\(\int \cos (c+d x) (b \cos (c+d x))^n (A+C \cos ^2(c+d x)) \, dx\) [184]

Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C (b \cos (c+d x))^{2+n} \sin (c+d x)}{b^2 d (3+n)}-\frac {(C (2+n)+A (3+n)) (b \cos (c+d x))^{2+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^2 d (2+n) (3+n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

C*(b*cos(d*x+c))^(2+n)*sin(d*x+c)/b^2/d/(3+n)-(C*(2+n)+A*(3+n))*(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)/(3+n)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3093, 2722} \[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C \sin (c+d x) (b \cos (c+d x))^{n+2}}{b^2 d (n+3)}-\frac {(A (n+3)+C (n+2)) \sin (c+d x) (b \cos (c+d x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\cos ^2(c+d x)\right )}{b^2 d (n+2) (n+3) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

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 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 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \cos (c+d x))^{1+n} \left (A+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 {\left (A+\frac {C (2+n)}{3+n}\right ) \int (b \cos (c+d x))^{1+n} \, dx}{b} \\ & = \frac {C (b \cos (c+d x))^{2+n} \sin (c+d x)}{b^2 d (3+n)}-\frac {\left (A+\frac {C (2+n)}{3+n}\right ) (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 = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {\cos (c+d x) (b \cos (c+d x))^n \cot (c+d x) \left (A (4+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\cos ^2(c+d x)\right )+C (2+n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (2+n) (4+n)} \]

[In]

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

[Out]

-((Cos[c + d*x]*(b*Cos[c + d*x])^n*Cot[c + d*x]*(A*(4 + n)*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Cos[c
+ d*x]^2] + C*(2 + n)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (4 + n)/2, (6 + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[c
+ d*x]^2])/(d*(2 + n)*(4 + n)))

Maple [F]

\[\int \cos \left (d x +c \right ) \left (\cos \left (d x +c \right ) b \right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

integral((C*cos(d*x + c)^3 + A*cos(d*x + c))*(b*cos(d*x + c))^n, x)

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n \,d x \]

[In]

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

[Out]

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