3.404 \(\int \cos ^6(c+d x) (a+b \cos (c+d x)) \, dx\)

Optimal. Leaf size=128 \[ \frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16}-\frac {b \sin ^7(c+d x)}{7 d}+\frac {3 b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{d}+\frac {b \sin (c+d x)}{d} \]

[Out]

5/16*a*x+b*sin(d*x+c)/d+5/16*a*cos(d*x+c)*sin(d*x+c)/d+5/24*a*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a*cos(d*x+c)^5*sin
(d*x+c)/d-b*sin(d*x+c)^3/d+3/5*b*sin(d*x+c)^5/d-1/7*b*sin(d*x+c)^7/d

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Rubi [A]  time = 0.09, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2748, 2635, 8, 2633} \[ \frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16}-\frac {b \sin ^7(c+d x)}{7 d}+\frac {3 b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^3(c+d x)}{d}+\frac {b \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + b*Cos[c + d*x]),x]

[Out]

(5*a*x)/16 + (b*Sin[c + d*x])/d + (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a*Cos[c + d*x]^3*Sin[c + d*x])/(
24*d) + (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (b*Sin[c + d*x]^3)/d + (3*b*Sin[c + d*x]^5)/(5*d) - (b*Sin[c +
d*x]^7)/(7*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2748

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]

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+b \cos (c+d x)) \, dx &=a \int \cos ^6(c+d x) \, dx+b \int \cos ^7(c+d x) \, dx\\ &=\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} (5 a) \int \cos ^4(c+d x) \, dx-\frac {b \operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {b \sin (c+d x)}{d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {b \sin ^3(c+d x)}{d}+\frac {3 b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^7(c+d x)}{7 d}+\frac {1}{8} (5 a) \int \cos ^2(c+d x) \, dx\\ &=\frac {b \sin (c+d x)}{d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {b \sin ^3(c+d x)}{d}+\frac {3 b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^7(c+d x)}{7 d}+\frac {1}{16} (5 a) \int 1 \, dx\\ &=\frac {5 a x}{16}+\frac {b \sin (c+d x)}{d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {b \sin ^3(c+d x)}{d}+\frac {3 b \sin ^5(c+d x)}{5 d}-\frac {b \sin ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 89, normalized size = 0.70 \[ \frac {35 a (45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))+60 c+60 d x)-960 b \sin ^7(c+d x)+4032 b \sin ^5(c+d x)-6720 b \sin ^3(c+d x)+6720 b \sin (c+d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + b*Cos[c + d*x]),x]

[Out]

(6720*b*Sin[c + d*x] - 6720*b*Sin[c + d*x]^3 + 4032*b*Sin[c + d*x]^5 - 960*b*Sin[c + d*x]^7 + 35*a*(60*c + 60*
d*x + 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d*x)] + Sin[6*(c + d*x)]))/(6720*d)

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fricas [A]  time = 0.91, size = 86, normalized size = 0.67 \[ \frac {525 \, a d x + {\left (240 \, b \cos \left (d x + c\right )^{6} + 280 \, a \cos \left (d x + c\right )^{5} + 288 \, b \cos \left (d x + c\right )^{4} + 350 \, a \cos \left (d x + c\right )^{3} + 384 \, b \cos \left (d x + c\right )^{2} + 525 \, a \cos \left (d x + c\right ) + 768 \, b\right )} \sin \left (d x + c\right )}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/1680*(525*a*d*x + (240*b*cos(d*x + c)^6 + 280*a*cos(d*x + c)^5 + 288*b*cos(d*x + c)^4 + 350*a*cos(d*x + c)^3
+ 384*b*cos(d*x + c)^2 + 525*a*cos(d*x + c) + 768*b)*sin(d*x + c))/d

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giac [A]  time = 0.60, size = 107, normalized size = 0.84 \[ \frac {5}{16} \, a x + \frac {b \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {7 \, b \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {3 \, a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {7 \, b \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} + \frac {15 \, a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {35 \, b \sin \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

5/16*a*x + 1/448*b*sin(7*d*x + 7*c)/d + 1/192*a*sin(6*d*x + 6*c)/d + 7/320*b*sin(5*d*x + 5*c)/d + 3/64*a*sin(4
*d*x + 4*c)/d + 7/64*b*sin(3*d*x + 3*c)/d + 15/64*a*sin(2*d*x + 2*c)/d + 35/64*b*sin(d*x + c)/d

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maple [A]  time = 0.04, size = 90, normalized size = 0.70 \[ \frac {\frac {b \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}+a \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+b*cos(d*x+c)),x)

[Out]

1/d*(1/7*b*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c)+a*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c
)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

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maxima [A]  time = 0.46, size = 94, normalized size = 0.73 \[ -\frac {35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a + 192 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} b}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

-1/6720*(35*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a + 192*(5*sin(d
*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*b)/d

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mupad [B]  time = 3.25, size = 154, normalized size = 1.20 \[ \frac {5\,a\,x}{16}+\frac {\left (2\,b-\frac {11\,a}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (4\,b-\frac {7\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {86\,b}{5}-\frac {85\,a}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {424\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{35}+\left (\frac {85\,a}{24}+\frac {86\,b}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {7\,a}{6}+4\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,a}{8}+2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(a + b*cos(c + d*x)),x)

[Out]

(5*a*x)/16 + (tan(c/2 + (d*x)/2)*((11*a)/8 + 2*b) + tan(c/2 + (d*x)/2)^3*((7*a)/6 + 4*b) - tan(c/2 + (d*x)/2)^
11*((7*a)/6 - 4*b) - tan(c/2 + (d*x)/2)^13*((11*a)/8 - 2*b) + tan(c/2 + (d*x)/2)^5*((85*a)/24 + (86*b)/5) - ta
n(c/2 + (d*x)/2)^9*((85*a)/24 - (86*b)/5) + (424*b*tan(c/2 + (d*x)/2)^7)/35)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 5.40, size = 238, normalized size = 1.86 \[ \begin {cases} \frac {5 a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {16 b \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 b \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 b \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {b \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right ) \cos ^{6}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+b*cos(d*x+c)),x)

[Out]

Piecewise((5*a*x*sin(c + d*x)**6/16 + 15*a*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a*x*sin(c + d*x)**2*cos(c
+ d*x)**4/16 + 5*a*x*cos(c + d*x)**6/16 + 5*a*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a*sin(c + d*x)**3*cos(c
+ d*x)**3/(6*d) + 11*a*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 16*b*sin(c + d*x)**7/(35*d) + 8*b*sin(c + d*x)**
5*cos(c + d*x)**2/(5*d) + 2*b*sin(c + d*x)**3*cos(c + d*x)**4/d + b*sin(c + d*x)*cos(c + d*x)**6/d, Ne(d, 0)),
(x*(a + b*cos(c))*cos(c)**6, True))

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