3.12.57 \(\int \cos ^m(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [1157]

3.12.57.1 Optimal result

Integrand size = 39, antiderivative size = 235 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(b B+a C) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}+\frac {b C \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m)}-\frac {((b B+a C) (1+m)+a A (2+m)) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {(b C (2+m)+A b (3+m)+a B (3+m)) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) (3+m) \sqrt {\sin ^2(c+d x)}} \]

output

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

3.12.57.2 Mathematica [A] (verified)

Time = 1.91 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.87 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\cos ^{1+m}(c+d x) \csc (c+d x) \left (-\frac {a A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )}{1+m}+\cos (c+d x) \left (-\frac {(A b+a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right )}{2+m}+\cos (c+d x) \left (-\frac {(b B+a C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(c+d x)\right )}{3+m}-\frac {b C \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\cos ^2(c+d x)\right )}{4+m}\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{d} \]

input

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

output

(Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(-((a*A*Hypergeometric2F1[1/2, (1 + m)/ 
2, (3 + m)/2, Cos[c + d*x]^2])/(1 + m)) + Cos[c + d*x]*(-(((A*b + a*B)*Hyp 
ergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2])/(2 + m)) + Cos[ 
c + d*x]*(-(((b*B + a*C)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[ 
c + d*x]^2])/(3 + m)) - (b*C*Cos[c + d*x]*Hypergeometric2F1[1/2, (4 + m)/2 
, (6 + m)/2, Cos[c + d*x]^2])/(4 + m))))*Sqrt[Sin[c + d*x]^2])/d
 

3.12.57.3 Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {3042, 3512, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\int \cos ^m(c+d x) ((m+3) ((b B+a C) (m+1)+a A (m+2))+(m+2) (b C (m+2)+A b (m+3)+a B (m+3)) \cos (c+d x))dx}{m+2}+\frac {(m+3) (a C+b B) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((m+3) ((b B+a C) (m+1)+a A (m+2))+(m+2) (b C (m+2)+A b (m+3)+a B (m+3)) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{m+2}+\frac {(m+3) (a C+b B) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {(m+2) (a B (m+3)+A b (m+3)+b C (m+2)) \int \cos ^{m+1}(c+d x)dx+(m+3) (a A (m+2)+(m+1) (a C+b B)) \int \cos ^m(c+d x)dx}{m+2}+\frac {(m+3) (a C+b B) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(m+3) (a A (m+2)+(m+1) (a C+b B)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx+(m+2) (a B (m+3)+A b (m+3)+b C (m+2)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx}{m+2}+\frac {(m+3) (a C+b B) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {b C \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3122

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

input

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

output

(b*C*Cos[c + d*x]^(2 + m)*Sin[c + d*x])/(d*(3 + m)) + (((b*B + a*C)*(3 + m 
)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) + (-(((3 + m)*((b*B + a*C 
)*(1 + m) + a*A*(2 + m))*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + 
m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + m)*Sqrt[Sin[c + d*x 
]^2])) - ((b*C*(2 + m) + A*b*(3 + m) + a*B*(3 + m))*Cos[c + d*x]^(2 + m)*H 
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/ 
(d*Sqrt[Sin[c + d*x]^2]))/(2 + m))/(3 + m)
 

3.12.57.3.1 Defintions of rubi rules used

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]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

3.12.57.4 Maple [F]

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

input

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

output

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

3.12.57.5 Fricas [F]

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

input

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

output

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

3.12.57.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.12.57.7 Maxima [F]

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

input

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

3.12.57.8 Giac [F]

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

input

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

3.12.57.9 Mupad [F(-1)]

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

input

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

output

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