Integrand size = 29, antiderivative size = 161 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {29 x}{128 a^3}-\frac {4 \cos ^3(c+d x)}{3 a^3 d}+\frac {7 \cos ^5(c+d x)}{5 a^3 d}-\frac {3 \cos ^7(c+d x)}{7 a^3 d}-\frac {29 \cos (c+d x) \sin (c+d x)}{128 a^3 d}+\frac {29 \cos ^3(c+d x) \sin (c+d x)}{64 a^3 d}+\frac {29 \cos ^3(c+d x) \sin ^3(c+d x)}{48 a^3 d}+\frac {\cos ^3(c+d x) \sin ^5(c+d x)}{8 a^3 d} \] Output:
-29/128*x/a^3-4/3*cos(d*x+c)^3/a^3/d+7/5*cos(d*x+c)^5/a^3/d-3/7*cos(d*x+c) ^7/a^3/d-29/128*cos(d*x+c)*sin(d*x+c)/a^3/d+29/64*cos(d*x+c)^3*sin(d*x+c)/ a^3/d+29/48*cos(d*x+c)^3*sin(d*x+c)^3/a^3/d+1/8*cos(d*x+c)^3*sin(d*x+c)^5/ a^3/d
Leaf count is larger than twice the leaf count of optimal. \(482\) vs. \(2(161)=322\).
Time = 5.09 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.99 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {84 (-7+12870 c-580 d x) \cos \left (\frac {c}{2}\right )-38640 \cos \left (\frac {c}{2}+d x\right )-38640 \cos \left (\frac {3 c}{2}+d x\right )+6720 \cos \left (\frac {3 c}{2}+2 d x\right )-6720 \cos \left (\frac {5 c}{2}+2 d x\right )-3920 \cos \left (\frac {5 c}{2}+3 d x\right )-3920 \cos \left (\frac {7 c}{2}+3 d x\right )+5880 \cos \left (\frac {7 c}{2}+4 d x\right )-5880 \cos \left (\frac {9 c}{2}+4 d x\right )+4368 \cos \left (\frac {9 c}{2}+5 d x\right )+4368 \cos \left (\frac {11 c}{2}+5 d x\right )-2240 \cos \left (\frac {11 c}{2}+6 d x\right )+2240 \cos \left (\frac {13 c}{2}+6 d x\right )-720 \cos \left (\frac {13 c}{2}+7 d x\right )-720 \cos \left (\frac {15 c}{2}+7 d x\right )+105 \cos \left (\frac {15 c}{2}+8 d x\right )-105 \cos \left (\frac {17 c}{2}+8 d x\right )-998928 \sin \left (\frac {c}{2}\right )+1081080 c \sin \left (\frac {c}{2}\right )-48720 d x \sin \left (\frac {c}{2}\right )+38640 \sin \left (\frac {c}{2}+d x\right )-38640 \sin \left (\frac {3 c}{2}+d x\right )+6720 \sin \left (\frac {3 c}{2}+2 d x\right )+6720 \sin \left (\frac {5 c}{2}+2 d x\right )+3920 \sin \left (\frac {5 c}{2}+3 d x\right )-3920 \sin \left (\frac {7 c}{2}+3 d x\right )+5880 \sin \left (\frac {7 c}{2}+4 d x\right )+5880 \sin \left (\frac {9 c}{2}+4 d x\right )-4368 \sin \left (\frac {9 c}{2}+5 d x\right )+4368 \sin \left (\frac {11 c}{2}+5 d x\right )-2240 \sin \left (\frac {11 c}{2}+6 d x\right )-2240 \sin \left (\frac {13 c}{2}+6 d x\right )+720 \sin \left (\frac {13 c}{2}+7 d x\right )-720 \sin \left (\frac {15 c}{2}+7 d x\right )+105 \sin \left (\frac {15 c}{2}+8 d x\right )+105 \sin \left (\frac {17 c}{2}+8 d x\right )}{215040 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:
Integrate[(Cos[c + d*x]^8*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
(84*(-7 + 12870*c - 580*d*x)*Cos[c/2] - 38640*Cos[c/2 + d*x] - 38640*Cos[( 3*c)/2 + d*x] + 6720*Cos[(3*c)/2 + 2*d*x] - 6720*Cos[(5*c)/2 + 2*d*x] - 39 20*Cos[(5*c)/2 + 3*d*x] - 3920*Cos[(7*c)/2 + 3*d*x] + 5880*Cos[(7*c)/2 + 4 *d*x] - 5880*Cos[(9*c)/2 + 4*d*x] + 4368*Cos[(9*c)/2 + 5*d*x] + 4368*Cos[( 11*c)/2 + 5*d*x] - 2240*Cos[(11*c)/2 + 6*d*x] + 2240*Cos[(13*c)/2 + 6*d*x] - 720*Cos[(13*c)/2 + 7*d*x] - 720*Cos[(15*c)/2 + 7*d*x] + 105*Cos[(15*c)/ 2 + 8*d*x] - 105*Cos[(17*c)/2 + 8*d*x] - 998928*Sin[c/2] + 1081080*c*Sin[c /2] - 48720*d*x*Sin[c/2] + 38640*Sin[c/2 + d*x] - 38640*Sin[(3*c)/2 + d*x] + 6720*Sin[(3*c)/2 + 2*d*x] + 6720*Sin[(5*c)/2 + 2*d*x] + 3920*Sin[(5*c)/ 2 + 3*d*x] - 3920*Sin[(7*c)/2 + 3*d*x] + 5880*Sin[(7*c)/2 + 4*d*x] + 5880* Sin[(9*c)/2 + 4*d*x] - 4368*Sin[(9*c)/2 + 5*d*x] + 4368*Sin[(11*c)/2 + 5*d *x] - 2240*Sin[(11*c)/2 + 6*d*x] - 2240*Sin[(13*c)/2 + 6*d*x] + 720*Sin[(1 3*c)/2 + 7*d*x] - 720*Sin[(15*c)/2 + 7*d*x] + 105*Sin[(15*c)/2 + 8*d*x] + 105*Sin[(17*c)/2 + 8*d*x])/(215040*a^3*d*(Cos[c/2] + Sin[c/2]))
Time = 0.76 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^8(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^8}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) \sin ^3(c+d x) (a-a \sin (c+d x))^3dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos (c+d x)^2 \sin (c+d x)^3 (a-a \sin (c+d x))^3dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (-a^3 \cos ^2(c+d x) \sin ^6(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^5(c+d x)-3 a^3 \cos ^2(c+d x) \sin ^4(c+d x)+a^3 \cos ^2(c+d x) \sin ^3(c+d x)\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {3 a^3 \cos ^7(c+d x)}{7 d}+\frac {7 a^3 \cos ^5(c+d x)}{5 d}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \sin ^5(c+d x) \cos ^3(c+d x)}{8 d}+\frac {29 a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{48 d}+\frac {29 a^3 \sin (c+d x) \cos ^3(c+d x)}{64 d}-\frac {29 a^3 \sin (c+d x) \cos (c+d x)}{128 d}-\frac {29 a^3 x}{128}}{a^6}\) |
Input:
Int[(Cos[c + d*x]^8*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
((-29*a^3*x)/128 - (4*a^3*Cos[c + d*x]^3)/(3*d) + (7*a^3*Cos[c + d*x]^5)/( 5*d) - (3*a^3*Cos[c + d*x]^7)/(7*d) - (29*a^3*Cos[c + d*x]*Sin[c + d*x])/( 128*d) + (29*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) + (29*a^3*Cos[c + d*x ]^3*Sin[c + d*x]^3)/(48*d) + (a^3*Cos[c + d*x]^3*Sin[c + d*x]^5)/(8*d))/a^ 6
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Time = 1.93 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.62
| method | result | size |
| parallelrisch | \(\frac {-24360 d x -720 \cos \left (7 d x +7 c \right )+4368 \cos \left (5 d x +5 c \right )-3920 \cos \left (3 d x +3 c \right )-38640 \cos \left (d x +c \right )+105 \sin \left (8 d x +8 c \right )-2240 \sin \left (6 d x +6 c \right )+5880 \sin \left (4 d x +4 c \right )+6720 \sin \left (2 d x +2 c \right )-38912}{107520 d \,a^{3}}\) | \(100\) |
| risch | \(-\frac {29 x}{128 a^{3}}-\frac {23 \cos \left (d x +c \right )}{64 a^{3} d}+\frac {\sin \left (8 d x +8 c \right )}{1024 d \,a^{3}}-\frac {3 \cos \left (7 d x +7 c \right )}{448 d \,a^{3}}-\frac {\sin \left (6 d x +6 c \right )}{48 d \,a^{3}}+\frac {13 \cos \left (5 d x +5 c \right )}{320 a^{3} d}+\frac {7 \sin \left (4 d x +4 c \right )}{128 d \,a^{3}}-\frac {7 \cos \left (3 d x +3 c \right )}{192 a^{3} d}+\frac {\sin \left (2 d x +2 c \right )}{16 d \,a^{3}}\) | \(141\) |
| derivativedivides | \(\frac {\frac {16 \left (-\frac {19}{420}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1024}-\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{105}+\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3072}-\frac {61 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{60}-\frac {1465 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3072}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15}-\frac {5117 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3072}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{12}+\frac {5117 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3072}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3}+\frac {1465 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3072}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}-\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3072}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{1024}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}-\frac {29 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{d \,a^{3}}\) | \(220\) |
| default | \(\frac {\frac {16 \left (-\frac {19}{420}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1024}-\frac {38 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{105}+\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3072}-\frac {61 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{60}-\frac {1465 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3072}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15}-\frac {5117 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3072}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{12}+\frac {5117 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3072}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3}+\frac {1465 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3072}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}-\frac {667 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3072}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{1024}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}-\frac {29 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{d \,a^{3}}\) | \(220\) |
Input:
int(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/107520*(-24360*d*x-720*cos(7*d*x+7*c)+4368*cos(5*d*x+5*c)-3920*cos(3*d*x +3*c)-38640*cos(d*x+c)+105*sin(8*d*x+8*c)-2240*sin(6*d*x+6*c)+5880*sin(4*d *x+4*c)+6720*sin(2*d*x+2*c)-38912)/d/a^3
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.56 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {5760 \, \cos \left (d x + c\right )^{7} - 18816 \, \cos \left (d x + c\right )^{5} + 17920 \, \cos \left (d x + c\right )^{3} + 3045 \, d x - 35 \, {\left (48 \, \cos \left (d x + c\right )^{7} - 328 \, \cos \left (d x + c\right )^{5} + 454 \, \cos \left (d x + c\right )^{3} - 87 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{13440 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/13440*(5760*cos(d*x + c)^7 - 18816*cos(d*x + c)^5 + 17920*cos(d*x + c)^ 3 + 3045*d*x - 35*(48*cos(d*x + c)^7 - 328*cos(d*x + c)^5 + 454*cos(d*x + c)^3 - 87*cos(d*x + c))*sin(d*x + c))/(a^3*d)
Timed out. \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**8*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (145) = 290\).
Time = 0.12 (sec) , antiderivative size = 499, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/6720*((3045*sin(d*x + c)/(cos(d*x + c) + 1) - 38912*sin(d*x + c)^2/(cos( d*x + c) + 1)^2 + 23345*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 109312*sin(d *x + c)^4/(cos(d*x + c) + 1)^4 - 51275*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 14336*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 179095*sin(d*x + c)^7/(cos( d*x + c) + 1)^7 - 170240*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 179095*sin( d*x + c)^9/(cos(d*x + c) + 1)^9 - 286720*sin(d*x + c)^10/(cos(d*x + c) + 1 )^10 + 51275*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 26880*sin(d*x + c)^12 /(cos(d*x + c) + 1)^12 - 23345*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 304 5*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 - 4864)/(a^3 + 8*a^3*sin(d*x + c)^ 2/(cos(d*x + c) + 1)^2 + 28*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 56*a ^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 56*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a^3*sin(d*x + c)^14/(cos(d*x + c) + 1)^1 4 + a^3*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 3045*arctan(sin(d*x + c)/ (cos(d*x + c) + 1))/a^3)/d
Time = 0.16 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {3045 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (3045 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 23345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 26880 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 51275 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 286720 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 179095 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 170240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 179095 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 14336 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 51275 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 109312 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 23345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 38912 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3045 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4864\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8} a^{3}}}{13440 \, d} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-1/13440*(3045*(d*x + c)/a^3 + 2*(3045*tan(1/2*d*x + 1/2*c)^15 + 23345*tan (1/2*d*x + 1/2*c)^13 + 26880*tan(1/2*d*x + 1/2*c)^12 - 51275*tan(1/2*d*x + 1/2*c)^11 + 286720*tan(1/2*d*x + 1/2*c)^10 - 179095*tan(1/2*d*x + 1/2*c)^ 9 + 170240*tan(1/2*d*x + 1/2*c)^8 + 179095*tan(1/2*d*x + 1/2*c)^7 - 14336* tan(1/2*d*x + 1/2*c)^6 + 51275*tan(1/2*d*x + 1/2*c)^5 + 109312*tan(1/2*d*x + 1/2*c)^4 - 23345*tan(1/2*d*x + 1/2*c)^3 + 38912*tan(1/2*d*x + 1/2*c)^2 - 3045*tan(1/2*d*x + 1/2*c) + 4864)/((tan(1/2*d*x + 1/2*c)^2 + 1)^8*a^3))/ d
Time = 34.60 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {29\,x}{128\,a^3}-\frac {\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {667\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {1465\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{192}+\frac {128\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}-\frac {5117\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {76\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {5117\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{192}-\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {1465\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}+\frac {244\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-\frac {667\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {608\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{105}-\frac {29\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {76}{105}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \] Input:
int((cos(c + d*x)^8*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)
Output:
- (29*x)/(128*a^3) - ((608*tan(c/2 + (d*x)/2)^2)/105 - (29*tan(c/2 + (d*x) /2))/64 - (667*tan(c/2 + (d*x)/2)^3)/192 + (244*tan(c/2 + (d*x)/2)^4)/15 + (1465*tan(c/2 + (d*x)/2)^5)/192 - (32*tan(c/2 + (d*x)/2)^6)/15 + (5117*ta n(c/2 + (d*x)/2)^7)/192 + (76*tan(c/2 + (d*x)/2)^8)/3 - (5117*tan(c/2 + (d *x)/2)^9)/192 + (128*tan(c/2 + (d*x)/2)^10)/3 - (1465*tan(c/2 + (d*x)/2)^1 1)/192 + 4*tan(c/2 + (d*x)/2)^12 + (667*tan(c/2 + (d*x)/2)^13)/192 + (29*t an(c/2 + (d*x)/2)^15)/64 + 76/105)/(a^3*d*(tan(c/2 + (d*x)/2)^2 + 1)^8)
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^8(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-1680 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+5760 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-6440 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+1536 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+2030 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-2432 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+3045 \cos \left (d x +c \right ) \sin \left (d x +c \right )-4864 \cos \left (d x +c \right )-3045 d x +4864}{13440 a^{3} d} \] Input:
int(cos(d*x+c)^8*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
Output:
( - 1680*cos(c + d*x)*sin(c + d*x)**7 + 5760*cos(c + d*x)*sin(c + d*x)**6 - 6440*cos(c + d*x)*sin(c + d*x)**5 + 1536*cos(c + d*x)*sin(c + d*x)**4 + 2030*cos(c + d*x)*sin(c + d*x)**3 - 2432*cos(c + d*x)*sin(c + d*x)**2 + 30 45*cos(c + d*x)*sin(c + d*x) - 4864*cos(c + d*x) - 3045*d*x + 4864)/(13440 *a**3*d)