\(\int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx\) [398]

Optimal result

Integrand size = 21, antiderivative size = 232 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=\frac {a^3 (7+2 m) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m)}+\frac {\cos ^{1+m}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m)}-\frac {a^3 (5+4 m) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {a^3 (11+4 m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) (3+m) \sqrt {\sin ^2(c+d x)}} \] Output:

a^3*(7+2*m)*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(2+m)/(3+m)+cos(d*x+c)^(1+m)*(a^ 
3+a^3*cos(d*x+c))*sin(d*x+c)/d/(3+m)-a^3*(5+4*m)*cos(d*x+c)^(1+m)*hypergeo 
m([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)-a^3*(11+4*m)*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)/(3+m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.72 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=-\frac {a^3 \cos ^{1+m}(c+d x) \sin (c+d x) \left (\frac {\left (15+17 m+4 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )}{\sqrt {\sin ^2(c+d x)}}-(1+m) \left (3 (3+m)+(2+m) \cos (c+d x)-(11+4 m) \cot (c+d x) \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (1+m) (2+m) (3+m)} \] Input:

Integrate[Cos[c + d*x]^m*(a + a*Cos[c + d*x])^3,x]
 

Output:

-((a^3*Cos[c + d*x]^(1 + m)*Sin[c + d*x]*(((15 + 17*m + 4*m^2)*Hypergeomet 
ric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2])/Sqrt[Sin[c + d*x]^2] - 
(1 + m)*(3*(3 + m) + (2 + m)*Cos[c + d*x] - (11 + 4*m)*Cot[c + d*x]*Csc[c 
+ d*x]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sqrt[S 
in[c + d*x]^2])))/(d*(1 + m)*(2 + m)*(3 + m)))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3242, 3042, 3447, 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.

\(\displaystyle \int (a \cos (c+d x)+a)^3 \cos ^m(c+d x) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

\(\displaystyle \frac {\int \cos ^m(c+d x) (\cos (c+d x) a+a) \left (2 (m+2) a^2+(2 m+7) \cos (c+d x) a^2\right )dx}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 (m+2) a^2+(2 m+7) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\int \cos ^m(c+d x) \left ((2 m+7) \cos ^2(c+d x) a^3+2 (m+2) a^3+\left (2 (m+2) a^3+(2 m+7) a^3\right ) \cos (c+d x)\right )dx}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((2 m+7) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+2 (m+2) a^3+\left (2 (m+2) a^3+(2 m+7) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\int \cos ^m(c+d x) \left ((m+3) (4 m+5) a^3+(m+2) (4 m+11) \cos (c+d x) a^3\right )dx}{m+2}+\frac {a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((m+3) (4 m+5) a^3+(m+2) (4 m+11) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx}{m+2}+\frac {a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {a^3 (m+2) (4 m+11) \int \cos ^{m+1}(c+d x)dx+a^3 (m+3) (4 m+5) \int \cos ^m(c+d x)dx}{m+2}+\frac {a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^3 (m+3) (4 m+5) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx+a^3 (m+2) (4 m+11) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx}{m+2}+\frac {a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {-\frac {a^3 (m+3) (4 m+5) \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {a^3 (4 m+11) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{m+2}+\frac {a^3 (2 m+7) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {\sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right ) \cos ^{m+1}(c+d x)}{d (m+3)}\)

Input:

Int[Cos[c + d*x]^m*(a + a*Cos[c + d*x])^3,x]
 

Output:

(Cos[c + d*x]^(1 + m)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(d*(3 + m)) + 
 ((a^3*(7 + 2*m)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)) + (-((a^3* 
(3 + m)*(5 + 4*m)*Cos[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])) 
- (a^3*(11 + 4*m)*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)) 
/(3 + 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]))
 

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

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]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{m} \left (a +a \cos \left (d x +c \right )\right )^{3}d x\]

Input:

int(cos(d*x+c)^m*(a+a*cos(d*x+c))^3,x)
 

Output:

int(cos(d*x+c)^m*(a+a*cos(d*x+c))^3,x)
 

Fricas [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(a+a*cos(d*x+c))^3,x, algorithm="fricas")
 

Output:

integral((a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + 
 a^3)*cos(d*x + c)^m, x)
 

Sympy [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=a^{3} \left (\int 3 \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx + \int 3 \cos ^{2}{\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx + \int \cos ^{m}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cos(d*x+c)**m*(a+a*cos(d*x+c))**3,x)
 

Output:

a**3*(Integral(3*cos(c + d*x)*cos(c + d*x)**m, x) + Integral(3*cos(c + d*x 
)**2*cos(c + d*x)**m, x) + Integral(cos(c + d*x)**3*cos(c + d*x)**m, x) + 
Integral(cos(c + d*x)**m, x))
 

Maxima [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(a+a*cos(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + a)^3*cos(d*x + c)^m, x)
 

Giac [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(a+a*cos(d*x+c))^3,x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + a)^3*cos(d*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3 \,d x \] Input:

int(cos(c + d*x)^m*(a + a*cos(c + d*x))^3,x)
 

Output:

int(cos(c + d*x)^m*(a + a*cos(c + d*x))^3, x)
 

Reduce [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^3 \, dx=a^{3} \left (\int \cos \left (d x +c \right )^{m}d x +3 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )d x \right )+\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{3}d x +3 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{2}d x \right )\right ) \] Input:

int(cos(d*x+c)^m*(a+a*cos(d*x+c))^3,x)
                                                                                    
                                                                                    
 

Output:

a**3*(int(cos(c + d*x)**m,x) + 3*int(cos(c + d*x)**m*cos(c + d*x),x) + int 
(cos(c + d*x)**m*cos(c + d*x)**3,x) + 3*int(cos(c + d*x)**m*cos(c + d*x)** 
2,x))