\(\int \cos ^m(c+d x) (a+a \cos (c+d x))^4 \, dx\) [397]

Optimal result

Integrand size = 21, antiderivative size = 302 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^4 \, dx=\frac {a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) (3+m) (4+m)}+\frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}-\frac {a^4 \left (35+40 m+8 m^2\right ) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) (2+m) (4+m) \sqrt {\sin ^2(c+d x)}}-\frac {4 a^4 (5+2 m) \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) (3+m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

a^4*(4*m^2+29*m+55)*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(4+m)/(m^2+5*m+6)+cos(d*x+c)^(1+m)*(a^2+a^2*cos(d*x+c))^2*si
n(d*x+c)/d/(4+m)+2*(5+m)*cos(d*x+c)^(1+m)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d/(3+m)/(4+m)-a^4*(8*m^2+40*m+35)*co
s(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/(m^3+7*m^2+14*m+8)/(sin(d*x+c
)^2)^(1/2)-4*a^4*(5+2*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)

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2842, 3055, 3047, 3102, 2827, 2722} \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^4 \, dx=-\frac {a^4 \left (8 m^2+40 m+35\right ) \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}-\frac {4 a^4 (2 m+5) \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {a^4 \left (4 m^2+29 m+55\right ) \sin (c+d x) \cos ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac {2 (m+5) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right ) \cos ^{m+1}(c+d x)}{d (m+3) (m+4)}+\frac {\sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2 \cos ^{m+1}(c+d x)}{d (m+4)} \]

[In]

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

[Out]

(a^4*(55 + 29*m + 4*m^2)*Cos[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)*(4 + m)) + (Cos[c + d*x]^(1 + m
)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*(4 + m)) + (2*(5 + m)*Cos[c + d*x]^(1 + m)*(a^4 + a^4*Cos[c + d*
x])*Sin[c + d*x])/(d*(3 + m)*(4 + m)) - (a^4*(35 + 40*m + 8*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)*(2 + m)*(4 + m)*Sqrt[Sin[c + d*x]^2]) - (4*a^4*(
5 + 2*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*(2
+ m)*(3 + m)*Sqrt[Sin[c + d*x]^2])

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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

Rule 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] + Dist[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 3047

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 3055

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

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {\int \cos ^m(c+d x) (a+a \cos (c+d x))^2 \left (a^2 (5+2 m)+2 a^2 (5+m) \cos (c+d x)\right ) \, dx}{4+m} \\ & = \frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac {\int \cos ^m(c+d x) (a+a \cos (c+d x)) \left (a^3 \left (25+23 m+4 m^2\right )+a^3 \left (55+29 m+4 m^2\right ) \cos (c+d x)\right ) \, dx}{12+7 m+m^2} \\ & = \frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac {\int \cos ^m(c+d x) \left (a^4 \left (25+23 m+4 m^2\right )+\left (a^4 \left (25+23 m+4 m^2\right )+a^4 \left (55+29 m+4 m^2\right )\right ) \cos (c+d x)+a^4 \left (55+29 m+4 m^2\right ) \cos ^2(c+d x)\right ) \, dx}{12+7 m+m^2} \\ & = \frac {a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac {\int \cos ^m(c+d x) \left (a^4 (3+m) \left (35+40 m+8 m^2\right )+4 a^4 (2+m) (4+m) (5+2 m) \cos (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3} \\ & = \frac {a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}+\frac {\left (4 a^4 (5+2 m)\right ) \int \cos ^{1+m}(c+d x) \, dx}{3+m}+\frac {\left (a^4 \left (35+40 m+8 m^2\right )\right ) \int \cos ^m(c+d x) \, dx}{8+6 m+m^2} \\ & = \frac {a^4 \left (55+29 m+4 m^2\right ) \cos ^{1+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\cos ^{1+m}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{d (4+m)}+\frac {2 (5+m) \cos ^{1+m}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{d (3+m) (4+m)}-\frac {a^4 \left (35+40 m+8 m^2\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) \left (8+6 m+m^2\right ) \sqrt {\sin ^2(c+d x)}}-\frac {4 a^4 (5+2 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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.93 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x))^4 \, dx=\frac {a^2 \cos ^{1+m}(c+d x) \csc (c+d x) \left ((a+a \cos (c+d x))^2 \sin ^2(c+d x)-\frac {a^2 (5+2 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{1+m}-\frac {2 a^2 (10+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{2+m}-\frac {a^2 (25+6 m) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{3+m}-\frac {2 a^2 (5+m) \cos ^3(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+m}{2},\frac {6+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{4+m}\right )}{d (4+m)} \]

[In]

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

[Out]

(a^2*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*((a + a*Cos[c + d*x])^2*Sin[c + d*x]^2 - (a^2*(5 + 2*m)*Hypergeometric2
F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/(1 + m) - (2*a^2*(10 + 3*m)*Cos[c + d*x]*H
ypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/(2 + m) - (a^2*(25 + 6*m)*Co
s[c + d*x]^2*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/(3 + m) - (2*a
^2*(5 + m)*Cos[c + d*x]^3*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/(
4 + m)))/(d*(4 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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