ^2(c+d x)\right )}{d (3 m+5) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+\frac {8}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+8),\frac {1}{6} (3 m+14),\cos ^2(c+d x)\right )}{d (3 m+8) \sqrt {\sin ^2(c+d x)}}\right )}{\cos ^{\frac {2}{3}}(c+d x)}\)

Input:

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

Output:

((b*Cos[c + d*x])^(2/3)*((-3*A*Cos[c + d*x]^(5/3 + m)*Hypergeometric2F1[1/ 
2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 + 3*m)*S 
qrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(8/3 + m)*Hypergeometric2F1[1/2, 
(8 + 3*m)/6, (14 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(8 + 3*m)*Sqrt 
[Sin[c + d*x]^2])))/Cos[c + d*x]^(2/3)
 

Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((b*cos(c + d*x))**(2/3)*(A + B*cos(c + d*x))*cos(c + d*x)**m, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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