3.638 \(\int (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx\)

Optimal. Leaf size=327 \[ -\frac {c^4 \sin (e+f x) \left (a^2 B (3-m)+2 a A b (3-m)+b^2 B (2-m)\right ) (c \sec (e+f x))^{m-4} \, _2F_1\left (\frac {1}{2},\frac {4-m}{2};\frac {6-m}{2};\cos ^2(e+f x)\right )}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {c^3 \sin (e+f x) \left (a^2 A (2-m)+2 a b B (1-m)+A b^2 (1-m)\right ) (c \sec (e+f x))^{m-3} \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (3-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^3 \tan (e+f x) (a B (1-m)-A b m) (c \sec (e+f x))^{m-3}}{f (1-m) (2-m)}-\frac {a A c^3 \tan (e+f x) (a \sec (e+f x)+b) (c \sec (e+f x))^{m-3}}{f (1-m)} \]

[Out]

-c^4*(b^2*B*(2-m)+2*a*A*b*(3-m)+a^2*B*(3-m))*hypergeom([1/2, 2-1/2*m],[3-1/2*m],cos(f*x+e)^2)*(c*sec(f*x+e))^(
-4+m)*sin(f*x+e)/f/(m^2-6*m+8)/(sin(f*x+e)^2)^(1/2)-c^3*(A*b^2*(1-m)+2*a*b*B*(1-m)+a^2*A*(2-m))*hypergeom([1/2
, 3/2-1/2*m],[5/2-1/2*m],cos(f*x+e)^2)*(c*sec(f*x+e))^(-3+m)*sin(f*x+e)/f/(m^2-4*m+3)/(sin(f*x+e)^2)^(1/2)-a*c
^3*(a*B*(1-m)-A*b*m)*(c*sec(f*x+e))^(-3+m)*tan(f*x+e)/f/(m^2-3*m+2)-a*A*c^3*(c*sec(f*x+e))^(-3+m)*(b+a*sec(f*x
+e))*tan(f*x+e)/f/(1-m)

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Rubi [A]  time = 0.64, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2960, 4026, 4047, 3772, 2643, 4046} \[ -\frac {c^4 \sin (e+f x) \left (a^2 B (3-m)+2 a A b (3-m)+b^2 B (2-m)\right ) (c \sec (e+f x))^{m-4} \, _2F_1\left (\frac {1}{2},\frac {4-m}{2};\frac {6-m}{2};\cos ^2(e+f x)\right )}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {c^3 \sin (e+f x) \left (a^2 A (2-m)+2 a b B (1-m)+A b^2 (1-m)\right ) (c \sec (e+f x))^{m-3} \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (3-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^3 \tan (e+f x) (a B (1-m)-A b m) (c \sec (e+f x))^{m-3}}{f (1-m) (2-m)}-\frac {a A c^3 \tan (e+f x) (a \sec (e+f x)+b) (c \sec (e+f x))^{m-3}}{f (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[e + f*x])^2*(A + B*Cos[e + f*x])*(c*Sec[e + f*x])^m,x]

[Out]

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

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]

Rule 2960

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 3772

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 4026

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 4046

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 4047

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]

Rubi steps

\begin {align*} \int (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx &=c^3 \int (c \sec (e+f x))^{-3+m} (b+a \sec (e+f x))^2 (B+A \sec (e+f x)) \, dx\\ &=-\frac {a A c^3 (c \sec (e+f x))^{-3+m} (b+a \sec (e+f x)) \tan (e+f x)}{f (1-m)}-\frac {c^3 \int (c \sec (e+f x))^{-3+m} \left (-b (b B (1-m)+a A (3-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 m) \sec ^2(e+f x)\right ) \, dx}{1-m}\\ &=-\frac {a A c^3 (c \sec (e+f x))^{-3+m} (b+a \sec (e+f x)) \tan (e+f x)}{f (1-m)}-\frac {c^3 \int (c \sec (e+f x))^{-3+m} \left (-b (b B (1-m)+a A (3-m))-a (a B (1-m)-A b m) \sec ^2(e+f x)\right ) \, dx}{1-m}+\frac {\left (c^2 \left (A b^2 (1-m)+2 a b B (1-m)+a^2 A (2-m)\right )\right ) \int (c \sec (e+f x))^{-2+m} \, dx}{1-m}\\ &=-\frac {a c^3 (a B (1-m)-A b m) (c \sec (e+f x))^{-3+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^3 (c \sec (e+f x))^{-3+m} (b+a \sec (e+f x)) \tan (e+f x)}{f (1-m)}+\frac {\left (c^3 \left (b^2 B (2-m)+2 a A b (3-m)+a^2 B (3-m)\right )\right ) \int (c \sec (e+f x))^{-3+m} \, dx}{2-m}+\frac {\left (c^2 \left (A b^2 (1-m)+2 a b B (1-m)+a^2 A (2-m)\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 )^{2-m} \, dx}{1-m}\\ &=-\frac {\left (A b^2 (1-m)+2 a b B (1-m)+a^2 A (2-m)\right ) \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (1-m) (3-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^3 (a B (1-m)-A b m) (c \sec (e+f x))^{-3+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^3 (c \sec (e+f x))^{-3+m} (b+a \sec (e+f x)) \tan (e+f x)}{f (1-m)}+\frac {\left (c^3 \left (b^2 B (2-m)+2 a A b (3-m)+a^2 B (3-m)\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 )^{3-m} \, dx}{2-m}\\ &=-\frac {\left (A b^2 (1-m)+2 a b B (1-m)+a^2 A (2-m)\right ) \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2},\frac {3-m}{2};\frac {5-m}{2};\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (1-m) (3-m) \sqrt {\sin ^2(e+f x)}}-\frac {\left (b^2 B (2-m)+2 a A b (3-m)+a^2 B (3-m)\right ) \cos ^4(e+f x) \, _2F_1\left (\frac {1}{2},\frac {4-m}{2};\frac {6-m}{2};\cos ^2(e+f x)\right ) (c \sec (e+f x))^m \sin (e+f x)}{f (2-m) (4-m) \sqrt {\sin ^2(e+f x)}}-\frac {a c^3 (a B (1-m)-A b m) (c \sec (e+f x))^{-3+m} \tan (e+f x)}{f (1-m) (2-m)}-\frac {a A c^3 (c \sec (e+f x))^{-3+m} (b+a \sec (e+f x)) \tan (e+f x)}{f (1-m)}\\ \end {align*}

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Mathematica [A]  time = 0.99, size = 205, normalized size = 0.63 \[ \frac {\sqrt {-\tan ^2(e+f x)} \cot (e+f x) (c \sec (e+f x))^m \left (\frac {b (2 a B+A b) \cos ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {m-2}{2};\frac {m}{2};\sec ^2(e+f x)\right )}{m-2}+a \left (\frac {(a B+2 A b) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {m-1}{2};\frac {m+1}{2};\sec ^2(e+f x)\right )}{m-1}+\frac {a A \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\sec ^2(e+f x)\right )}{m}\right )+\frac {b^2 B \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2},\frac {m-3}{2};\frac {m-1}{2};\sec ^2(e+f x)\right )}{m-3}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[e + f*x])^2*(A + B*Cos[e + f*x])*(c*Sec[e + f*x])^m,x]

[Out]

(Cot[e + f*x]*((b^2*B*Cos[e + f*x]^3*Hypergeometric2F1[1/2, (-3 + m)/2, (-1 + m)/2, Sec[e + f*x]^2])/(-3 + m)
+ (b*(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*(((2*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*Hypergeo
metric2F1[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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{2} \cos \left (f x + e\right )^{3} + A a^{2} + {\left (2 \, B a b + A b^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (f x + e\right )\right )} \left (c \sec \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{2} \left (c \sec \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(f*x+e))^2*(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)^2*(c*sec(f*x + e))^m, x)

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maple [F]  time = 1.80, size = 0, normalized size = 0.00 \[ \int \left (a +b \cos \left (f x +e \right )\right )^{2} \left (A +B \cos \left (f x +e \right )\right ) \left (c \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{2} \left (c \sec \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(f*x+e))^2*(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)^2*(c*sec(f*x + e))^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sec {\left (e + f x \right )}\right )^{m} \left (A + B \cos {\left (e + f x \right )}\right ) \left (a + b \cos {\left (e + f x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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