\(\int (c \cos (e+f x))^m (a+b \cos (e+f x)) (A+B \cos (e+f x)) \, dx\) [453]

Optimal result

Integrand size = 31, antiderivative size = 196 \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x)) (A+B \cos (e+f x)) \, dx=\frac {b B (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m)}-\frac {(b B (1+m)+a A (2+m)) (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c f (1+m) (2+m) \sqrt {\sin ^2(e+f x)}}-\frac {(A b+a B) (c \cos (e+f x))^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c^2 f (2+m) \sqrt {\sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3047, 3102, 2827, 2722} \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x)) (A+B \cos (e+f x)) \, dx=-\frac {(a B+A b) \sin (e+f x) (c \cos (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(e+f x)\right )}{c^2 f (m+2) \sqrt {\sin ^2(e+f x)}}-\frac {\sin (e+f x) (a A (m+2)+b B (m+1)) (c \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{c f (m+1) (m+2) \sqrt {\sin ^2(e+f x)}}+\frac {b B \sin (e+f x) (c \cos (e+f x))^{m+1}}{c f (m+2)} \]

[In]

Int[(c*Cos[e + f*x])^m*(a + b*Cos[e + f*x])*(A + B*Cos[e + f*x]),x]

[Out]

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

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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

Rule 3047

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

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \int (c \cos (e+f x))^m \left (a A+(A b+a B) \cos (e+f x)+b B \cos ^2(e+f x)\right ) \, dx \\ & = \frac {b B (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m)}+\frac {\int (c \cos (e+f x))^m (c (b B (1+m)+a A (2+m))+(A b+a B) c (2+m) \cos (e+f x)) \, dx}{c (2+m)} \\ & = \frac {b B (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m)}+\frac {(A b+a B) \int (c \cos (e+f x))^{1+m} \, dx}{c}+\left (a A+\frac {b B (1+m)}{2+m}\right ) \int (c \cos (e+f x))^m \, dx \\ & = \frac {b B (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m)}-\frac {\left (a A+\frac {b B (1+m)}{2+m}\right ) (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c f (1+m) \sqrt {\sin ^2(e+f x)}}-\frac {(A b+a B) (c \cos (e+f x))^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c^2 f (2+m) \sqrt {\sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.83 \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x)) (A+B \cos (e+f x)) \, dx=-\frac {(c \cos (e+f x))^m \left ((b B (1+m)+a A (2+m)) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sqrt {\sin ^2(e+f x)}+(A b+a B) (1+m) \cos (e+f x) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right ) \sqrt {\sin ^2(e+f x)}-\frac {1}{2} b B (1+m) \sin (2 (e+f x))\right )}{f (1+m) (2+m)} \]

[In]

Integrate[(c*Cos[e + f*x])^m*(a + b*Cos[e + f*x])*(A + B*Cos[e + f*x]),x]

[Out]

-(((c*Cos[e + f*x])^m*((b*B*(1 + m) + a*A*(2 + m))*Cot[e + f*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, C
os[e + f*x]^2]*Sqrt[Sin[e + f*x]^2] + (A*b + a*B)*(1 + m)*Cos[e + f*x]*Cot[e + f*x]*Hypergeometric2F1[1/2, (2
+ m)/2, (4 + m)/2, Cos[e + f*x]^2]*Sqrt[Sin[e + f*x]^2] - (b*B*(1 + m)*Sin[2*(e + f*x)])/2))/(f*(1 + m)*(2 + m
)))

Maple [F]

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

[In]

int((c*cos(f*x+e))^m*(a+b*cos(f*x+e))*(A+cos(f*x+e)*B),x)

[Out]

int((c*cos(f*x+e))^m*(a+b*cos(f*x+e))*(A+cos(f*x+e)*B),x)

Fricas [F]

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

[In]

integrate((c*cos(f*x+e))^m*(a+b*cos(f*x+e))*(A+B*cos(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*b*cos(f*x + e)^2 + A*a + (B*a + A*b)*cos(f*x + e))*(c*cos(f*x + e))^m, x)

Sympy [F]

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

[In]

integrate((c*cos(f*x+e))**m*(a+b*cos(f*x+e))*(A+B*cos(f*x+e)),x)

[Out]

Integral((c*cos(e + f*x))**m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x)), x)

Maxima [F]

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

[In]

integrate((c*cos(f*x+e))^m*(a+b*cos(f*x+e))*(A+B*cos(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*cos(f*x + e) + A)*(b*cos(f*x + e) + a)*(c*cos(f*x + e))^m, x)

Giac [F]

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

[In]

integrate((c*cos(f*x+e))^m*(a+b*cos(f*x+e))*(A+B*cos(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*cos(f*x + e) + A)*(b*cos(f*x + e) + a)*(c*cos(f*x + e))^m, x)

Mupad [F(-1)]

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

[In]

int((c*cos(e + f*x))^m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x)),x)

[Out]

int((c*cos(e + f*x))^m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x)), x)