Integrand size = 29, antiderivative size = 165 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {13 a^2 x}{256}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^9(c+d x)}{9 d}+\frac {13 a^2 \cos (c+d x) \sin (c+d x)}{256 d}+\frac {13 a^2 \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {13 a^2 \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {13 a^2 \cos ^7(c+d x) \sin (c+d x)}{80 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d} \] Output:
13/256*a^2*x-2/7*a^2*cos(d*x+c)^7/d+2/9*a^2*cos(d*x+c)^9/d+13/256*a^2*cos( d*x+c)*sin(d*x+c)/d+13/384*a^2*cos(d*x+c)^3*sin(d*x+c)/d+13/480*a^2*cos(d* x+c)^5*sin(d*x+c)/d-13/80*a^2*cos(d*x+c)^7*sin(d*x+c)/d-1/10*a^2*cos(d*x+c )^7*sin(d*x+c)^3/d
Time = 0.55 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.64 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (12600 c+32760 d x-30240 \cos (c+d x)-13440 \cos (3 (c+d x))+2160 \cos (7 (c+d x))+560 \cos (9 (c+d x))+11340 \sin (2 (c+d x))-7560 \sin (4 (c+d x))-3990 \sin (6 (c+d x))-315 \sin (8 (c+d x))+126 \sin (10 (c+d x)))}{645120 d} \] Input:
Integrate[Cos[c + d*x]^6*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(12600*c + 32760*d*x - 30240*Cos[c + d*x] - 13440*Cos[3*(c + d*x)] + 2160*Cos[7*(c + d*x)] + 560*Cos[9*(c + d*x)] + 11340*Sin[2*(c + d*x)] - 75 60*Sin[4*(c + d*x)] - 3990*Sin[6*(c + d*x)] - 315*Sin[8*(c + d*x)] + 126*S in[10*(c + d*x)]))/(645120*d)
Time = 0.56 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3352, 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.
\(\displaystyle \int \sin ^2(c+d x) \cos ^6(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^6 (a \sin (c+d x)+a)^2dx\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \int \left (a^2 \sin ^4(c+d x) \cos ^6(c+d x)+2 a^2 \sin ^3(c+d x) \cos ^6(c+d x)+a^2 \sin ^2(c+d x) \cos ^6(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^2 \cos ^9(c+d x)}{9 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{10 d}-\frac {13 a^2 \sin (c+d x) \cos ^7(c+d x)}{80 d}+\frac {13 a^2 \sin (c+d x) \cos ^5(c+d x)}{480 d}+\frac {13 a^2 \sin (c+d x) \cos ^3(c+d x)}{384 d}+\frac {13 a^2 \sin (c+d x) \cos (c+d x)}{256 d}+\frac {13 a^2 x}{256}\) |
Input:
Int[Cos[c + d*x]^6*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
Output:
(13*a^2*x)/256 - (2*a^2*Cos[c + d*x]^7)/(7*d) + (2*a^2*Cos[c + d*x]^9)/(9* d) + (13*a^2*Cos[c + d*x]*Sin[c + d*x])/(256*d) + (13*a^2*Cos[c + d*x]^3*S in[c + d*x])/(384*d) + (13*a^2*Cos[c + d*x]^5*Sin[c + d*x])/(480*d) - (13* a^2*Cos[c + d*x]^7*Sin[c + d*x])/(80*d) - (a^2*Cos[c + d*x]^7*Sin[c + d*x] ^3)/(10*d)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12
\[\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{10}-\frac {3 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{80}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+2 a^{2} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )+a^{2} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )}{d}\]
Input:
int(cos(d*x+c)^6*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
1/d*(a^2*(-1/10*sin(d*x+c)^3*cos(d*x+c)^7-3/80*cos(d*x+c)^7*sin(d*x+c)+1/1 60*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+3/256*d*x+3/ 256*c)+2*a^2*(-1/9*cos(d*x+c)^7*sin(d*x+c)^2-2/63*cos(d*x+c)^7)+a^2*(-1/8* cos(d*x+c)^7*sin(d*x+c)+1/48*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c ))*sin(d*x+c)+5/128*d*x+5/128*c))
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.67 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {17920 \, a^{2} \cos \left (d x + c\right )^{9} - 23040 \, a^{2} \cos \left (d x + c\right )^{7} + 4095 \, a^{2} d x + 21 \, {\left (384 \, a^{2} \cos \left (d x + c\right )^{9} - 1008 \, a^{2} \cos \left (d x + c\right )^{7} + 104 \, a^{2} \cos \left (d x + c\right )^{5} + 130 \, a^{2} \cos \left (d x + c\right )^{3} + 195 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80640 \, d} \] Input:
integrate(cos(d*x+c)^6*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/80640*(17920*a^2*cos(d*x + c)^9 - 23040*a^2*cos(d*x + c)^7 + 4095*a^2*d* x + 21*(384*a^2*cos(d*x + c)^9 - 1008*a^2*cos(d*x + c)^7 + 104*a^2*cos(d*x + c)^5 + 130*a^2*cos(d*x + c)^3 + 195*a^2*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (158) = 316\).
Time = 1.35 (sec) , antiderivative size = 529, normalized size of antiderivative = 3.21 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{10}{\left (c + d x \right )}}{256} + \frac {15 a^{2} x \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{256} + \frac {5 a^{2} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{128} + \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{256} + \frac {5 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 a^{2} x \cos ^{10}{\left (c + d x \right )}}{256} + \frac {5 a^{2} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 a^{2} \sin ^{9}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{256 d} + \frac {7 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {5 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{10 d} + \frac {55 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{384 d} - \frac {7 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} + \frac {73 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{384 d} - \frac {2 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {3 a^{2} \sin {\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{256 d} - \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {4 a^{2} \cos ^{9}{\left (c + d x \right )}}{63 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{6}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**6*sin(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((3*a**2*x*sin(c + d*x)**10/256 + 15*a**2*x*sin(c + d*x)**8*cos(c + d*x)**2/256 + 5*a**2*x*sin(c + d*x)**8/128 + 15*a**2*x*sin(c + d*x)**6* cos(c + d*x)**4/128 + 5*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 15*a** 2*x*sin(c + d*x)**4*cos(c + d*x)**6/128 + 15*a**2*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 15*a**2*x*sin(c + d*x)**2*cos(c + d*x)**8/256 + 5*a**2*x*si n(c + d*x)**2*cos(c + d*x)**6/32 + 3*a**2*x*cos(c + d*x)**10/256 + 5*a**2* x*cos(c + d*x)**8/128 + 3*a**2*sin(c + d*x)**9*cos(c + d*x)/(256*d) + 7*a* *2*sin(c + d*x)**7*cos(c + d*x)**3/(128*d) + 5*a**2*sin(c + d*x)**7*cos(c + d*x)/(128*d) + a**2*sin(c + d*x)**5*cos(c + d*x)**5/(10*d) + 55*a**2*sin (c + d*x)**5*cos(c + d*x)**3/(384*d) - 7*a**2*sin(c + d*x)**3*cos(c + d*x) **7/(128*d) + 73*a**2*sin(c + d*x)**3*cos(c + d*x)**5/(384*d) - 2*a**2*sin (c + d*x)**2*cos(c + d*x)**7/(7*d) - 3*a**2*sin(c + d*x)*cos(c + d*x)**9/( 256*d) - 5*a**2*sin(c + d*x)*cos(c + d*x)**7/(128*d) - 4*a**2*cos(c + d*x) **9/(63*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**2*cos(c)**6, True))
Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.78 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {20480 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{2} + 63 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 210 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{645120 \, d} \] Input:
integrate(cos(d*x+c)^6*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/645120*(20480*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^2 + 63*(32*sin(2*d *x + 2*c)^5 + 120*d*x + 120*c + 5*sin(8*d*x + 8*c) - 40*sin(4*d*x + 4*c))* a^2 + 210*(64*sin(2*d*x + 2*c)^3 + 120*d*x + 120*c - 3*sin(8*d*x + 8*c) - 24*sin(4*d*x + 4*c))*a^2)/d
Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {13}{256} \, a^{2} x + \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{1152 \, d} + \frac {3 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{896 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, a^{2} \cos \left (d x + c\right )}{64 \, d} + \frac {a^{2} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{2048 \, d} - \frac {19 \, a^{2} \sin \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {3 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{256 \, d} + \frac {9 \, a^{2} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} \] Input:
integrate(cos(d*x+c)^6*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
13/256*a^2*x + 1/1152*a^2*cos(9*d*x + 9*c)/d + 3/896*a^2*cos(7*d*x + 7*c)/ d - 1/48*a^2*cos(3*d*x + 3*c)/d - 3/64*a^2*cos(d*x + c)/d + 1/5120*a^2*sin (10*d*x + 10*c)/d - 1/2048*a^2*sin(8*d*x + 8*c)/d - 19/3072*a^2*sin(6*d*x + 6*c)/d - 3/256*a^2*sin(4*d*x + 4*c)/d + 9/512*a^2*sin(2*d*x + 2*c)/d
Time = 36.25 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.84 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:
int(cos(c + d*x)^6*sin(c + d*x)^2*(a + a*sin(c + d*x))^2,x)
Output:
(13*a^2*x)/256 - ((13*a^2*(c + d*x))/256 - (647*a^2*tan(c/2 + (d*x)/2)^3)/ 384 - (2311*a^2*tan(c/2 + (d*x)/2)^5)/480 + (457*a^2*tan(c/2 + (d*x)/2)^7) /32 - (2169*a^2*tan(c/2 + (d*x)/2)^9)/64 + (2169*a^2*tan(c/2 + (d*x)/2)^11 )/64 - (457*a^2*tan(c/2 + (d*x)/2)^13)/32 + (2311*a^2*tan(c/2 + (d*x)/2)^1 5)/480 + (647*a^2*tan(c/2 + (d*x)/2)^17)/384 - (13*a^2*tan(c/2 + (d*x)/2)^ 19)/128 - (a^2*(4095*c + 4095*d*x - 10240))/80640 + tan(c/2 + (d*x)/2)^2*( (65*a^2*(c + d*x))/128 - (a^2*(40950*c + 40950*d*x - 102400))/80640) + tan (c/2 + (d*x)/2)^4*((585*a^2*(c + d*x))/256 - (a^2*(184275*c + 184275*d*x + 184320))/80640) + tan(c/2 + (d*x)/2)^16*((585*a^2*(c + d*x))/256 - (a^2*( 184275*c + 184275*d*x - 645120))/80640) + tan(c/2 + (d*x)/2)^14*((195*a^2* (c + d*x))/32 - (a^2*(491400*c + 491400*d*x + 430080))/80640) + tan(c/2 + (d*x)/2)^6*((195*a^2*(c + d*x))/32 - (a^2*(491400*c + 491400*d*x - 1658880 ))/80640) + tan(c/2 + (d*x)/2)^10*((819*a^2*(c + d*x))/64 - (a^2*(1031940* c + 1031940*d*x - 1290240))/80640) + tan(c/2 + (d*x)/2)^12*((1365*a^2*(c + d*x))/128 - (a^2*(859950*c + 859950*d*x - 2150400))/80640) + (13*a^2*tan( c/2 + (d*x)/2))/128)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^10)
Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.99 \[ \int \cos ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (8064 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9}+17920 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}-11088 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}-48640 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-12936 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+38400 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+24150 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-2560 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-4095 \cos \left (d x +c \right ) \sin \left (d x +c \right )-5120 \cos \left (d x +c \right )+4095 d x +5120\right )}{80640 d} \] Input:
int(cos(d*x+c)^6*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*(8064*cos(c + d*x)*sin(c + d*x)**9 + 17920*cos(c + d*x)*sin(c + d*x) **8 - 11088*cos(c + d*x)*sin(c + d*x)**7 - 48640*cos(c + d*x)*sin(c + d*x) **6 - 12936*cos(c + d*x)*sin(c + d*x)**5 + 38400*cos(c + d*x)*sin(c + d*x) **4 + 24150*cos(c + d*x)*sin(c + d*x)**3 - 2560*cos(c + d*x)*sin(c + d*x)* *2 - 4095*cos(c + d*x)*sin(c + d*x) - 5120*cos(c + d*x) + 4095*d*x + 5120) )/(80640*d)