\(\int (b \cos (c+d x))^m (A+C \cos ^2(c+d x)) \, dx\) [34]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int (b \cos (c+d x))^m \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C (b \cos (c+d x))^{1+m} \sin (c+d x)}{b d (2+m)}-\frac {(C (1+m)+A (2+m)) (b \cos (c+d x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3093, 2722} \[ \int (b \cos (c+d x))^m \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C \sin (c+d x) (b \cos (c+d x))^{m+1}}{b d (m+2)}-\frac {(A (m+2)+C (m+1)) \sin (c+d x) (b \cos (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{b d (m+1) (m+2) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(C*(b*Cos[c + d*x])^(1 + m)*Sin[c + d*x])/(b*d*(2 + m)) - ((C*(1 + m) + A*(2 + m))*(b*Cos[c + d*x])^(1 + m)*Hy
pergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(1 + m)*(2 + m)*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 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 {C (b \cos (c+d x))^{1+m} \sin (c+d x)}{b d (2+m)}+\left (A+\frac {C (1+m)}{2+m}\right ) \int (b \cos (c+d x))^m \, dx \\ & = \frac {C (b \cos (c+d x))^{1+m} \sin (c+d x)}{b d (2+m)}-\frac {\left (A+\frac {C (1+m)}{2+m}\right ) (b \cos (c+d x))^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1+m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int (b \cos (c+d x))^m \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {(b \cos (c+d x))^m \cot (c+d x) \left (A (3+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )+C (1+m) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+m) (3+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^m, x)

Sympy [F]

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

[In]

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

[Out]

Integral((b*cos(c + d*x))**m*(A + C*cos(c + d*x)**2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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