3.182 \(\int (a \cos (c+d x))^m (b \cos (c+d x))^n (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=144 \[ \frac{C \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n}{a d (m+n+2)}-\frac{(A (m+n+2)+C (m+n+1)) \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\cos ^2(c+d x)\right )}{a d (m+n+1) (m+n+2) \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.111634, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {20, 3014, 2643} \[ \frac{C \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n}{a d (m+n+2)}-\frac{(A (m+n+2)+C (m+n+1)) \sin (c+d x) (a \cos (c+d x))^{m+1} (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\cos ^2(c+d x)\right )}{a d (m+n+1) (m+n+2) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 20

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

Rule 3014

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]

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

Mathematica [A]  time = 0.239387, size = 132, normalized size = 0.92 \[ -\frac{\sqrt{\sin ^2(c+d x)} \cot (c+d x) (a \cos (c+d x))^m (b \cos (c+d x))^n \left (A (m+n+3) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\cos ^2(c+d x)\right )+C (m+n+1) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m+n+3);\frac{1}{2} (m+n+5);\cos ^2(c+d x)\right )\right )}{d (m+n+1) (m+n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 2.474, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) a \right ) ^{m} \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( A+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)*a)^m*(b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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