\(\int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx\) [452]

Optimal result

Integrand size = 33, antiderivative size = 287 \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\frac {b (A b (3+m)+a B (4+m)) (c \cos (e+f x))^{1+m} \sin (e+f x)}{c f (2+m) (3+m)}+\frac {b B (c \cos (e+f x))^{1+m} (a+b \cos (e+f x)) \sin (e+f x)}{c f (3+m)}-\frac {\left (A b^2 (1+m)+2 a b B (1+m)+a^2 A (2+m)\right ) (c \cos (e+f x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c f (1+m) (2+m) \sqrt {\sin ^2(e+f x)}}-\frac {\left (b^2 B (2+m)+a (2 A b+a B) (3+m)\right ) (c \cos (e+f x))^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right ) \sin (e+f x)}{c^2 f (2+m) (3+m) \sqrt {\sin ^2(e+f x)}} \]

[Out]

b*(A*b*(3+m)+a*B*(4+m))*(c*cos(f*x+e))^(1+m)*sin(f*x+e)/c/f/(2+m)/(3+m)+b*B*(c*cos(f*x+e))^(1+m)*(a+b*cos(f*x+
e))*sin(f*x+e)/c/f/(3+m)-(A*b^2*(1+m)+2*a*b*B*(1+m)+a^2*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)-(b^2*B*(2+m)+a*(2*A*b+B*a)*(3+m))
*(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)/(sin(f*x+e
)^2)^(1/2)

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3069, 3102, 2827, 2722} \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=-\frac {\sin (e+f x) \left (a^2 A (m+2)+2 a b B (m+1)+A b^2 (m+1)\right ) (c \cos (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(e+f x)\right )}{c f (m+1) (m+2) \sqrt {\sin ^2(e+f x)}}-\frac {\sin (e+f x) \left (a (m+3) (a B+2 A b)+b^2 B (m+2)\right ) (c \cos (e+f x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(e+f x)\right )}{c^2 f (m+2) (m+3) \sqrt {\sin ^2(e+f x)}}+\frac {b \sin (e+f x) (a B (m+4)+A b (m+3)) (c \cos (e+f x))^{m+1}}{c f (m+2) (m+3)}+\frac {b B \sin (e+f x) (a+b \cos (e+f x)) (c \cos (e+f x))^{m+1}}{c f (m+3)} \]

[In]

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

[Out]

(b*(A*b*(3 + m) + a*B*(4 + m))*(c*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(c*f*(2 + m)*(3 + m)) + (b*B*(c*Cos[e +
f*x])^(1 + m)*(a + b*Cos[e + f*x])*Sin[e + f*x])/(c*f*(3 + m)) - ((A*b^2*(1 + m) + 2*a*b*B*(1 + m) + a^2*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]) - ((b^2*B*(2 + m) + a*(2*A*b + a*B)*(3 + m))*(c*Cos[e + f*x])^(2 + m)*H
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(c^2*f*(2 + m)*(3 + 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 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]

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

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.74 \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\frac {(c \cos (e+f x))^m \cot (e+f x) \left (-\frac {a^2 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(e+f x)\right )}{1+m}+\cos (e+f x) \left (-\frac {a (2 A b+a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(e+f x)\right )}{2+m}+b \cos (e+f x) \left (-\frac {(A b+2 a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(e+f x)\right )}{3+m}-\frac {b B \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\cos ^2(e+f x)\right )}{4+m}\right )\right )\right ) \sqrt {\sin ^2(e+f x)}}{f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (c \cos \left (f x +e \right )\right )^{m} \left (a +b \cos \left (f x +e \right )\right )^{2} \left (A +\cos \left (f x +e \right ) B \right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{2} \left (c \cos \left (f x + e\right )\right )^{m} \,d x } \]

[In]

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

Sympy [F(-1)]

Timed out. \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{2} \left (c \cos \left (f x + e\right )\right )^{m} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{2} \left (c \cos \left (f x + e\right )\right )^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c \cos (e+f x))^m (a+b \cos (e+f x))^2 (A+B \cos (e+f x)) \, dx=\int {\left (c\,\cos \left (e+f\,x\right )\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^2 \,d x \]

[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)