\(\int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx\) [450]

Optimal result

Integrand size = 33, antiderivative size = 595 \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\frac {b \left (A b^3 \left (15+8 m+m^2\right )+4 a b^2 B \left (15+8 m+m^2\right )+2 a^3 B \left (28+10 m+m^2\right )+a^2 A b \left (110+47 m+5 m^2\right )\right ) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m) (4+m) (5+m)}+\frac {b^2 \left (b^2 B (4+m)^2+2 a A b (5+m)^2+a^2 B \left (36+11 m+m^2\right )\right ) \cos (e+f x) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (3+m) (4+m) (5+m)}+\frac {b (A b (5+m)+a B (8+m)) (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^2 \sin (e+f x)}{c f (4+m) (5+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}-\frac {\left (A b^4 \left (3+4 m+m^2\right )+4 a b^3 B \left (3+4 m+m^2\right )+6 a^2 A b^2 \left (4+5 m+m^2\right )+4 a^3 b B \left (4+5 m+m^2\right )+a^4 A \left (8+6 m+m^2\right )\right ) (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c f (1+m) (2+m) (4+m) \sqrt {\sin ^2(e+f x)}}-\frac {\left (b^4 B \left (8+6 m+m^2\right )+4 a A b^3 \left (10+7 m+m^2\right )+6 a^2 b^2 B \left (10+7 m+m^2\right )+4 a^3 A b \left (15+8 m+m^2\right )+a^4 B \left (15+8 m+m^2\right )\right ) (c \cos (e+f x))^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c^2 f (2+m) (3+m) (5+m) \sqrt {\sin ^2(e+f x)}} \]

[Out]

b*(A*b^3*(m^2+8*m+15)+4*a*b^2*B*(m^2+8*m+15)+2*a^3*B*(m^2+10*m+28)+a^2*A*b*(5*m^2+47*m+110))*(c*cos(f*x+e))^(1
+m)*sin(f*x+e)/c/f/(5+m)/(m^2+6*m+8)+b^2*(b^2*B*(4+m)^2+2*a*A*b*(5+m)^2+a^2*B*(m^2+11*m+36))*cos(f*x+e)*(c*cos
(f*x+e))^(1+m)*sin(f*x+e)/c/f/(3+m)/(4+m)/(5+m)+b*(A*b*(5+m)+a*B*(8+m))*(c*cos(f*x+e))^(1+m)*(a+b*cos(f*x+e))^
2*sin(f*x+e)/c/f/(4+m)/(5+m)+b*B*(c*cos(f*x+e))^(1+m)*(a+b*cos(f*x+e))^3*sin(f*x+e)/c/f/(5+m)-(A*b^4*(m^2+4*m+
3)+4*a*b^3*B*(m^2+4*m+3)+6*a^2*A*b^2*(m^2+5*m+4)+4*a^3*b*B*(m^2+5*m+4)+a^4*A*(m^2+6*m+8))*(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/(4+m)/(m^2+3*m+2)/(sin(f*x+e)^2)^(1/2)-(b
^4*B*(m^2+6*m+8)+4*a*A*b^3*(m^2+7*m+10)+6*a^2*b^2*B*(m^2+7*m+10)+4*a^3*A*b*(m^2+8*m+15)+a^4*B*(m^2+8*m+15))*(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)/(3+m)/(5+m)/(sin(f*
x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 2.14 (sec) , antiderivative size = 595, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3069, 3128, 3112, 3102, 2827, 2722} \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\frac {b^2 \sin (e+f x) \cos (e+f x) \left (a^2 B \left (m^2+11 m+36\right )+2 a A b (m+5)^2+b^2 B (m+4)^2\right ) (c \cos (e+f x))^{m+1}}{c f (m+3) (m+4) (m+5)}+\frac {b \sin (e+f x) \left (2 a^3 B \left (m^2+10 m+28\right )+a^2 A b \left (5 m^2+47 m+110\right )+4 a b^2 B \left (m^2+8 m+15\right )+A b^3 \left (m^2+8 m+15\right )\right ) (c \cos (e+f x))^{m+1}}{c f (m+2) (m+4) (m+5)}-\frac {\sin (e+f x) \left (a^4 B \left (m^2+8 m+15\right )+4 a^3 A b \left (m^2+8 m+15\right )+6 a^2 b^2 B \left (m^2+7 m+10\right )+4 a A b^3 \left (m^2+7 m+10\right )+b^4 B \left (m^2+6 m+8\right )\right ) (c \cos (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(e+f x)\right )}{c^2 f (m+2) (m+3) (m+5) \sqrt {\sin ^2(e+f x)}}-\frac {\sin (e+f x) \left (a^4 A \left (m^2+6 m+8\right )+4 a^3 b B \left (m^2+5 m+4\right )+6 a^2 A b^2 \left (m^2+5 m+4\right )+4 a b^3 B \left (m^2+4 m+3\right )+A b^4 \left (m^2+4 m+3\right )\right ) (c \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{c f (m+1) (m+2) (m+4) \sqrt {\sin ^2(e+f x)}}+\frac {b \sin (e+f x) (a B (m+8)+A b (m+5)) (a+b \cos (e+f x))^2 (c \cos (e+f x))^{m+1}}{c f (m+4) (m+5)}+\frac {b B \sin (e+f x) (a+b \cos (e+f x))^3 (c \cos (e+f x))^{m+1}}{c f (m+5)} \]

[In]

Int[(c*Cos[e + f*x])^m*(a + b*Cos[e + f*x])^4*(A + B*Cos[e + f*x]),x]

[Out]

(b*(A*b^3*(15 + 8*m + m^2) + 4*a*b^2*B*(15 + 8*m + m^2) + 2*a^3*B*(28 + 10*m + m^2) + a^2*A*b*(110 + 47*m + 5*
m^2))*(c*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(c*f*(2 + m)*(4 + m)*(5 + m)) + (b^2*(b^2*B*(4 + m)^2 + 2*a*A*b*(
5 + m)^2 + a^2*B*(36 + 11*m + m^2))*Cos[e + f*x]*(c*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(c*f*(3 + m)*(4 + m)*(
5 + m)) + (b*(A*b*(5 + m) + a*B*(8 + m))*(c*Cos[e + f*x])^(1 + m)*(a + b*Cos[e + f*x])^2*Sin[e + f*x])/(c*f*(4
+ m)*(5 + m)) + (b*B*(c*Cos[e + f*x])^(1 + m)*(a + b*Cos[e + f*x])^3*Sin[e + f*x])/(c*f*(5 + m)) - ((A*b^4*(3
+ 4*m + m^2) + 4*a*b^3*B*(3 + 4*m + m^2) + 6*a^2*A*b^2*(4 + 5*m + m^2) + 4*a^3*b*B*(4 + 5*m + m^2) + a^4*A*(8
+ 6*m + m^2))*(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)*(4 + m)*Sqrt[Sin[e + f*x]^2]) - ((b^4*B*(8 + 6*m + m^2) + 4*a*A*b^3*(10 + 7*m + m^2)
+ 6*a^2*b^2*B*(10 + 7*m + m^2) + 4*a^3*A*b*(15 + 8*m + m^2) + a^4*B*(15 + 8*m + m^2))*(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)*(3 + m)*(5 + m)*Sq
rt[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 3069

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

Rule 3112

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*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[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]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}+\frac {\int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 \left (a c (b B (1+m)+a A (5+m))+c \left (b^2 B (4+m)+a (2 A b+a B) (5+m)\right ) \cos (e+f x)+b c (A b (5+m)+a B (8+m)) \cos ^2(e+f x)\right ) \, dx}{c (5+m)} \\ & = \frac {b (A b (5+m)+a B (8+m)) (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^2 \sin (e+f x)}{c f (4+m) (5+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}+\frac {\int (c \cos (e+f x))^m (a+b \cos (e+f x)) \left (a c^2 (a (4+m) (b B (1+m)+a A (5+m))+b (1+m) (A b (5+m)+a B (8+m)))+c^2 \left (b^2 (3+m) (A b (5+m)+a B (8+m))+a (4+m) \left (3 a A b (5+m)+a^2 B (5+m)+b^2 B (5+2 m)\right )\right ) \cos (e+f x)+b c^2 \left (b^2 B (4+m)^2+2 a A b (5+m)^2+a^2 B \left (36+11 m+m^2\right )\right ) \cos ^2(e+f x)\right ) \, dx}{c^2 (4+m) (5+m)} \\ & = \frac {b^2 \left (b^2 B (4+m)^2+2 a A b (5+m)^2+a^2 B \left (36+11 m+m^2\right )\right ) \cos (e+f x) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (3+m) (4+m) (5+m)}+\frac {b (A b (5+m)+a B (8+m)) (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^2 \sin (e+f x)}{c f (4+m) (5+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}+\frac {\int (c \cos (e+f x))^m \left (a^2 c^3 (3+m) (a (4+m) (b B (1+m)+a A (5+m))+b (1+m) (A b (5+m)+a B (8+m)))+c^3 (4+m) \left (b^4 B \left (8+6 m+m^2\right )+4 a A b^3 \left (10+7 m+m^2\right )+6 a^2 b^2 B \left (10+7 m+m^2\right )+4 a^3 A b \left (15+8 m+m^2\right )+a^4 B \left (15+8 m+m^2\right )\right ) \cos (e+f x)+b c^3 (3+m) \left (A b^3 \left (15+8 m+m^2\right )+4 a b^2 B \left (15+8 m+m^2\right )+2 a^3 B \left (28+10 m+m^2\right )+a^2 A b \left (110+47 m+5 m^2\right )\right ) \cos ^2(e+f x)\right ) \, dx}{c^3 (3+m) (4+m) (5+m)} \\ & = \frac {b \left (A b^3 \left (15+8 m+m^2\right )+4 a b^2 B \left (15+8 m+m^2\right )+2 a^3 B \left (28+10 m+m^2\right )+a^2 A b \left (110+47 m+5 m^2\right )\right ) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m) (4+m) (5+m)}+\frac {b^2 \left (b^2 B (4+m)^2+2 a A b (5+m)^2+a^2 B \left (36+11 m+m^2\right )\right ) \cos (e+f x) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (3+m) (4+m) (5+m)}+\frac {b (A b (5+m)+a B (8+m)) (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^2 \sin (e+f x)}{c f (4+m) (5+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}+\frac {\int (c \cos (e+f x))^m \left (c^4 (3+m) \left (A b^4 \left (15+23 m+9 m^2+m^3\right )+4 a b^3 B \left (15+23 m+9 m^2+m^3\right )+6 a^2 A b^2 \left (20+29 m+10 m^2+m^3\right )+4 a^3 b B \left (20+29 m+10 m^2+m^3\right )+a^4 A \left (40+38 m+11 m^2+m^3\right )\right )+c^4 (2+m) (4+m) \left (b^4 B \left (8+6 m+m^2\right )+4 a A b^3 \left (10+7 m+m^2\right )+6 a^2 b^2 B \left (10+7 m+m^2\right )+4 a^3 A b \left (15+8 m+m^2\right )+a^4 B \left (15+8 m+m^2\right )\right ) \cos (e+f x)\right ) \, dx}{c^4 (2+m) (3+m) (4+m) (5+m)} \\ & = \frac {b \left (A b^3 \left (15+8 m+m^2\right )+4 a b^2 B \left (15+8 m+m^2\right )+2 a^3 B \left (28+10 m+m^2\right )+a^2 A b \left (110+47 m+5 m^2\right )\right ) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m) (4+m) (5+m)}+\frac {b^2 \left (b^2 B (4+m)^2+2 a A b (5+m)^2+a^2 B \left (36+11 m+m^2\right )\right ) \cos (e+f x) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (3+m) (4+m) (5+m)}+\frac {b (A b (5+m)+a B (8+m)) (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^2 \sin (e+f x)}{c f (4+m) (5+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}+\frac {\left (A b^4 \left (3+4 m+m^2\right )+4 a b^3 B \left (3+4 m+m^2\right )+6 a^2 A b^2 \left (4+5 m+m^2\right )+4 a^3 b B \left (4+5 m+m^2\right )+a^4 A \left (8+6 m+m^2\right )\right ) \int (c \cos (e+f x))^m \, dx}{(2+m) (4+m)}+\frac {\left (b^4 B \left (8+6 m+m^2\right )+4 a A b^3 \left (10+7 m+m^2\right )+6 a^2 b^2 B \left (10+7 m+m^2\right )+4 a^3 A b \left (15+8 m+m^2\right )+a^4 B \left (15+8 m+m^2\right )\right ) \int (c \cos (e+f x))^{1+m} \, dx}{c (3+m) (5+m)} \\ & = \frac {b \left (A b^3 \left (15+8 m+m^2\right )+4 a b^2 B \left (15+8 m+m^2\right )+2 a^3 B \left (28+10 m+m^2\right )+a^2 A b \left (110+47 m+5 m^2\right )\right ) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m) (4+m) (5+m)}+\frac {b^2 \left (b^2 B (4+m)^2+2 a A b (5+m)^2+a^2 B \left (36+11 m+m^2\right )\right ) \cos (e+f x) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (3+m) (4+m) (5+m)}+\frac {b (A b (5+m)+a B (8+m)) (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^2 \sin (e+f x)}{c f (4+m) (5+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x))^3 \sin (e+f x)}{c f (5+m)}-\frac {\left (A b^4 \left (3+4 m+m^2\right )+4 a b^3 B \left (3+4 m+m^2\right )+6 a^2 A b^2 \left (4+5 m+m^2\right )+4 a^3 b B \left (4+5 m+m^2\right )+a^4 A \left (8+6 m+m^2\right )\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) (2+m) (4+m) \sqrt {\sin ^2(e+f x)}}-\frac {\left (b^4 B \left (8+6 m+m^2\right )+4 a A b^3 \left (10+7 m+m^2\right )+6 a^2 b^2 B \left (10+7 m+m^2\right )+4 a^3 A b \left (15+8 m+m^2\right )+a^4 B \left (15+8 m+m^2\right )\right ) (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) (3+m) (5+m) \sqrt {\sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.73 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.54 \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\frac {(c \cos (e+f x))^m \cot (e+f x) \left (-\frac {a^4 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right )}{1+m}+\cos (e+f x) \left (-\frac {a^3 (4 A b+a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right )}{2+m}+b \cos (e+f x) \left (-\frac {2 a^2 (3 A b+2 a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(e+f x)\right )}{3+m}+b \cos (e+f x) \left (-\frac {2 a (2 A b+3 a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\cos ^2(e+f x)\right )}{4+m}+b \cos (e+f x) \left (-\frac {(A b+4 a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},\cos ^2(e+f x)\right )}{5+m}-\frac {b B \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {6+m}{2},\frac {8+m}{2},\cos ^2(e+f x)\right )}{6+m}\right )\right )\right )\right )\right ) \sqrt {\sin ^2(e+f x)}}{f} \]

[In]

Integrate[(c*Cos[e + f*x])^m*(a + b*Cos[e + f*x])^4*(A + B*Cos[e + f*x]),x]

[Out]

((c*Cos[e + f*x])^m*Cot[e + f*x]*(-((a^4*A*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[e + f*x]^2])/(1 +
m)) + Cos[e + f*x]*(-((a^3*(4*A*b + a*B)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[e + f*x]^2])/(2 + m)
) + b*Cos[e + f*x]*((-2*a^2*(3*A*b + 2*a*B)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[e + f*x]^2])/(3 +
m) + b*Cos[e + f*x]*((-2*a*(2*A*b + 3*a*B)*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[e + f*x]^2])/(4 +
m) + b*Cos[e + f*x]*(-(((A*b + 4*a*B)*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, Cos[e + f*x]^2])/(5 + m))
- (b*B*Cos[e + f*x]*Hypergeometric2F1[1/2, (6 + m)/2, (8 + m)/2, Cos[e + f*x]^2])/(6 + m))))))*Sqrt[Sin[e + f*
x]^2])/f

Maple [F]

\[\int \left (c \cos \left (f x +e \right )\right )^{m} \left (a +b \cos \left (f x +e \right )\right )^{4} \left (A +\cos \left (f x +e \right ) B \right )d x\]

[In]

int((c*cos(f*x+e))^m*(a+b*cos(f*x+e))^4*(A+cos(f*x+e)*B),x)

[Out]

int((c*cos(f*x+e))^m*(a+b*cos(f*x+e))^4*(A+cos(f*x+e)*B),x)

Fricas [F]

\[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{4} \left (c \cos \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((c*cos(f*x+e))^m*(a+b*cos(f*x+e))^4*(A+B*cos(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*b^4*cos(f*x + e)^5 + A*a^4 + (4*B*a*b^3 + A*b^4)*cos(f*x + e)^4 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*cos(
f*x + e)^3 + 2*(2*B*a^3*b + 3*A*a^2*b^2)*cos(f*x + e)^2 + (B*a^4 + 4*A*a^3*b)*cos(f*x + e))*(c*cos(f*x + e))^m
, x)

Sympy [F(-1)]

Timed out. \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\text {Timed out} \]

[In]

integrate((c*cos(f*x+e))**m*(a+b*cos(f*x+e))**4*(A+B*cos(f*x+e)),x)

[Out]

Timed out

Maxima [F]

\[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{4} \left (c \cos \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((c*cos(f*x+e))^m*(a+b*cos(f*x+e))^4*(A+B*cos(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*cos(f*x + e) + A)*(b*cos(f*x + e) + a)^4*(c*cos(f*x + e))^m, x)

Giac [F]

\[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{4} \left (c \cos \left (f x + e\right )\right )^{m} \,d x } \]

[In]

integrate((c*cos(f*x+e))^m*(a+b*cos(f*x+e))^4*(A+B*cos(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*cos(f*x + e) + A)*(b*cos(f*x + e) + a)^4*(c*cos(f*x + e))^m, x)

Mupad [F(-1)]

Timed out. \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx=\int {\left (c\,\cos \left (e+f\,x\right )\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^4 \,d x \]

[In]

int((c*cos(e + f*x))^m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x))^4,x)

[Out]

int((c*cos(e + f*x))^m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x))^4, x)