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

Optimal. Leaf size=141 \[ -\frac{A \sin (c+d x) (b \cos (c+d x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt{\sin ^2(c+d x)}}-\frac{B \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) \sqrt{\sin ^2(c+d x)}} \]

[Out]

-((A*(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)*Sqrt[Sin[c + d*x]^2])) - (B*(b*Cos[c + d*x])^(3 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Co
s[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 = 0.0948405, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {16, 2748, 2643} \[ -\frac{A \sin (c+d x) (b \cos (c+d x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt{\sin ^2(c+d x)}}-\frac{B \sin (c+d x) (b \cos (c+d x))^{n+3} \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\cos ^2(c+d x)\right )}{b^3 d (n+3) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*(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)*Sqrt[Sin[c + d*x]^2])) - (B*(b*Cos[c + d*x])^(3 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Co
s[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 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 (A+B \cos (c+d x)) \, dx &=\frac{\int (b \cos (c+d x))^{1+n} (A+B \cos (c+d x)) \, dx}{b}\\ &=\frac{A \int (b \cos (c+d x))^{1+n} \, dx}{b}+\frac{B \int (b \cos (c+d x))^{2+n} \, dx}{b^2}\\ &=-\frac{A (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) \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 = 0.224801, size = 118, normalized size = 0.84 \[ -\frac{\sqrt{\sin ^2(c+d x)} \cos (c+d x) \cot (c+d x) (b \cos (c+d x))^n \left (A (n+3) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\cos ^2(c+d x)\right )+B (n+2) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\cos ^2(c+d x)\right )\right )}{d (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.497, 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 ) \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)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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)),x, algorithm="maxima")

[Out]

integrate((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 (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)),x, algorithm="fricas")

[Out]

integral((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)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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)),x, algorithm="giac")

[Out]

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