\(\int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx\) [399]

Optimal result

Integrand size = 21, antiderivative size = 173 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {a^2 \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m)}-\frac {a^2 (3+2 m) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a^2 \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) \sqrt {\sin ^2(c+d x)}} \] Output:

a^2*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(2+m)-a^2*(3+2*m)*cos(d*x+c)^(1+m)*hyper 
geom([1/2, 1/2+1/2*m],[3/2+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(1+m)/(2+m)/( 
sin(d*x+c)^2)^(1/2)-2*a^2*cos(d*x+c)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2 
*m],cos(d*x+c)^2)*sin(d*x+c)/d/(2+m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.80 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=\frac {a^2 \cos ^{1+m}(c+d x) \csc (c+d x) \left (-\left ((3+2 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )+(1+m) \left (\sin ^2(c+d x)-2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (1+m) (2+m)} \] Input:

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

Output:

(a^2*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(-((3 + 2*m)*Hypergeometric2F1[1/2, 
 (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2]) + (1 + m)*(Si 
n[c + d*x]^2 - 2*Cos[c + d*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, 
 Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])))/(d*(1 + m)*(2 + m))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3242, 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.

\(\displaystyle \int (a \cos (c+d x)+a)^2 \cos ^m(c+d x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \sin \left (c+d x+\frac {\pi }{2}\right )^mdx\)

\(\Big \downarrow \) 3242

\(\displaystyle \frac {\int \cos ^m(c+d x) \left ((2 m+3) a^2+2 (m+2) \cos (c+d x) a^2\right )dx}{m+2}+\frac {a^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((2 m+3) a^2+2 (m+2) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{m+2}+\frac {a^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {2 a^2 (m+2) \int \cos ^{m+1}(c+d x)dx+a^2 (2 m+3) \int \cos ^m(c+d x)dx}{m+2}+\frac {a^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 (2 m+3) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx+2 a^2 (m+2) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx}{m+2}+\frac {a^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {-\frac {a^2 (2 m+3) \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {2 a^2 \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{m+2}+\frac {a^2 \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}\)

Input:

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

Output:

(a^2*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) + (-((a^2*(3 + 2*m)*Co 
s[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d* 
x]^2]*Sin[c + d*x])/(d*(1 + m)*Sqrt[Sin[c + d*x]^2])) - (2*a^2*Cos[c + d*x 
]^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sin 
[c + d*x])/(d*Sqrt[Sin[c + d*x]^2]))/(2 + m)
 

Defintions of rubi rules used

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 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

Maple [F]

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

Input:

int(cos(d*x+c)^m*(a+a*cos(d*x+c))^2,x)
 

Output:

int(cos(d*x+c)^m*(a+a*cos(d*x+c))^2,x)
 

Fricas [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integral((a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^2)*cos(d*x + c)^m, x 
)
 

Sympy [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=a^{2} \left (\int 2 \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx + \int \cos ^{2}{\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx + \int \cos ^{m}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cos(d*x+c)**m*(a+a*cos(d*x+c))**2,x)
 

Output:

a**2*(Integral(2*cos(c + d*x)*cos(c + d*x)**m, x) + Integral(cos(c + d*x)* 
*2*cos(c + d*x)**m, x) + Integral(cos(c + d*x)**m, x))
 

Maxima [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integrate((a*cos(d*x + c) + a)^2*cos(d*x + c)^m, x)
 

Giac [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integrate((a*cos(d*x + c) + a)^2*cos(d*x + c)^m, x)
 

Mupad [F(-1)]

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

int(cos(c + d*x)^m*(a + a*cos(c + d*x))^2,x)
 

Output:

int(cos(c + d*x)^m*(a + a*cos(c + d*x))^2, x)
 

Reduce [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \, dx=a^{2} \left (\int \cos \left (d x +c \right )^{m}d x +2 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )d x \right )+\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{2}d x \right ) \] Input:

int(cos(d*x+c)^m*(a+a*cos(d*x+c))^2,x)
 

Output:

a**2*(int(cos(c + d*x)**m,x) + 2*int(cos(c + d*x)**m*cos(c + d*x),x) + int 
(cos(c + d*x)**m*cos(c + d*x)**2,x))