\(\int (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx\) [636]

Optimal result

Integrand size = 33, antiderivative size = 644 \[ \int (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=-\frac {c^6 \left (4 a^3 A b \left (15-8 m+m^2\right )+a^4 B \left (15-8 m+m^2\right )+4 a A b^3 \left (10-7 m+m^2\right )+6 a^2 b^2 B \left (10-7 m+m^2\right )+b^4 B \left (8-6 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {6-m}{2},\frac {8-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^{-6+m} \sin (e+f x)}{f (2-m) (4-m) (6-m) \sqrt {\sin ^2(e+f x)}}-\frac {c^5 \left (a^4 A \left (8-6 m+m^2\right )+6 a^2 A b^2 \left (4-5 m+m^2\right )+4 a^3 b B \left (4-5 m+m^2\right )+A b^4 \left (3-4 m+m^2\right )+4 a b^3 B \left (3-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^{-5+m} \sin (e+f x)}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^5 \left (4 a^2 A b \left (3-4 m+m^2\right )+a^3 B \left (3-4 m+m^2\right )+2 A b^3 \left (4-2 m+m^2\right )+a b^2 B \left (8-13 m+5 m^2\right )\right ) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (1-m) (2-m) (4-m)}-\frac {a^2 c^5 \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec (e+f x) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (1-m) (2-m) (3-m)}-\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)} \]

[Out]

-c^6*(4*a^3*A*b*(m^2-8*m+15)+a^4*B*(m^2-8*m+15)+4*a*A*b^3*(m^2-7*m+10)+6*a^2*b^2*B*(m^2-7*m+10)+b^4*B*(m^2-6*m
+8))*hypergeom([1/2, 3-1/2*m],[4-1/2*m],cos(f*x+e)^2)*(c*sec(f*x+e))^(-6+m)*sin(f*x+e)/f/(-m^3+12*m^2-44*m+48)
/(sin(f*x+e)^2)^(1/2)-c^5*(a^4*A*(m^2-6*m+8)+6*a^2*A*b^2*(m^2-5*m+4)+4*a^3*b*B*(m^2-5*m+4)+A*b^4*(m^2-4*m+3)+4
*a*b^3*B*(m^2-4*m+3))*hypergeom([1/2, 5/2-1/2*m],[7/2-1/2*m],cos(f*x+e)^2)*(c*sec(f*x+e))^(-5+m)*sin(f*x+e)/f/
(1-m)/(m^2-8*m+15)/(sin(f*x+e)^2)^(1/2)-a*c^5*(4*a^2*A*b*(m^2-4*m+3)+a^3*B*(m^2-4*m+3)+2*A*b^3*(m^2-2*m+4)+a*b
^2*B*(5*m^2-13*m+8))*(c*sec(f*x+e))^(-5+m)*tan(f*x+e)/f/(1-m)/(m^2-6*m+8)-a^2*c^5*(2*a*b*B*(1-m)^2+a^2*A*(2-m)
^2+A*b^2*(m^2-m+6))*sec(f*x+e)*(c*sec(f*x+e))^(-5+m)*tan(f*x+e)/f/(-m^3+6*m^2-11*m+6)-a*c^5*(a*B*(1-m)-A*b*(2+
m))*(c*sec(f*x+e))^(-5+m)*(b+a*sec(f*x+e))^2*tan(f*x+e)/f/(m^2-3*m+2)-a*A*c^5*(c*sec(f*x+e))^(-5+m)*(b+a*sec(f
*x+e))^3*tan(f*x+e)/f/(1-m)

Rubi [A] (verified)

Time = 2.17 (sec) , antiderivative size = 644, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3039, 4111, 4181, 4161, 4132, 3857, 2722, 4131} \[ \int (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=-\frac {a^2 c^5 \tan (e+f x) \sec (e+f x) \left (a^2 A (2-m)^2+2 a b B (1-m)^2+A b^2 \left (m^2-m+6\right )\right ) (c \sec (e+f x))^{m-5}}{f (1-m) (2-m) (3-m)}-\frac {a c^5 \tan (e+f x) \left (a^3 B \left (m^2-4 m+3\right )+4 a^2 A b \left (m^2-4 m+3\right )+a b^2 B \left (5 m^2-13 m+8\right )+2 A b^3 \left (m^2-2 m+4\right )\right ) (c \sec (e+f x))^{m-5}}{f (1-m) (2-m) (4-m)}-\frac {c^6 \sin (e+f x) \left (a^4 B \left (m^2-8 m+15\right )+4 a^3 A b \left (m^2-8 m+15\right )+6 a^2 b^2 B \left (m^2-7 m+10\right )+4 a A b^3 \left (m^2-7 m+10\right )+b^4 B \left (m^2-6 m+8\right )\right ) (c \sec (e+f x))^{m-6} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {6-m}{2},\frac {8-m}{2},\cos ^2(e+f x)\right )}{f (2-m) (4-m) (6-m) \sqrt {\sin ^2(e+f x)}}-\frac {c^5 \sin (e+f x) \left (a^4 A \left (m^2-6 m+8\right )+4 a^3 b B \left (m^2-5 m+4\right )+6 a^2 A b^2 \left (m^2-5 m+4\right )+4 a b^3 B \left (m^2-4 m+3\right )+A b^4 \left (m^2-4 m+3\right )\right ) (c \sec (e+f x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right )}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^5 \tan (e+f x) (a B (1-m)-A b (m+2)) (a \sec (e+f x)+b)^2 (c \sec (e+f x))^{m-5}}{f (1-m) (2-m)}-\frac {a A c^5 \tan (e+f x) (a \sec (e+f x)+b)^3 (c \sec (e+f x))^{m-5}}{f (1-m)} \]

[In]

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

[Out]

-((c^6*(4*a^3*A*b*(15 - 8*m + m^2) + a^4*B*(15 - 8*m + m^2) + 4*a*A*b^3*(10 - 7*m + m^2) + 6*a^2*b^2*B*(10 - 7
*m + m^2) + b^4*B*(8 - 6*m + m^2))*Hypergeometric2F1[1/2, (6 - m)/2, (8 - m)/2, Cos[e + f*x]^2]*(c*Sec[e + f*x
])^(-6 + m)*Sin[e + f*x])/(f*(2 - m)*(4 - m)*(6 - m)*Sqrt[Sin[e + f*x]^2])) - (c^5*(a^4*A*(8 - 6*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*b^4*(3 - 4*m + m^2) + 4*a*b^3*B*(3 - 4*m + m^2))*Hyp
ergeometric2F1[1/2, (5 - m)/2, (7 - m)/2, Cos[e + f*x]^2]*(c*Sec[e + f*x])^(-5 + m)*Sin[e + f*x])/(f*(1 - m)*(
3 - m)*(5 - m)*Sqrt[Sin[e + f*x]^2]) - (a*c^5*(4*a^2*A*b*(3 - 4*m + m^2) + a^3*B*(3 - 4*m + m^2) + 2*A*b^3*(4
- 2*m + m^2) + a*b^2*B*(8 - 13*m + 5*m^2))*(c*Sec[e + f*x])^(-5 + m)*Tan[e + f*x])/(f*(1 - m)*(2 - m)*(4 - m))
- (a^2*c^5*(2*a*b*B*(1 - m)^2 + a^2*A*(2 - m)^2 + A*b^2*(6 - m + m^2))*Sec[e + f*x]*(c*Sec[e + f*x])^(-5 + m)
*Tan[e + f*x])/(f*(1 - m)*(2 - m)*(3 - m)) - (a*c^5*(a*B*(1 - m) - A*b*(2 + m))*(c*Sec[e + f*x])^(-5 + m)*(b +
a*Sec[e + f*x])^2*Tan[e + f*x])/(f*(1 - m)*(2 - m)) - (a*A*c^5*(c*Sec[e + f*x])^(-5 + m)*(b + a*Sec[e + f*x])
^3*Tan[e + f*x])/(f*(1 - m))

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 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3857

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 4111

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4161

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f
*x])^n/(f*(n + 2))), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1
) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C
, n}, x] &&  !LtQ[n, -1]

Rule 4181

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(
(d*Csc[e + f*x])^n/(f*(m + n + 1))), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x]
)^n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a
*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] &&
!LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = c^5 \int (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^4 (B+A \sec (e+f x)) \, dx \\ & = -\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)}-\frac {c^5 \int (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \left (-b (b B (1-m)+a A (5-m))-\left (b (A b+2 a B) (1-m)+a^2 A (2-m)\right ) \sec (e+f x)-a (a B (1-m)-A b (2+m)) \sec ^2(e+f x)\right ) \, dx}{1-m} \\ & = -\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)}+\frac {c^5 \int (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x)) \left (-b \left (2 a A b (5-m) m-a^2 B \left (5-6 m+m^2\right )-b^2 B \left (2-3 m+m^2\right )\right )+\left (b \left (a^2 A (7-2 m)+A b^2 (1-m)+3 a b B (1-m)\right ) (2-m)+a^2 (3-m) (a B (1-m)-A b (2+m))\right ) \sec (e+f x)+a \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec ^2(e+f x)\right ) \, dx}{2-3 m+m^2} \\ & = -\frac {a^2 c^5 \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec (e+f x) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (3-m) \left (2-3 m+m^2\right )}-\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)}+\frac {c^5 \int (c \sec (e+f x))^{-5+m} \left (b^2 (3-m) \left (2 a A b (5-m) m-a^2 B \left (5-6 m+m^2\right )-b^2 B \left (2-3 m+m^2\right )\right )-(2-m) \left (a^4 A \left (8-6 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 b^4 \left (3-4 m+m^2\right )+4 a b^3 B \left (3-4 m+m^2\right )\right ) \sec (e+f x)-a (3-m) \left (4 a^2 A b \left (3-4 m+m^2\right )+a^3 B \left (3-4 m+m^2\right )+2 A b^3 \left (4-2 m+m^2\right )+a b^2 B \left (8-13 m+5 m^2\right )\right ) \sec ^2(e+f x)\right ) \, dx}{(-3+m) \left (2-3 m+m^2\right )} \\ & = -\frac {a^2 c^5 \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec (e+f x) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (3-m) \left (2-3 m+m^2\right )}-\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)}+\frac {c^5 \int (c \sec (e+f x))^{-5+m} \left (b^2 (3-m) \left (2 a A b (5-m) m-a^2 B \left (5-6 m+m^2\right )-b^2 B \left (2-3 m+m^2\right )\right )-a (3-m) \left (4 a^2 A b \left (3-4 m+m^2\right )+a^3 B \left (3-4 m+m^2\right )+2 A b^3 \left (4-2 m+m^2\right )+a b^2 B \left (8-13 m+5 m^2\right )\right ) \sec ^2(e+f x)\right ) \, dx}{(-3+m) \left (2-3 m+m^2\right )}+\frac {\left (c^4 \left (a^4 A \left (8-6 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 b^4 \left (3-4 m+m^2\right )+4 a b^3 B \left (3-4 m+m^2\right )\right )\right ) \int (c \sec (e+f x))^{-4+m} \, dx}{(1-m) (3-m)} \\ & = -\frac {a c^5 \left (4 a^2 A b \left (3-4 m+m^2\right )+a^3 B \left (3-4 m+m^2\right )+2 A b^3 \left (4-2 m+m^2\right )+a b^2 B \left (8-13 m+5 m^2\right )\right ) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (4-m) \left (2-3 m+m^2\right )}-\frac {a^2 c^5 \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec (e+f x) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (3-m) \left (2-3 m+m^2\right )}-\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)}+\frac {\left (c^5 \left (4 a^3 A b \left (15-8 m+m^2\right )+a^4 B \left (15-8 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 )+b^4 B \left (8-6 m+m^2\right )\right )\right ) \int (c \sec (e+f x))^{-5+m} \, dx}{(2-m) (4-m)}+\frac {\left (c^4 \left (a^4 A \left (8-6 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 b^4 \left (3-4 m+m^2\right )+4 a b^3 B \left (3-4 m+m^2\right )\right ) \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{c}\right )^{4-m} \, dx}{(1-m) (3-m)} \\ & = -\frac {\left (a^4 A \left (8-6 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 b^4 \left (3-4 m+m^2\right )+4 a b^3 B \left (3-4 m+m^2\right )\right ) \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^5 \left (4 a^2 A b \left (3-4 m+m^2\right )+a^3 B \left (3-4 m+m^2\right )+2 A b^3 \left (4-2 m+m^2\right )+a b^2 B \left (8-13 m+5 m^2\right )\right ) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (4-m) \left (2-3 m+m^2\right )}-\frac {a^2 c^5 \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec (e+f x) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (3-m) \left (2-3 m+m^2\right )}-\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)}+\frac {\left (c^5 \left (4 a^3 A b \left (15-8 m+m^2\right )+a^4 B \left (15-8 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 )+b^4 B \left (8-6 m+m^2\right )\right ) \left (\frac {\cos (e+f x)}{c}\right )^m (c \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{c}\right )^{5-m} \, dx}{(2-m) (4-m)} \\ & = -\frac {\left (a^4 A \left (8-6 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 b^4 \left (3-4 m+m^2\right )+4 a b^3 B \left (3-4 m+m^2\right )\right ) \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5-m}{2},\frac {7-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (1-m) (3-m) (5-m) \sqrt {\sin ^2(e+f x)}}-\frac {\left (4 a^3 A b \left (15-8 m+m^2\right )+a^4 B \left (15-8 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 )+b^4 B \left (8-6 m+m^2\right )\right ) \cos ^6(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {6-m}{2},\frac {8-m}{2},\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (2-m) (4-m) (6-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^5 \left (4 a^2 A b \left (3-4 m+m^2\right )+a^3 B \left (3-4 m+m^2\right )+2 A b^3 \left (4-2 m+m^2\right )+a b^2 B \left (8-13 m+5 m^2\right )\right ) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (4-m) \left (2-3 m+m^2\right )}-\frac {a^2 c^5 \left (2 a b B (1-m)^2+a^2 A (2-m)^2+A b^2 \left (6-m+m^2\right )\right ) \sec (e+f x) (c \sec (e+f x))^{-5+m} \tan (e+f x)}{f (3-m) \left (2-3 m+m^2\right )}-\frac {a c^5 (a B (1-m)-A b (2+m)) (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^2 \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^5 (c \sec (e+f x))^{-5+m} (b+a \sec (e+f x))^3 \tan (e+f x)}{f (1-m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.66 (sec) , antiderivative size = 317, normalized size of antiderivative = 0.49 \[ \int (a+b \cos (e+f x))^4 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\frac {\cot (e+f x) \left (\frac {b^4 B \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sec ^2(e+f x)\right )}{-5+m}+\frac {b^3 (A b+4 a B) \cos ^4(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-4+m),\frac {1}{2} (-2+m),\sec ^2(e+f x)\right )}{-4+m}+a \left (\frac {2 b^2 (2 A b+3 a B) \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sec ^2(e+f x)\right )}{-3+m}+a \left (\frac {2 b (3 A b+2 a B) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+m),\frac {m}{2},\sec ^2(e+f x)\right )}{-2+m}+a \left (\frac {(4 A b+a B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sec ^2(e+f x)\right )}{-1+m}+\frac {a A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right )}{m}\right )\right )\right )\right ) (c \sec (e+f x))^m \sqrt {-\tan ^2(e+f x)}}{f} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)