\(\int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx\) [914]

Optimal result

Integrand size = 21, antiderivative size = 141 \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=-\frac {A (b \cos (c+d x))^{1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^{2+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 d (2+n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2827, 2722} \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=-\frac {A \sin (c+d x) (b \cos (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\cos ^2(c+d x)\right )}{b d (n+1) \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

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

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]

Rubi steps \begin{align*} \text {integral}& = A \int (b \cos (c+d x))^n \, dx+\frac {B \int (b \cos (c+d x))^{1+n} \, dx}{b} \\ & = -\frac {A (b \cos (c+d x))^{1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^{2+n} \operatorname {Hypergeometric2F1}\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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=-\frac {(b \cos (c+d x))^n \cot (c+d x) \left (A (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(c+d x)\right )+B (1+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+n) (2+n)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int \left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )}\right )\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \,d x } \]

[In]

integrate((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, x)

Giac [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \,d x } \]

[In]

integrate((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, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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