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

Optimal. Leaf size=595 \[ -\frac{\sin (e+f x) \left (4 a^3 A b \left (m^2+8 m+15\right )+6 a^2 b^2 B \left (m^2+7 m+10\right )+a^4 B \left (m^2+8 m+15\right )+4 a A b^3 \left (m^2+7 m+10\right )+b^4 B \left (m^2+6 m+8\right )\right ) (c \cos (e+f x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(e+f x)\right )}{c^2 f (m+2) (m+3) (m+5) \sqrt{\sin ^2(e+f x)}}-\frac{\sin (e+f x) \left (6 a^2 A b^2 \left (m^2+5 m+4\right )+a^4 A \left (m^2+6 m+8\right )+4 a^3 b B \left (m^2+5 m+4\right )+4 a b^3 B \left (m^2+4 m+3\right )+A b^4 \left (m^2+4 m+3\right )\right ) (c \cos (e+f x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{c f (m+1) (m+2) (m+4) \sqrt{\sin ^2(e+f x)}}+\frac{b \sin (e+f x) \left (a^2 A b \left (5 m^2+47 m+110\right )+2 a^3 B \left (m^2+10 m+28\right )+4 a b^2 B \left (m^2+8 m+15\right )+A b^3 \left (m^2+8 m+15\right )\right ) (c \cos (e+f x))^{m+1}}{c f (m+2) (m+4) (m+5)}+\frac{b^2 \sin (e+f x) \cos (e+f x) \left (a^2 B \left (m^2+11 m+36\right )+2 a A b (m+5)^2+b^2 B (m+4)^2\right ) (c \cos (e+f x))^{m+1}}{c f (m+3) (m+4) (m+5)}+\frac{b \sin (e+f x) (a B (m+8)+A b (m+5)) (a+b \cos (e+f x))^2 (c \cos (e+f x))^{m+1}}{c f (m+4) (m+5)}+\frac{b B \sin (e+f x) (a+b \cos (e+f x))^3 (c \cos (e+f x))^{m+1}}{c f (m+5)} \]

[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])

________________________________________________________________________________________

Rubi [A]  time = 1.98357, antiderivative size = 595, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2990, 3049, 3033, 3023, 2748, 2643} \[ -\frac{\sin (e+f x) \left (4 a^3 A b \left (m^2+8 m+15\right )+6 a^2 b^2 B \left (m^2+7 m+10\right )+a^4 B \left (m^2+8 m+15\right )+4 a A b^3 \left (m^2+7 m+10\right )+b^4 B \left (m^2+6 m+8\right )\right ) (c \cos (e+f x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(e+f x)\right )}{c^2 f (m+2) (m+3) (m+5) \sqrt{\sin ^2(e+f x)}}-\frac{\sin (e+f x) \left (6 a^2 A b^2 \left (m^2+5 m+4\right )+a^4 A \left (m^2+6 m+8\right )+4 a^3 b B \left (m^2+5 m+4\right )+4 a b^3 B \left (m^2+4 m+3\right )+A b^4 \left (m^2+4 m+3\right )\right ) (c \cos (e+f x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{c f (m+1) (m+2) (m+4) \sqrt{\sin ^2(e+f x)}}+\frac{b \sin (e+f x) \left (a^2 A b \left (5 m^2+47 m+110\right )+2 a^3 B \left (m^2+10 m+28\right )+4 a b^2 B \left (m^2+8 m+15\right )+A b^3 \left (m^2+8 m+15\right )\right ) (c \cos (e+f x))^{m+1}}{c f (m+2) (m+4) (m+5)}+\frac{b^2 \sin (e+f x) \cos (e+f x) \left (a^2 B \left (m^2+11 m+36\right )+2 a A b (m+5)^2+b^2 B (m+4)^2\right ) (c \cos (e+f x))^{m+1}}{c f (m+3) (m+4) (m+5)}+\frac{b \sin (e+f x) (a B (m+8)+A b (m+5)) (a+b \cos (e+f x))^2 (c \cos (e+f x))^{m+1}}{c f (m+4) (m+5)}+\frac{b B \sin (e+f x) (a+b \cos (e+f x))^3 (c \cos (e+f x))^{m+1}}{c f (m+5)} \]

Antiderivative was successfully verified.

[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 2990

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] /; F
reeQ[{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 3049

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])))

Rule 3033

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 3023

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 2748

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int (c \cos (e+f x))^m (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) \, dx &=\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} \, _2F_1\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} \, _2F_1\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]  time = 6.19836, size = 487, normalized size = 0.82 \[ -\frac{a^3 (a B+4 A b) \sin (e+f x) \cos ^2(e+f x) (c \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(e+f x)\right )}{f (m+2) \sqrt{\sin ^2(e+f x)}}-\frac{2 a^2 b (2 a B+3 A b) \sin (e+f x) \cos ^3(e+f x) (c \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};\cos ^2(e+f x)\right )}{f (m+3) \sqrt{\sin ^2(e+f x)}}-\frac{a^4 A \sin (e+f x) \cos (e+f x) (c \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(e+f x)\right )}{f (m+1) \sqrt{\sin ^2(e+f x)}}-\frac{2 a b^2 (3 a B+2 A b) \sin (e+f x) \cos ^4(e+f x) (c \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+4}{2};\frac{m+6}{2};\cos ^2(e+f x)\right )}{f (m+4) \sqrt{\sin ^2(e+f x)}}-\frac{b^3 (4 a B+A b) \sin (e+f x) \cos ^5(e+f x) (c \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+5}{2};\frac{m+7}{2};\cos ^2(e+f x)\right )}{f (m+5) \sqrt{\sin ^2(e+f x)}}-\frac{b^4 B \sin (e+f x) \cos ^6(e+f x) (c \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+6}{2};\frac{m+8}{2};\cos ^2(e+f x)\right )}{f (m+6) \sqrt{\sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^4*A*Cos[e + f*x]*(c*Cos[e + f*x])^m*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[e + f*x]^2]*Sin[e +
f*x])/(f*(1 + m)*Sqrt[Sin[e + f*x]^2])) - (a^3*(4*A*b + a*B)*Cos[e + f*x]^2*(c*Cos[e + f*x])^m*Hypergeometric2
F1[1/2, (2 + m)/2, (4 + m)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(2 + m)*Sqrt[Sin[e + f*x]^2]) - (2*a^2*b*(3*A*b
 + 2*a*B)*Cos[e + f*x]^3*(c*Cos[e + f*x])^m*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[e + f*x]^2]*Sin[e
 + f*x])/(f*(3 + m)*Sqrt[Sin[e + f*x]^2]) - (2*a*b^2*(2*A*b + 3*a*B)*Cos[e + f*x]^4*(c*Cos[e + f*x])^m*Hyperge
ometric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(4 + m)*Sqrt[Sin[e + f*x]^2]) - (b^3*(A
*b + 4*a*B)*Cos[e + f*x]^5*(c*Cos[e + f*x])^m*Hypergeometric2F1[1/2, (5 + m)/2, (7 + m)/2, Cos[e + f*x]^2]*Sin
[e + f*x])/(f*(5 + m)*Sqrt[Sin[e + f*x]^2]) - (b^4*B*Cos[e + f*x]^6*(c*Cos[e + f*x])^m*Hypergeometric2F1[1/2,
(6 + m)/2, (8 + m)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(6 + m)*Sqrt[Sin[e + f*x]^2])

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Maple [F]  time = 2.33, size = 0, normalized size = 0. \begin{align*} \int \left ( c\cos \left ( fx+e \right ) \right ) ^{m} \left ( a+b\cos \left ( fx+e \right ) \right ) ^{4} \left ( A+B\cos \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{4} \cos \left (f x + e\right )^{5} + A a^{4} +{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (f x + e\right )\right )} \left (c \cos \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)