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

Optimal. Leaf size=187 \[ -\frac{(A (n+3)+C (n+2)) \sin (c+d x) (b \cos (c+d x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) (n+3) \sqrt{\sin ^2(c+d x)}}-\frac{B \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) \sqrt{\sin ^2(c+d x)}}+\frac{C \sin (c+d x) (b \cos (c+d x))^{n+2}}{b^2 d (n+3)} \]

[Out]

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

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Rubi [A]  time = 0.213301, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {16, 3023, 2748, 2643} \[ -\frac{(A (n+3)+C (n+2)) \sin (c+d x) (b \cos (c+d x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) (n+3) \sqrt{\sin ^2(c+d x)}}-\frac{B \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) \sqrt{\sin ^2(c+d x)}}+\frac{C \sin (c+d x) (b \cos (c+d x))^{n+2}}{b^2 d (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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]

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 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]

Rubi steps

\begin{align*} \int \cos (c+d x) (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{\int (b \cos (c+d x))^{1+n} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{b}\\ &=\frac{C (b \cos (c+d x))^{2+n} \sin (c+d x)}{b^2 d (3+n)}+\frac{\int (b \cos (c+d x))^{1+n} (b (C (2+n)+A (3+n))+b B (3+n) \cos (c+d x)) \, dx}{b^2 (3+n)}\\ &=\frac{C (b \cos (c+d x))^{2+n} \sin (c+d x)}{b^2 d (3+n)}+\frac{B \int (b \cos (c+d x))^{2+n} \, dx}{b^2}+\frac{(C (2+n)+A (3+n)) \int (b \cos (c+d x))^{1+n} \, dx}{b (3+n)}\\ &=\frac{C (b \cos (c+d x))^{2+n} \sin (c+d x)}{b^2 d (3+n)}-\frac{(C (2+n)+A (3+n)) (b \cos (c+d x))^{2+n} \, _2F_1\left (\frac{1}{2},\frac{2+n}{2};\frac{4+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 d (2+n) (3+n) \sqrt{\sin ^2(c+d x)}}-\frac{B (b \cos (c+d x))^{3+n} \, _2F_1\left (\frac{1}{2},\frac{3+n}{2};\frac{5+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{b^3 d (3+n) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.330168, size = 144, normalized size = 0.77 \[ -\frac{\sin (c+d x) \cos ^2(c+d x) (b \cos (c+d x))^n \left ((A (n+3)+C (n+2)) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\cos ^2(c+d x)\right )+(n+2) \left (B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\cos ^2(c+d x)\right )-C \sqrt{\sin ^2(c+d x)}\right )\right )}{d (n+2) (n+3) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.522, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( dx+c \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(d*x+c)*(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., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{3} + B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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)^3 + B*cos(d*x + c)^2 + A*cos(d*x + c))*(b*cos(d*x + c))^n, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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