3.4.69 \(\int (a \cos (c+d x))^m (b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [369]

3.4.69.1 Optimal result

Integrand size = 41, antiderivative size = 227 \[ \int (a \cos (c+d x))^m (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {C (a \cos (c+d x))^{1+m} (b \cos (c+d x))^n \sin (c+d x)}{a d (2+m+n)}-\frac {(C (1+m+n)+A (2+m+n)) (a \cos (c+d x))^{1+m} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\cos ^2(c+d x)\right ) \sin (c+d x)}{a d (1+m+n) (2+m+n) \sqrt {\sin ^2(c+d x)}}-\frac {B (a \cos (c+d x))^{2+m} (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+m+n),\frac {1}{2} (4+m+n),\cos ^2(c+d x)\right ) \sin (c+d x)}{a^2 d (2+m+n) \sqrt {\sin ^2(c+d x)}} \]

output

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

3.4.69.2 Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.69 \[ \int (a \cos (c+d x))^m (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(a \cos (c+d x))^m (b \cos (c+d x))^n \cot (c+d x) \left (C \sin ^2(c+d x)-\frac {(C (1+m+n)+A (2+m+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{1+m+n}-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+m+n),\frac {1}{2} (4+m+n),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{d (2+m+n)} \]

input

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

output

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

3.4.69.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2034, 3042, 3502, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

input

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

output

((b*Cos[c + d*x])^n*((C*(a*Cos[c + d*x])^(1 + m + n)*Sin[c + d*x])/(a*d*(2 
 + m + n)) + (-(((C*(1 + m + n) + A*(2 + m + n))*(a*Cos[c + d*x])^(1 + m + 
 n)*Hypergeometric2F1[1/2, (1 + m + n)/2, (3 + m + n)/2, Cos[c + d*x]^2]*S 
in[c + d*x])/(d*(1 + m + n)*Sqrt[Sin[c + d*x]^2])) - (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*Sqrt[Sin[c + d*x]^2]))/(a*(2 + m + n))))/(a*Cos[c 
 + d*x])^n
 

3.4.69.3.1 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 3502

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

3.4.69.4 Maple [F]

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

input

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

output

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

3.4.69.5 Fricas [F]

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

input

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(A+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)*(a*cos(d*x + c))^m*(b*cos 
(d*x + c))^n, x)
 

3.4.69.6 Sympy [F]

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

input

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

3.4.69.7 Maxima [F]

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

input

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(A+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)*(a*cos(d*x + c))^m*(b*co 
s(d*x + c))^n, x)
 

3.4.69.8 Giac [F]

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

input

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(A+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)*(a*cos(d*x + c))^m*(b*co 
s(d*x + c))^n, x)
 

3.4.69.9 Mupad [F(-1)]

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

input

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