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

Optimal. Leaf size=167 \[ -\frac {C \sin (c+d x) (a \cos (c+d x))^{m+3} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+3);\frac {1}{2} (m+n+5);\cos ^2(c+d x)\right )}{a^3 d (m+n+3) \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (a \cos (c+d x))^{m+2} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\cos ^2(c+d x)\right )}{a^2 d (m+n+2) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-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)-C*(a*cos(d*x+c))^(3+m)*(b*cos(d*x+c))^n*hypergeom([1/2, 3/2+1/2*m+1/2*n]
,[5/2+1/2*m+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/a^3/d/(3+m+n)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {20, 3010, 2748, 2643} \[ -\frac {B \sin (c+d x) (a \cos (c+d x))^{m+2} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\cos ^2(c+d x)\right )}{a^2 d (m+n+2) \sqrt {\sin ^2(c+d x)}}-\frac {C \sin (c+d x) (a \cos (c+d x))^{m+3} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+3);\frac {1}{2} (m+n+5);\cos ^2(c+d x)\right )}{a^3 d (m+n+3) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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])) - (C*(a*Cos[c + d*x])^(3 + m)*(b*Cos[c + d*x])
^n*Hypergeometric2F1[1/2, (3 + m + n)/2, (5 + m + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a^3*d*(3 + 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 3010

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

Rubi steps

\begin {align*} \int (a \cos (c+d x))^m (b \cos (c+d x))^n \left (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 (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {\left ((a \cos (c+d x))^{-n} (b \cos (c+d x))^n\right ) \int (a \cos (c+d x))^{1+m+n} (B+C \cos (c+d x)) \, dx}{a}\\ &=\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 (a \cos (c+d x))^{-n} (b \cos (c+d x))^n\right ) \int (a \cos (c+d x))^{2+m+n} \, dx}{a^2}\\ &=-\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)}}-\frac {C (a \cos (c+d x))^{3+m} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (3+m+n);\frac {1}{2} (5+m+n);\cos ^2(c+d x)\right ) \sin (c+d x)}{a^3 d (3+m+n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 136, normalized size = 0.81 \[ -\frac {\sqrt {\sin ^2(c+d x)} \cos (c+d x) \cot (c+d x) (a \cos (c+d x))^m (b \cos (c+d x))^n \left (B (m+n+3) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+2);\frac {1}{2} (m+n+4);\cos ^2(c+d x)\right )+C (m+n+2) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m+n+3);\frac {1}{2} (m+n+5);\cos ^2(c+d x)\right )\right )}{d (m+n+2) (m+n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x])^m*(b*Cos[c + d*x])^n*(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*Cot[c + d*x]*(B*(3 + m + n)*Hypergeometric2F1[1/2, (2 +
m + n)/2, (4 + m + n)/2, Cos[c + d*x]^2] + C*(2 + m + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (3 + m + n)/2, (5
+ m + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(2 + m + n)*(3 + m + n)))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (a \cos \left (d x + c\right )\right )^{m} \left (b \cos \left (d x + c\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(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*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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (a \cos \left (d x + c\right )\right )^{m} \left (b \cos \left (d x + c\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(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*cos(d*x + c))^m*(b*cos(d*x + c))^n, x)

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maple [F]  time = 2.14, size = 0, normalized size = 0.00 \[ \int \left (a \cos \left (d x +c \right )\right )^{m} \left (b \cos \left (d x +c \right )\right )^{n} \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(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*(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (a \cos \left (d x + c\right )\right )^{m} \left (b \cos \left (d x + c\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c))^m*(b*cos(d*x+c))^n*(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*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 = -0.01 \[ \int {\left (a\,\cos \left (c+d\,x\right )\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x))^m*(b*cos(c + d*x))^n*(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*(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 \[ \int \left (a \cos {\left (c + d x \right )}\right )^{m} \left (b \cos {\left (c + d x \right )}\right )^{n} \left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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