Integrand size = 18, antiderivative size = 61 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {64 \sin ^7(a+b x)}{7 b}-\frac {64 \sin ^9(a+b x)}{3 b}+\frac {192 \sin ^{11}(a+b x)}{11 b}-\frac {64 \sin ^{13}(a+b x)}{13 b} \] Output:
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
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {2 (5230+6377 \cos (2 (a+b x))+1890 \cos (4 (a+b x))+231 \cos (6 (a+b x))) \sin ^7(a+b x)}{3003 b} \] Input:
Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^6,x]
Output:
(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)
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4775, 3042, 3044, 244, 2009}
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.
\(\displaystyle \int \sin ^6(2 a+2 b x) \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (2 a+2 b x)^6 \cos (a+b x)dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle 64 \int \cos ^7(a+b x) \sin ^6(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 64 \int \cos (a+b x)^7 \sin (a+b x)^6dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {64 \int \sin ^6(a+b x) \left (1-\sin ^2(a+b x)\right )^3d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {64 \int \left (-\sin ^{12}(a+b x)+3 \sin ^{10}(a+b x)-3 \sin ^8(a+b x)+\sin ^6(a+b x)\right )d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {64 \left (-\frac {1}{13} \sin ^{13}(a+b x)+\frac {3}{11} \sin ^{11}(a+b x)-\frac {1}{3} \sin ^9(a+b x)+\frac {1}{7} \sin ^7(a+b x)\right )}{b}\) |
Input:
Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^6,x]
Output:
(64*(Sin[a + b*x]^7/7 - Sin[a + b*x]^9/3 + (3*Sin[a + b*x]^11)/11 - Sin[a + b*x]^13/13))/b
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[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] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[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] && I ntegerQ[p]
Time = 14.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.33
method | result | size |
parallelrisch | \(\frac {60060 \sin \left (b x +a \right )-15015 \sin \left (3 b x +3 a \right )-9009 \sin \left (5 b x +5 a \right )+2574 \sin \left (7 b x +7 a \right )+2002 \sin \left (9 b x +9 a \right )-273 \sin \left (11 b x +11 a \right )-231 \sin \left (13 b x +13 a \right )}{192192 b}\) | \(81\) |
default | \(\frac {5 \sin \left (b x +a \right )}{16 b}-\frac {5 \sin \left (3 b x +3 a \right )}{64 b}-\frac {3 \sin \left (5 b x +5 a \right )}{64 b}+\frac {3 \sin \left (7 b x +7 a \right )}{224 b}+\frac {\sin \left (9 b x +9 a \right )}{96 b}-\frac {\sin \left (11 b x +11 a \right )}{704 b}-\frac {\sin \left (13 b x +13 a \right )}{832 b}\) | \(97\) |
risch | \(\frac {5 \sin \left (b x +a \right )}{16 b}-\frac {5 \sin \left (3 b x +3 a \right )}{64 b}-\frac {3 \sin \left (5 b x +5 a \right )}{64 b}+\frac {3 \sin \left (7 b x +7 a \right )}{224 b}+\frac {\sin \left (9 b x +9 a \right )}{96 b}-\frac {\sin \left (11 b x +11 a \right )}{704 b}-\frac {\sin \left (13 b x +13 a \right )}{832 b}\) | \(97\) |
orering | \(\text {Expression too large to display}\) | \(1215\) |
Input:
int(cos(b*x+a)*sin(2*b*x+2*a)^6,x,method=_RETURNVERBOSE)
Output:
1/192192*(60060*sin(b*x+a)-15015*sin(3*b*x+3*a)-9009*sin(5*b*x+5*a)+2574*s in(7*b*x+7*a)+2002*sin(9*b*x+9*a)-273*sin(11*b*x+11*a)-231*sin(13*b*x+13*a ))/b
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {64 \, {\left (231 \, \cos \left (b x + a\right )^{12} - 567 \, \cos \left (b x + a\right )^{10} + 371 \, \cos \left (b x + a\right )^{8} - 5 \, \cos \left (b x + a\right )^{6} - 6 \, \cos \left (b x + a\right )^{4} - 8 \, \cos \left (b x + a\right )^{2} - 16\right )} \sin \left (b x + a\right )}{3003 \, b} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="fricas")
Output:
-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
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (53) = 106\).
Time = 11.12 (sec) , antiderivative size = 233, normalized size of antiderivative = 3.82 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\begin {cases} \frac {835 \sin {\left (a + b x \right )} \sin ^{6}{\left (2 a + 2 b x \right )}}{3003 b} + \frac {2776 \sin {\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{3003 b} + \frac {2944 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{3003 b} + \frac {1024 \sin {\left (a + b x \right )} \cos ^{6}{\left (2 a + 2 b x \right )}}{3003 b} - \frac {1084 \sin ^{5}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{3003 b} - \frac {64 \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{143 b} - \frac {512 \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{5}{\left (2 a + 2 b x \right )}}{3003 b} & \text {for}\: b \neq 0 \\x \sin ^{6}{\left (2 a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)**6,x)
Output:
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)*si n(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/(1 43*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))
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {231 \, \sin \left (13 \, b x + 13 \, a\right ) + 273 \, \sin \left (11 \, b x + 11 \, a\right ) - 2002 \, \sin \left (9 \, b x + 9 \, a\right ) - 2574 \, \sin \left (7 \, b x + 7 \, a\right ) + 9009 \, \sin \left (5 \, b x + 5 \, a\right ) + 15015 \, \sin \left (3 \, b x + 3 \, a\right ) - 60060 \, \sin \left (b x + a\right )}{192192 \, b} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="maxima")
Output:
-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) + 9009*sin(5*b*x + 5*a) + 15015*sin(3*b*x + 3*a) - 60060*sin(b*x + a))/b
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=-\frac {231 \, \sin \left (13 \, b x + 13 \, a\right ) + 273 \, \sin \left (11 \, b x + 11 \, a\right ) - 2002 \, \sin \left (9 \, b x + 9 \, a\right ) - 2574 \, \sin \left (7 \, b x + 7 \, a\right ) + 9009 \, \sin \left (5 \, b x + 5 \, a\right ) + 15015 \, \sin \left (3 \, b x + 3 \, a\right ) - 60060 \, \sin \left (b x + a\right )}{192192 \, b} \] Input:
integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="giac")
Output:
-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) + 9009*sin(5*b*x + 5*a) + 15015*sin(3*b*x + 3*a) - 60060*sin(b*x + a))/b
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {-\frac {64\,{\sin \left (a+b\,x\right )}^{13}}{13}+\frac {192\,{\sin \left (a+b\,x\right )}^{11}}{11}-\frac {64\,{\sin \left (a+b\,x\right )}^9}{3}+\frac {64\,{\sin \left (a+b\,x\right )}^7}{7}}{b} \] Input:
int(cos(a + b*x)*sin(2*a + 2*b*x)^6,x)
Output:
((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
Time = 0.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.51 \[ \int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx=\frac {-252 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{5}-320 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3}-512 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )-21 \sin \left (2 b x +2 a \right )^{6} \sin \left (b x +a \right )-40 \sin \left (2 b x +2 a \right )^{4} \sin \left (b x +a \right )-128 \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )+1024 \sin \left (b x +a \right )}{3003 b} \] Input:
int(cos(b*x+a)*sin(2*b*x+2*a)^6,x)
Output:
( - 252*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**5 - 320*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x)**3 - 512*cos(2*a + 2*b*x)*cos(a + b*x )*sin(2*a + 2*b*x) - 21*sin(2*a + 2*b*x)**6*sin(a + b*x) - 40*sin(2*a + 2* b*x)**4*sin(a + b*x) - 128*sin(2*a + 2*b*x)**2*sin(a + b*x) + 1024*sin(a + b*x))/(3003*b)