3.4.70 \(\int \cos ^2(c+d x) (b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [370]

3.4.70.1 Optimal result

Integrand size = 39, antiderivative size = 187 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {C (b \cos (c+d x))^{3+n} \sin (c+d x)}{b^3 d (4+n)}-\frac {(C (3+n)+A (4+n)) (b \cos (c+d x))^{3+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^3 d (3+n) (4+n) \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^{4+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{b^4 d (4+n) \sqrt {\sin ^2(c+d x)}} \]

output

C*(b*cos(d*x+c))^(3+n)*sin(d*x+c)/b^3/d/(4+n)-(C*(3+n)+A*(4+n))*(b*cos(d*x 
+c))^(3+n)*hypergeom([1/2, 3/2+1/2*n],[5/2+1/2*n],cos(d*x+c)^2)*sin(d*x+c) 
/b^3/d/(3+n)/(4+n)/(sin(d*x+c)^2)^(1/2)-B*(b*cos(d*x+c))^(4+n)*hypergeom([ 
1/2, 2+1/2*n],[3+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/b^4/d/(4+n)/(sin(d*x+c)^2 
)^(1/2)
 

3.4.70.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.82 \[ \int \cos ^2(c+d x) (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {\cos ^2(c+d x) (b \cos (c+d x))^n \cot (c+d x) \left ((C (3+n)+A (4+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-(3+n) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (3+n) (4+n)} \]

input

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

output

-((Cos[c + d*x]^2*(b*Cos[c + d*x])^n*Cot[c + d*x]*((C*(3 + n) + A*(4 + n)) 
*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + 
 d*x]^2] - (3 + n)*(C*Sin[c + d*x]^2 - B*Cos[c + d*x]*Hypergeometric2F1[1/ 
2, (4 + n)/2, (6 + n)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])))/(d*(3 + n 
)*(4 + n)))
 

3.4.70.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2030, 3042, 3502, 3042, 3227, 3042, 3122}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^2(c+d x) (b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{n+2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{n+2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^2}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\int (b \cos (c+d x))^{n+2} (b (C (n+3)+A (n+4))+b B (n+4) \cos (c+d x))dx}{b (n+4)}+\frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b d (n+4)}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{n+2} \left (b (C (n+3)+A (n+4))+b B (n+4) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b (n+4)}+\frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b d (n+4)}}{b^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {b (A (n+4)+C (n+3)) \int (b \cos (c+d x))^{n+2}dx+B (n+4) \int (b \cos (c+d x))^{n+3}dx}{b (n+4)}+\frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b d (n+4)}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b (A (n+4)+C (n+3)) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{n+2}dx+B (n+4) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{n+3}dx}{b (n+4)}+\frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b d (n+4)}}{b^2}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {-\frac {(A (n+4)+C (n+3)) \sin (c+d x) (b \cos (c+d x))^{n+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+3}{2},\frac {n+5}{2},\cos ^2(c+d x)\right )}{d (n+3) \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^{n+4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+4}{2},\frac {n+6}{2},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{b (n+4)}+\frac {C \sin (c+d x) (b \cos (c+d x))^{n+3}}{b d (n+4)}}{b^2}\)

input

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

output

((C*(b*Cos[c + d*x])^(3 + n)*Sin[c + d*x])/(b*d*(4 + n)) + (-(((C*(3 + n) 
+ A*(4 + n))*(b*Cos[c + d*x])^(3 + n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 
 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 + n)*Sqrt[Sin[c + d*x]^2])) - 
 (B*(b*Cos[c + d*x])^(4 + n)*Hypergeometric2F1[1/2, (4 + n)/2, (6 + n)/2, 
Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d*x]^2]))/(b*(4 + n)))/b^2
 

3.4.70.3.1 Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

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]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502

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

3.4.70.4 Maple [F]

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

input

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

output

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

3.4.70.5 Fricas [F]

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

input

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

output

integral((C*cos(d*x + c)^4 + B*cos(d*x + c)^3 + A*cos(d*x + c)^2)*(b*cos(d 
*x + c))^n, x)
 

3.4.70.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.4.70.7 Maxima [F]

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

input

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

output

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

3.4.70.8 Giac [F]

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

input

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

output

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

3.4.70.9 Mupad [F(-1)]

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

input

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

output

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