\(\int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx\) [769]

Optimal result

Integrand size = 21, antiderivative size = 330 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {b^2 \left (b^2 (3+m)+a^2 (22+5 m)\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (4+m)}+\frac {2 a b^3 (5+m) \cos ^{2+m}(c+d x) \sin (c+d x)}{d (3+m) (4+m)}+\frac {b^2 \cos ^{1+m}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{d (4+m)}-\frac {\left (b^4 \left (3+4 m+m^2\right )+6 a^2 b^2 \left (4+5 m+m^2\right )+a^4 \left (8+6 m+m^2\right )\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) (4+m) \sqrt {\sin ^2(c+d x)}}-\frac {4 a b \left (b^2 (2+m)+a^2 (3+m)\right ) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) (3+m) \sqrt {\sin ^2(c+d x)}} \] Output:

b^2*(b^2*(3+m)+a^2*(22+5*m))*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(2+m)/(4+m)+2*a 
*b^3*(5+m)*cos(d*x+c)^(2+m)*sin(d*x+c)/d/(3+m)/(4+m)+b^2*cos(d*x+c)^(1+m)* 
(a+b*cos(d*x+c))^2*sin(d*x+c)/d/(4+m)-(b^4*(m^2+4*m+3)+6*a^2*b^2*(m^2+5*m+ 
4)+a^4*(m^2+6*m+8))*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)/d/(1+m)/(2+m)/(4+m)/(sin(d*x+c)^2)^(1/2)-4*a*b* 
(b^2*(2+m)+a^2*(3+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 = 1.19 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.73 \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\frac {\cos ^{1+m}(c+d x) \csc (c+d x) \left (-\frac {a^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )}{1+m}+b \cos (c+d x) \left (-\frac {4 a^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right )}{2+m}+b \cos (c+d x) \left (-\frac {6 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(c+d x)\right )}{3+m}+b \cos (c+d x) \left (-\frac {4 a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\cos ^2(c+d x)\right )}{4+m}-\frac {b \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},\cos ^2(c+d x)\right )}{5+m}\right )\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{d} \] Input:

Integrate[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^4,x]
 

Output:

(Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(-((a^4*Hypergeometric2F1[1/2, (1 + m)/ 
2, (3 + m)/2, Cos[c + d*x]^2])/(1 + m)) + b*Cos[c + d*x]*((-4*a^3*Hypergeo 
metric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2])/(2 + m) + b*Cos[c + 
d*x]*((-6*a^2*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2] 
)/(3 + m) + b*Cos[c + d*x]*((-4*a*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m 
)/2, Cos[c + d*x]^2])/(4 + m) - (b*Cos[c + d*x]*Hypergeometric2F1[1/2, (5 
+ m)/2, (7 + m)/2, Cos[c + d*x]^2])/(5 + m)))))*Sqrt[Sin[c + d*x]^2])/d
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3272, 3042, 3512, 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.

\(\displaystyle \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4dx\)

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {\frac {\int \cos ^m(c+d x) \left ((m+3) \left ((m+4) a^2+b^2 (m+1)\right ) a^2+4 b (m+4) \left ((m+3) a^2+b^2 (m+2)\right ) \cos (c+d x) a+b^2 (m+3) \left ((5 m+22) a^2+b^2 (m+3)\right ) \cos ^2(c+d x)\right )dx}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((m+3) \left ((m+4) a^2+b^2 (m+1)\right ) a^2+4 b (m+4) \left ((m+3) a^2+b^2 (m+2)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b^2 (m+3) \left ((5 m+22) a^2+b^2 (m+3)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\frac {\int \cos ^m(c+d x) \left ((m+3) \left (\left (m^2+6 m+8\right ) a^4+6 b^2 \left (m^2+5 m+4\right ) a^2+b^4 \left (m^2+4 m+3\right )\right )+4 a b (m+2) (m+4) \left ((m+3) a^2+b^2 (m+2)\right ) \cos (c+d x)\right )dx}{m+2}+\frac {b^2 (m+3) \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^m \left ((m+3) \left (\left (m^2+6 m+8\right ) a^4+6 b^2 \left (m^2+5 m+4\right ) a^2+b^4 \left (m^2+4 m+3\right )\right )+4 a b (m+2) (m+4) \left ((m+3) a^2+b^2 (m+2)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{m+2}+\frac {b^2 (m+3) \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {4 a b (m+2) (m+4) \left (a^2 (m+3)+b^2 (m+2)\right ) \int \cos ^{m+1}(c+d x)dx+(m+3) \left (a^4 \left (m^2+6 m+8\right )+6 a^2 b^2 \left (m^2+5 m+4\right )+b^4 \left (m^2+4 m+3\right )\right ) \int \cos ^m(c+d x)dx}{m+2}+\frac {b^2 (m+3) \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {4 a b (m+2) (m+4) \left (a^2 (m+3)+b^2 (m+2)\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx+(m+3) \left (a^4 \left (m^2+6 m+8\right )+6 a^2 b^2 \left (m^2+5 m+4\right )+b^4 \left (m^2+4 m+3\right )\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^mdx}{m+2}+\frac {b^2 (m+3) \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {\frac {b^2 (m+3) \left (a^2 (5 m+22)+b^2 (m+3)\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2)}+\frac {-\frac {4 a b (m+4) \left (a^2 (m+3)+b^2 (m+2)\right ) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}-\frac {(m+3) \left (a^4 \left (m^2+6 m+8\right )+6 a^2 b^2 \left (m^2+5 m+4\right )+b^4 \left (m^2+4 m+3\right )\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}}{m+2}}{m+3}+\frac {2 a b^3 (m+5) \sin (c+d x) \cos ^{m+2}(c+d x)}{d (m+3)}}{m+4}+\frac {b^2 \sin (c+d x) \cos ^{m+1}(c+d x) (a+b \cos (c+d x))^2}{d (m+4)}\)

Input:

Int[Cos[c + d*x]^m*(a + b*Cos[c + d*x])^4,x]
 

Output:

(b^2*Cos[c + d*x]^(1 + m)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*(4 + m)) 
 + ((2*a*b^3*(5 + m)*Cos[c + d*x]^(2 + m)*Sin[c + d*x])/(d*(3 + m)) + ((b^ 
2*(3 + m)*(b^2*(3 + m) + a^2*(22 + 5*m))*Cos[c + d*x]^(1 + m)*Sin[c + d*x] 
)/(d*(2 + m)) + (-(((3 + m)*(b^4*(3 + 4*m + m^2) + 6*a^2*b^2*(4 + 5*m + m^ 
2) + a^4*(8 + 6*m + m^2))*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])) - (4*a*b*(4 + m)*(b^2*(2 + m) + a^2*(3 + 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))/(4 + 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 3272

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 - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d 
*(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 
 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m 
] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] 
&& NeQ[c, 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]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{m} \left (a +\cos \left (d x +c \right ) b \right )^{4}d x\]

Input:

int(cos(d*x+c)^m*(a+cos(d*x+c)*b)^4,x)
 

Output:

int(cos(d*x+c)^m*(a+cos(d*x+c)*b)^4,x)
 

Fricas [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x, algorithm="fricas")
 

Output:

integral((b^4*cos(d*x + c)^4 + 4*a*b^3*cos(d*x + c)^3 + 6*a^2*b^2*cos(d*x 
+ c)^2 + 4*a^3*b*cos(d*x + c) + a^4)*cos(d*x + c)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**m*(a+b*cos(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x, algorithm="maxima")
 

Output:

integrate((b*cos(d*x + c) + a)^4*cos(d*x + c)^m, x)
 

Giac [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{4} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x, algorithm="giac")
 

Output:

integrate((b*cos(d*x + c) + a)^4*cos(d*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^4 \,d x \] Input:

int(cos(c + d*x)^m*(a + b*cos(c + d*x))^4,x)
 

Output:

int(cos(c + d*x)^m*(a + b*cos(c + d*x))^4, x)
 

Reduce [F]

\[ \int \cos ^m(c+d x) (a+b \cos (c+d x))^4 \, dx=\left (\int \cos \left (d x +c \right )^{m}d x \right ) a^{4}+4 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )d x \right ) a^{3} b +\left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{4}d x \right ) b^{4}+4 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{3}d x \right ) a \,b^{3}+6 \left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{2}d x \right ) a^{2} b^{2} \] Input:

int(cos(d*x+c)^m*(a+b*cos(d*x+c))^4,x)
                                                                                    
                                                                                    
 

Output:

int(cos(c + d*x)**m,x)*a**4 + 4*int(cos(c + d*x)**m*cos(c + d*x),x)*a**3*b 
 + int(cos(c + d*x)**m*cos(c + d*x)**4,x)*b**4 + 4*int(cos(c + d*x)**m*cos 
(c + d*x)**3,x)*a*b**3 + 6*int(cos(c + d*x)**m*cos(c + d*x)**2,x)*a**2*b** 
2