\(\int (a \cos (c+d x))^m (b \cos (c+d x))^n (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [216]

Optimal result

Integrand size = 40, antiderivative size = 167 \[ \int (a \cos (c+d x))^m (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {B (a \cos (c+d x))^{2+m} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+m+n),\frac {1}{2} (4+m+n),\cos ^2(c+d x)\right ) \sin (c+d x)}{a^2 d (2+m+n) \sqrt {\sin ^2(c+d x)}}-\frac {C (a \cos (c+d x))^{3+m} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (3+m+n),\frac {1}{2} (5+m+n),\cos ^2(c+d x)\right ) \sin (c+d x)}{a^3 d (3+m+n) \sqrt {\sin ^2(c+d x)}} \] Output:

-B*(a*cos(d*x+c))^(2+m)*(b*cos(d*x+c))^n*hypergeom([1/2, 1+1/2*m+1/2*n],[2 
+1/2*m+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/a^2/d/(2+m+n)/(sin(d*x+c)^2)^(1/2)- 
C*(a*cos(d*x+c))^(3+m)*(b*cos(d*x+c))^n*hypergeom([1/2, 3/2+1/2*m+1/2*n],[ 
5/2+1/2*m+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/a^3/d/(3+m+n)/(sin(d*x+c)^2)^(1/ 
2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.81 \[ \int (a \cos (c+d x))^m (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {\cos (c+d x) (a \cos (c+d x))^m (b \cos (c+d x))^n \cot (c+d x) \left (B (3+m+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+m+n),\frac {1}{2} (4+m+n),\cos ^2(c+d x)\right )+C (2+m+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (3+m+n),\frac {1}{2} (5+m+n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (2+m+n) (3+m+n)} \] Input:

Integrate[(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c 
+ d*x]^2),x]
 

Output:

-((Cos[c + d*x]*(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*Cot[c + d*x]*(B*(3 + 
 m + n)*Hypergeometric2F1[1/2, (2 + m + n)/2, (4 + m + n)/2, Cos[c + d*x]^ 
2] + C*(2 + m + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (3 + m + n)/2, (5 + 
 m + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(2 + m + n)*(3 + m + 
n)))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2034, 3042, 3489, 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.

\(\displaystyle \int (a \cos (c+d x))^m (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle (a \cos (c+d x))^{-n} (b \cos (c+d x))^n \int (a \cos (c+d x))^{m+n} \left (C \cos ^2(c+d x)+B \cos (c+d x)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (a \cos (c+d x))^{-n} (b \cos (c+d x))^n \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{m+n} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3489

\(\displaystyle \frac {(a \cos (c+d x))^{-n} (b \cos (c+d x))^n \int (a \cos (c+d x))^{m+n+1} (B+C \cos (c+d x))dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a \cos (c+d x))^{-n} (b \cos (c+d x))^n \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{m+n+1} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {(a \cos (c+d x))^{-n} (b \cos (c+d x))^n \left (B \int (a \cos (c+d x))^{m+n+1}dx+\frac {C \int (a \cos (c+d x))^{m+n+2}dx}{a}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a \cos (c+d x))^{-n} (b \cos (c+d x))^n \left (B \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{m+n+1}dx+\frac {C \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{m+n+2}dx}{a}\right )}{a}\)

\(\Big \downarrow \) 3122

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

Input:

Int[(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x] 
^2),x]
 

Output:

((b*Cos[c + d*x])^n*(-((B*(a*Cos[c + d*x])^(2 + m + n)*Hypergeometric2F1[1 
/2, (2 + m + n)/2, (4 + m + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a*d*(2 + 
m + n)*Sqrt[Sin[c + d*x]^2])) - (C*(a*Cos[c + d*x])^(3 + m + n)*Hypergeome 
tric2F1[1/2, (3 + m + n)/2, (5 + m + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/( 
a^2*d*(3 + m + n)*Sqrt[Sin[c + d*x]^2])))/(a*(a*Cos[c + d*x])^n)
 

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]
 

rule 3489

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

Maple [F]

\[\int \left (a \cos \left (d x +c \right )\right )^{m} \left (b \cos \left (d x +c \right )\right )^{n} \left (B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )d x\]

Input:

int((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

int((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Fricas [F]

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

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^2 + B*cos(d*x + c))*(a*cos(d*x + c))^m*(b*cos(d*x 
 + c))^n, x)
 

Sympy [F]

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

integrate((a*cos(d*x+c))**m*(b*cos(d*x+c))**n*(B*cos(d*x+c)+C*cos(d*x+c)** 
2),x)
 

Output:

Integral((a*cos(c + d*x))**m*(b*cos(c + d*x))**n*(B + C*cos(c + d*x))*cos( 
c + d*x), x)
 

Maxima [F]

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

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(a*cos(d*x + c))^m*(b*cos(d* 
x + c))^n, x)
 

Giac [F]

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

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(a*cos(d*x + c))^m*(b*cos(d* 
x + c))^n, x)
 

Mupad [F(-1)]

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

int((a*cos(c + d*x))^m*(b*cos(c + d*x))^n*(B*cos(c + d*x) + C*cos(c + d*x) 
^2),x)
 

Output:

int((a*cos(c + d*x))^m*(b*cos(c + d*x))^n*(B*cos(c + d*x) + C*cos(c + d*x) 
^2), x)
 

Reduce [F]

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

int((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

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