\(\int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx\) [130]

Optimal result

Integrand size = 18, antiderivative size = 61 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {64 \sin ^7(a+b x)}{7 b}-\frac {64 \sin ^9(a+b x)}{3 b}+\frac {192 \sin ^{11}(a+b x)}{11 b}-\frac {64 \sin ^{13}(a+b x)}{13 b} \]

[Out]

64/7*sin(b*x+a)^7/b-64/3*sin(b*x+a)^9/b+192/11*sin(b*x+a)^11/b-64/13*sin(b*x+a)^13/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4372, 2644, 276} \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {64 \sin ^{13}(a+b x)}{13 b}+\frac {192 \sin ^{11}(a+b x)}{11 b}-\frac {64 \sin ^9(a+b x)}{3 b}+\frac {64 \sin ^7(a+b x)}{7 b} \]

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^6,x]

[Out]

(64*Sin[a + b*x]^7)/(7*b) - (64*Sin[a + b*x]^9)/(3*b) + (192*Sin[a + b*x]^11)/(11*b) - (64*Sin[a + b*x]^13)/(1
3*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 4372

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = 64 \int \cos ^7(a+b x) \sin ^6(a+b x) \, dx \\ & = \frac {64 \text {Subst}\left (\int x^6 \left (1-x^2\right )^3 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {64 \text {Subst}\left (\int \left (x^6-3 x^8+3 x^{10}-x^{12}\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {64 \sin ^7(a+b x)}{7 b}-\frac {64 \sin ^9(a+b x)}{3 b}+\frac {192 \sin ^{11}(a+b x)}{11 b}-\frac {64 \sin ^{13}(a+b x)}{13 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {2 (5230+6377 \cos (2 (a+b x))+1890 \cos (4 (a+b x))+231 \cos (6 (a+b x))) \sin ^7(a+b x)}{3003 b} \]

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^6,x]

[Out]

(2*(5230 + 6377*Cos[2*(a + b*x)] + 1890*Cos[4*(a + b*x)] + 231*Cos[6*(a + b*x)])*Sin[a + b*x]^7)/(3003*b)

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {5 \sin \left (x b +a \right )}{16 b}-\frac {5 \sin \left (3 x b +3 a \right )}{64 b}-\frac {3 \sin \left (5 x b +5 a \right )}{64 b}+\frac {3 \sin \left (7 x b +7 a \right )}{224 b}+\frac {\sin \left (9 x b +9 a \right )}{96 b}-\frac {\sin \left (11 x b +11 a \right )}{704 b}-\frac {\sin \left (13 x b +13 a \right )}{832 b}\)	\(97\)
risch	\(\frac {5 \sin \left (x b +a \right )}{16 b}-\frac {5 \sin \left (3 x b +3 a \right )}{64 b}-\frac {3 \sin \left (5 x b +5 a \right )}{64 b}+\frac {3 \sin \left (7 x b +7 a \right )}{224 b}+\frac {\sin \left (9 x b +9 a \right )}{96 b}-\frac {\sin \left (11 x b +11 a \right )}{704 b}-\frac {\sin \left (13 x b +13 a \right )}{832 b}\)	\(97\)
parallelrisch	\(\frac {\frac {128 \left (-8 \tan \left (x b +a \right )^{11}-44 \tan \left (x b +a \right )^{9}-99 \tan \left (x b +a \right )^{7}+99 \tan \left (x b +a \right )^{5}+44 \tan \left (x b +a \right )^{3}+8 \tan \left (x b +a \right )\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}}{3003}+\frac {2048 \left (\tan \left (x b +a \right )^{8}+5 \tan \left (x b +a \right )^{6}+\frac {71 \tan \left (x b +a \right )^{4}}{8}+5 \tan \left (x b +a \right )^{2}+1\right ) \left (\tan \left (x b +a \right )^{4}+\frac {\tan \left (x b +a \right )^{2}}{2}+1\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{3003}+\frac {1024 \tan \left (x b +a \right )^{11}}{3003}+\frac {512 \tan \left (x b +a \right )^{9}}{273}+\frac {384 \tan \left (x b +a \right )^{7}}{91}-\frac {384 \tan \left (x b +a \right )^{5}}{91}-\frac {512 \tan \left (x b +a \right )^{3}}{273}-\frac {1024 \tan \left (x b +a \right )}{3003}}{b \left (1+\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}\right ) \left (1+\tan \left (x b +a \right )^{2}\right )^{6}}\)	\(234\)

[In]

int(cos(b*x+a)*sin(2*b*x+2*a)^6,x,method=_RETURNVERBOSE)

[Out]

5/16*sin(b*x+a)/b-5/64*sin(3*b*x+3*a)/b-3/64/b*sin(5*b*x+5*a)+3/224/b*sin(7*b*x+7*a)+1/96/b*sin(9*b*x+9*a)-1/7
04/b*sin(11*b*x+11*a)-1/832/b*sin(13*b*x+13*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {64 \, {\left (231 \, \cos \left (b x + a\right )^{12} - 567 \, \cos \left (b x + a\right )^{10} + 371 \, \cos \left (b x + a\right )^{8} - 5 \, \cos \left (b x + a\right )^{6} - 6 \, \cos \left (b x + a\right )^{4} - 8 \, \cos \left (b x + a\right )^{2} - 16\right )} \sin \left (b x + a\right )}{3003 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="fricas")

[Out]

-64/3003*(231*cos(b*x + a)^12 - 567*cos(b*x + a)^10 + 371*cos(b*x + a)^8 - 5*cos(b*x + a)^6 - 6*cos(b*x + a)^4
- 8*cos(b*x + a)^2 - 16)*sin(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (53) = 106\).

Time = 11.06 (sec) , antiderivative size = 233, normalized size of antiderivative = 3.82 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\begin {cases} \frac {835 \sin {\left (a + b x \right )} \sin ^{6}{\left (2 a + 2 b x \right )}}{3003 b} + \frac {2776 \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{3003 b} + \frac {2944 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{3003 b} + \frac {1024 \sin {\left (a + b x \right )} \cos ^{6}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {1084 \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{3003 b} - \frac {64 \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{143 b} - \frac {512 \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{3003 b} & \text {for}\: b \neq 0 \\x \sin ^{6}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)**6,x)

[Out]

Piecewise((835*sin(a + b*x)*sin(2*a + 2*b*x)**6/(3003*b) + 2776*sin(a + b*x)*sin(2*a + 2*b*x)**4*cos(2*a + 2*b
*x)**2/(3003*b) + 2944*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*x)**4/(3003*b) + 1024*sin(a + b*x)*cos(2
*a + 2*b*x)**6/(3003*b) - 1084*sin(2*a + 2*b*x)**5*cos(a + b*x)*cos(2*a + 2*b*x)/(3003*b) - 64*sin(2*a + 2*b*x
)**3*cos(a + b*x)*cos(2*a + 2*b*x)**3/(143*b) - 512*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**5/(3003*b)
, Ne(b, 0)), (x*sin(2*a)**6*cos(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {231 \, \sin \left (13 \, b x + 13 \, a\right ) + 273 \, \sin \left (11 \, b x + 11 \, a\right ) - 2002 \, \sin \left (9 \, b x + 9 \, a\right ) - 2574 \, \sin \left (7 \, b x + 7 \, a\right ) + 9009 \, \sin \left (5 \, b x + 5 \, a\right ) + 15015 \, \sin \left (3 \, b x + 3 \, a\right ) - 60060 \, \sin \left (b x + a\right )}{192192 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="maxima")

[Out]

-1/192192*(231*sin(13*b*x + 13*a) + 273*sin(11*b*x + 11*a) - 2002*sin(9*b*x + 9*a) - 2574*sin(7*b*x + 7*a) + 9
009*sin(5*b*x + 5*a) + 15015*sin(3*b*x + 3*a) - 60060*sin(b*x + a))/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {231 \, \sin \left (13 \, b x + 13 \, a\right ) + 273 \, \sin \left (11 \, b x + 11 \, a\right ) - 2002 \, \sin \left (9 \, b x + 9 \, a\right ) - 2574 \, \sin \left (7 \, b x + 7 \, a\right ) + 9009 \, \sin \left (5 \, b x + 5 \, a\right ) + 15015 \, \sin \left (3 \, b x + 3 \, a\right ) - 60060 \, \sin \left (b x + a\right )}{192192 \, b} \]

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="giac")

[Out]

-1/192192*(231*sin(13*b*x + 13*a) + 273*sin(11*b*x + 11*a) - 2002*sin(9*b*x + 9*a) - 2574*sin(7*b*x + 7*a) + 9
009*sin(5*b*x + 5*a) + 15015*sin(3*b*x + 3*a) - 60060*sin(b*x + a))/b

Mupad [B] (verification not implemented)

Time = 19.71 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {-\frac {64\,{\sin \left (a+b\,x\right )}^{13}}{13}+\frac {192\,{\sin \left (a+b\,x\right )}^{11}}{11}-\frac {64\,{\sin \left (a+b\,x\right )}^9}{3}+\frac {64\,{\sin \left (a+b\,x\right )}^7}{7}}{b} \]

[In]

int(cos(a + b*x)*sin(2*a + 2*b*x)^6,x)

[Out]

((64*sin(a + b*x)^7)/7 - (64*sin(a + b*x)^9)/3 + (192*sin(a + b*x)^11)/11 - (64*sin(a + b*x)^13)/13)/b