3.369 \(\int (a \cos (c+d x))^m (b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=227 \[ -\frac {B \sin (c+d x) (a \cos (c+d x))^{m+2} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\cos ^2(c+d x)\right )}{a^2 d (m+n+2) \sqrt {\sin ^2(c+d x)}}-\frac {(A (m+n+2)+C (m+n+1)) \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+1);\frac {1}{2} (m+n+3);\cos ^2(c+d x)\right )}{a d (m+n+1) (m+n+2) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n}{a d (m+n+2)} \]

[Out]

C*(a*cos(d*x+c))^(1+m)*(b*cos(d*x+c))^n*sin(d*x+c)/a/d/(2+m+n)-(C*(1+m+n)+A*(2+m+n))*(a*cos(d*x+c))^(1+m)*(b*c
os(d*x+c))^n*hypergeom([1/2, 1/2+1/2*m+1/2*n],[3/2+1/2*m+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/a/d/(1+m+n)/(2+m+n)/(
sin(d*x+c)^2)^(1/2)-B*(a*cos(d*x+c))^(2+m)*(b*cos(d*x+c))^n*hypergeom([1/2, 1+1/2*m+1/2*n],[2+1/2*m+1/2*n],cos
(d*x+c)^2)*sin(d*x+c)/a^2/d/(2+m+n)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {20, 3023, 2748, 2643} \[ -\frac {B \sin (c+d x) (a \cos (c+d x))^{m+2} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\cos ^2(c+d x)\right )}{a^2 d (m+n+2) \sqrt {\sin ^2(c+d x)}}-\frac {(A (m+n+2)+C (m+n+1)) \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+1);\frac {1}{2} (m+n+3);\cos ^2(c+d x)\right )}{a d (m+n+1) (m+n+2) \sqrt {\sin ^2(c+d x)}}+\frac {C \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n}{a d (m+n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(C*(a*Cos[c + d*x])^(1 + m)*(b*Cos[c + d*x])^n*Sin[c + d*x])/(a*d*(2 + m + n)) - ((C*(1 + m + n) + A*(2 + m +
n))*(a*Cos[c + d*x])^(1 + m)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (1 + m + n)/2, (3 + m + n)/2, Cos[c + d
*x]^2]*Sin[c + d*x])/(a*d*(1 + m + n)*(2 + m + n)*Sqrt[Sin[c + d*x]^2]) - (B*(a*Cos[c + d*x])^(2 + m)*(b*Cos[c
+ d*x])^n*Hypergeometric2F1[1/2, (2 + m + n)/2, (4 + m + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a^2*d*(2 + m +
n)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 161, normalized size = 0.71 \[ -\frac {\sin (c+d x) \cos (c+d x) (a \cos (c+d x))^m (b \cos (c+d x))^n \left ((A (m+n+2)+C (m+n+1)) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+1);\frac {1}{2} (m+n+3);\cos ^2(c+d x)\right )+(m+n+1) \left (B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\cos ^2(c+d x)\right )-C \sqrt {\sin ^2(c+d x)}\right )\right )}{d (m+n+1) (m+n+2) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

-((Cos[c + d*x]*(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*Sin[c + d*x]*((C*(1 + m + n) + A*(2 + m + n))*Hypergeome
tric2F1[1/2, (1 + m + n)/2, (3 + m + n)/2, Cos[c + d*x]^2] + (1 + m + n)*(B*Cos[c + d*x]*Hypergeometric2F1[1/2
, (2 + m + n)/2, (4 + m + n)/2, Cos[c + d*x]^2] - C*Sqrt[Sin[c + d*x]^2])))/(d*(1 + m + n)*(2 + m + n)*Sqrt[Si
n[c + d*x]^2]))

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fricas [F]  time = 1.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (a \cos \left (d x + c\right )\right )^{m} \left (b \cos \left (d x + c\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c))^m*(b*cos(d*x + c))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c))^m*(b*cos(d*x + c))^n, x)

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maple [F]  time = 2.50, size = 0, normalized size = 0.00 \[ \int \left (a \cos \left (d x +c \right )\right )^{m} \left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

int((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c))^m*(b*cos(d*x + c))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x))^m*(b*cos(c + d*x))^n*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((a*cos(c + d*x))^m*(b*cos(c + d*x))^n*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c))**m*(b*cos(d*x+c))**n*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Integral((a*cos(c + d*x))**m*(b*cos(c + d*x))**n*(A + B*cos(c + d*x) + C*cos(c + d*x)**2), x)

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