Integrand size = 29, antiderivative size = 141 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {11 x}{128 a^2}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {2 \cos ^7(c+d x)}{7 a^2 d}+\frac {11 \cos (c+d x) \sin (c+d x)}{128 a^2 d}+\frac {11 \cos ^3(c+d x) \sin (c+d x)}{192 a^2 d}-\frac {11 \cos ^5(c+d x) \sin (c+d x)}{48 a^2 d}-\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a^2 d} \] Output:
11/128*x/a^2+2/5*cos(d*x+c)^5/a^2/d-2/7*cos(d*x+c)^7/a^2/d+11/128*cos(d*x+ c)*sin(d*x+c)/a^2/d+11/192*cos(d*x+c)^3*sin(d*x+c)/a^2/d-11/48*cos(d*x+c)^ 5*sin(d*x+c)/a^2/d-1/8*cos(d*x+c)^5*sin(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(481\) vs. \(2(141)=282\).
Time = 4.98 (sec) , antiderivative size = 481, normalized size of antiderivative = 3.41 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {9240 (15 c+2 d x) \cos \left (\frac {c}{2}\right )+10080 \cos \left (\frac {c}{2}+d x\right )+10080 \cos \left (\frac {3 c}{2}+d x\right )+1680 \cos \left (\frac {3 c}{2}+2 d x\right )-1680 \cos \left (\frac {5 c}{2}+2 d x\right )+3360 \cos \left (\frac {5 c}{2}+3 d x\right )+3360 \cos \left (\frac {7 c}{2}+3 d x\right )-2520 \cos \left (\frac {7 c}{2}+4 d x\right )+2520 \cos \left (\frac {9 c}{2}+4 d x\right )-672 \cos \left (\frac {9 c}{2}+5 d x\right )-672 \cos \left (\frac {11 c}{2}+5 d x\right )-560 \cos \left (\frac {11 c}{2}+6 d x\right )+560 \cos \left (\frac {13 c}{2}+6 d x\right )-480 \cos \left (\frac {13 c}{2}+7 d x\right )-480 \cos \left (\frac {15 c}{2}+7 d x\right )+105 \cos \left (\frac {15 c}{2}+8 d x\right )-105 \cos \left (\frac {17 c}{2}+8 d x\right )-79800 \sin \left (\frac {c}{2}\right )+138600 c \sin \left (\frac {c}{2}\right )+18480 d x \sin \left (\frac {c}{2}\right )-10080 \sin \left (\frac {c}{2}+d x\right )+10080 \sin \left (\frac {3 c}{2}+d x\right )+1680 \sin \left (\frac {3 c}{2}+2 d x\right )+1680 \sin \left (\frac {5 c}{2}+2 d x\right )-3360 \sin \left (\frac {5 c}{2}+3 d x\right )+3360 \sin \left (\frac {7 c}{2}+3 d x\right )-2520 \sin \left (\frac {7 c}{2}+4 d x\right )-2520 \sin \left (\frac {9 c}{2}+4 d x\right )+672 \sin \left (\frac {9 c}{2}+5 d x\right )-672 \sin \left (\frac {11 c}{2}+5 d x\right )-560 \sin \left (\frac {11 c}{2}+6 d x\right )-560 \sin \left (\frac {13 c}{2}+6 d x\right )+480 \sin \left (\frac {13 c}{2}+7 d x\right )-480 \sin \left (\frac {15 c}{2}+7 d x\right )+105 \sin \left (\frac {15 c}{2}+8 d x\right )+105 \sin \left (\frac {17 c}{2}+8 d x\right )}{215040 a^2 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:
Integrate[(Cos[c + d*x]^8*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(9240*(15*c + 2*d*x)*Cos[c/2] + 10080*Cos[c/2 + d*x] + 10080*Cos[(3*c)/2 + d*x] + 1680*Cos[(3*c)/2 + 2*d*x] - 1680*Cos[(5*c)/2 + 2*d*x] + 3360*Cos[( 5*c)/2 + 3*d*x] + 3360*Cos[(7*c)/2 + 3*d*x] - 2520*Cos[(7*c)/2 + 4*d*x] + 2520*Cos[(9*c)/2 + 4*d*x] - 672*Cos[(9*c)/2 + 5*d*x] - 672*Cos[(11*c)/2 + 5*d*x] - 560*Cos[(11*c)/2 + 6*d*x] + 560*Cos[(13*c)/2 + 6*d*x] - 480*Cos[( 13*c)/2 + 7*d*x] - 480*Cos[(15*c)/2 + 7*d*x] + 105*Cos[(15*c)/2 + 8*d*x] - 105*Cos[(17*c)/2 + 8*d*x] - 79800*Sin[c/2] + 138600*c*Sin[c/2] + 18480*d* x*Sin[c/2] - 10080*Sin[c/2 + d*x] + 10080*Sin[(3*c)/2 + d*x] + 1680*Sin[(3 *c)/2 + 2*d*x] + 1680*Sin[(5*c)/2 + 2*d*x] - 3360*Sin[(5*c)/2 + 3*d*x] + 3 360*Sin[(7*c)/2 + 3*d*x] - 2520*Sin[(7*c)/2 + 4*d*x] - 2520*Sin[(9*c)/2 + 4*d*x] + 672*Sin[(9*c)/2 + 5*d*x] - 672*Sin[(11*c)/2 + 5*d*x] - 560*Sin[(1 1*c)/2 + 6*d*x] - 560*Sin[(13*c)/2 + 6*d*x] + 480*Sin[(13*c)/2 + 7*d*x] - 480*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^2*d*(Cos[c/2] + Sin[c/2]))
Time = 0.63 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^8(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^8}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {\int \cos ^4(c+d x) \sin ^2(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^4 \sin (c+d x)^2 (a-a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {\int \left (a^2 \sin ^4(c+d x) \cos ^4(c+d x)-2 a^2 \sin ^3(c+d x) \cos ^4(c+d x)+a^2 \sin ^2(c+d x) \cos ^4(c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {11 a^2 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {11 a^2 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {11 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {11 a^2 x}{128}}{a^4}\) |
Input:
Int[(Cos[c + d*x]^8*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
((11*a^2*x)/128 + (2*a^2*Cos[c + d*x]^5)/(5*d) - (2*a^2*Cos[c + d*x]^7)/(7 *d) + (11*a^2*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (11*a^2*Cos[c + d*x]^3* Sin[c + d*x])/(192*d) - (11*a^2*Cos[c + d*x]^5*Sin[c + d*x])/(48*d) - (a^2 *Cos[c + d*x]^5*Sin[c + d*x]^3)/(8*d))/a^4
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 = 1.69 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71
method | result | size |
parallelrisch | \(\frac {9240 d x -480 \cos \left (7 d x +7 c \right )-672 \cos \left (5 d x +5 c \right )+3360 \cos \left (3 d x +3 c \right )+10080 \cos \left (d x +c \right )+105 \sin \left (8 d x +8 c \right )-560 \sin \left (6 d x +6 c \right )-2520 \sin \left (4 d x +4 c \right )+1680 \sin \left (2 d x +2 c \right )+12288}{107520 d \,a^{2}}\) | \(100\) |
risch | \(\frac {11 x}{128 a^{2}}+\frac {3 \cos \left (d x +c \right )}{32 a^{2} d}+\frac {\sin \left (8 d x +8 c \right )}{1024 d \,a^{2}}-\frac {\cos \left (7 d x +7 c \right )}{224 d \,a^{2}}-\frac {\sin \left (6 d x +6 c \right )}{192 d \,a^{2}}-\frac {\cos \left (5 d x +5 c \right )}{160 a^{2} d}-\frac {3 \sin \left (4 d x +4 c \right )}{128 d \,a^{2}}+\frac {\cos \left (3 d x +3 c \right )}{32 d \,a^{2}}+\frac {\sin \left (2 d x +2 c \right )}{64 d \,a^{2}}\) | \(141\) |
derivativedivides | \(\frac {\frac {8 \left (\frac {1}{35}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{512}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35}+\frac {259 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{1536}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+\frac {1103 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{1536}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}-\frac {2261 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{1536}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {2261 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1536}-\frac {1103 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{1536}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {259 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{1536}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{512}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}+\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{d \,a^{2}}\) | \(203\) |
default | \(\frac {\frac {8 \left (\frac {1}{35}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{512}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35}+\frac {259 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{1536}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+\frac {1103 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{1536}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}-\frac {2261 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{1536}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {2261 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1536}-\frac {1103 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{1536}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {259 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{1536}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{512}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}+\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{d \,a^{2}}\) | \(203\) |
Input:
int(cos(d*x+c)^8*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/107520*(9240*d*x-480*cos(7*d*x+7*c)-672*cos(5*d*x+5*c)+3360*cos(3*d*x+3* c)+10080*cos(d*x+c)+105*sin(8*d*x+8*c)-560*sin(6*d*x+6*c)-2520*sin(4*d*x+4 *c)+1680*sin(2*d*x+2*c)+12288)/d/a^2
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3840 \, \cos \left (d x + c\right )^{7} - 5376 \, \cos \left (d x + c\right )^{5} - 1155 \, d x - 35 \, {\left (48 \, \cos \left (d x + c\right )^{7} - 136 \, \cos \left (d x + c\right )^{5} + 22 \, \cos \left (d x + c\right )^{3} + 33 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{13440 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/13440*(3840*cos(d*x + c)^7 - 5376*cos(d*x + c)^5 - 1155*d*x - 35*(48*co s(d*x + c)^7 - 136*cos(d*x + c)^5 + 22*cos(d*x + c)^3 + 33*cos(d*x + c))*s in(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 3934 vs. \(2 (134) = 268\).
Time = 139.07 (sec) , antiderivative size = 3934, normalized size of antiderivative = 27.90 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**8*sin(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((1155*d*x*tan(c/2 + d*x/2)**16/(13440*a**2*d*tan(c/2 + d*x/2)**1 6 + 107520*a**2*d*tan(c/2 + d*x/2)**14 + 376320*a**2*d*tan(c/2 + d*x/2)**1 2 + 752640*a**2*d*tan(c/2 + d*x/2)**10 + 940800*a**2*d*tan(c/2 + d*x/2)**8 + 752640*a**2*d*tan(c/2 + d*x/2)**6 + 376320*a**2*d*tan(c/2 + d*x/2)**4 + 107520*a**2*d*tan(c/2 + d*x/2)**2 + 13440*a**2*d) + 9240*d*x*tan(c/2 + d* x/2)**14/(13440*a**2*d*tan(c/2 + d*x/2)**16 + 107520*a**2*d*tan(c/2 + d*x/ 2)**14 + 376320*a**2*d*tan(c/2 + d*x/2)**12 + 752640*a**2*d*tan(c/2 + d*x/ 2)**10 + 940800*a**2*d*tan(c/2 + d*x/2)**8 + 752640*a**2*d*tan(c/2 + d*x/2 )**6 + 376320*a**2*d*tan(c/2 + d*x/2)**4 + 107520*a**2*d*tan(c/2 + d*x/2)* *2 + 13440*a**2*d) + 32340*d*x*tan(c/2 + d*x/2)**12/(13440*a**2*d*tan(c/2 + d*x/2)**16 + 107520*a**2*d*tan(c/2 + d*x/2)**14 + 376320*a**2*d*tan(c/2 + d*x/2)**12 + 752640*a**2*d*tan(c/2 + d*x/2)**10 + 940800*a**2*d*tan(c/2 + d*x/2)**8 + 752640*a**2*d*tan(c/2 + d*x/2)**6 + 376320*a**2*d*tan(c/2 + d*x/2)**4 + 107520*a**2*d*tan(c/2 + d*x/2)**2 + 13440*a**2*d) + 64680*d*x* tan(c/2 + d*x/2)**10/(13440*a**2*d*tan(c/2 + d*x/2)**16 + 107520*a**2*d*ta n(c/2 + d*x/2)**14 + 376320*a**2*d*tan(c/2 + d*x/2)**12 + 752640*a**2*d*ta n(c/2 + d*x/2)**10 + 940800*a**2*d*tan(c/2 + d*x/2)**8 + 752640*a**2*d*tan (c/2 + d*x/2)**6 + 376320*a**2*d*tan(c/2 + d*x/2)**4 + 107520*a**2*d*tan(c /2 + d*x/2)**2 + 13440*a**2*d) + 80850*d*x*tan(c/2 + d*x/2)**8/(13440*a**2 *d*tan(c/2 + d*x/2)**16 + 107520*a**2*d*tan(c/2 + d*x/2)**14 + 376320*a...
Leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (127) = 254\).
Time = 0.15 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.40 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/6720*((1155*sin(d*x + c)/(cos(d*x + c) + 1) - 12288*sin(d*x + c)^2/(cos (d*x + c) + 1)^2 - 9065*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 10752*sin(d* x + c)^4/(cos(d*x + c) + 1)^4 - 38605*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 86016*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 79135*sin(d*x + c)^7/(cos(d* x + c) + 1)^7 - 53760*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 79135*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 38605*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 53760*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 9065*sin(d*x + c)^13/(cos( d*x + c) + 1)^13 - 1155*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 - 1536)/(a^2 + 8*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^2*sin(d*x + c)^4/(cos( d*x + c) + 1)^4 + 56*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^2*sin( d*x + c)^8/(cos(d*x + c) + 1)^8 + 56*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1 )^10 + 28*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 8*a^2*sin(d*x + c)^1 4/(cos(d*x + c) + 1)^14 + a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 115 5*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.16 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.45 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {1155 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 9065 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 53760 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 38605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 79135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 53760 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 79135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 86016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 38605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10752 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9065 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12288 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1536\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8} a^{2}}}{13440 \, d} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/13440*(1155*(d*x + c)/a^2 + 2*(1155*tan(1/2*d*x + 1/2*c)^15 - 9065*tan(1 /2*d*x + 1/2*c)^13 + 53760*tan(1/2*d*x + 1/2*c)^12 - 38605*tan(1/2*d*x + 1 /2*c)^11 + 79135*tan(1/2*d*x + 1/2*c)^9 + 53760*tan(1/2*d*x + 1/2*c)^8 - 7 9135*tan(1/2*d*x + 1/2*c)^7 + 86016*tan(1/2*d*x + 1/2*c)^6 + 38605*tan(1/2 *d*x + 1/2*c)^5 - 10752*tan(1/2*d*x + 1/2*c)^4 + 9065*tan(1/2*d*x + 1/2*c) ^3 + 12288*tan(1/2*d*x + 1/2*c)^2 - 1155*tan(1/2*d*x + 1/2*c) + 1536)/((ta n(1/2*d*x + 1/2*c)^2 + 1)^8*a^2))/d
Time = 34.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {11\,x}{128\,a^2}+\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}-\frac {259\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {1103\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{192}+\frac {2261\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {2261\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{192}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {1103\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {259\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {8}{35}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \] Input:
int((cos(c + d*x)^8*sin(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)
Output:
(11*x)/(128*a^2) + ((64*tan(c/2 + (d*x)/2)^2)/35 - (11*tan(c/2 + (d*x)/2)) /64 + (259*tan(c/2 + (d*x)/2)^3)/192 - (8*tan(c/2 + (d*x)/2)^4)/5 + (1103* tan(c/2 + (d*x)/2)^5)/192 + (64*tan(c/2 + (d*x)/2)^6)/5 - (2261*tan(c/2 + (d*x)/2)^7)/192 + 8*tan(c/2 + (d*x)/2)^8 + (2261*tan(c/2 + (d*x)/2)^9)/192 - (1103*tan(c/2 + (d*x)/2)^11)/192 + 8*tan(c/2 + (d*x)/2)^12 - (259*tan(c /2 + (d*x)/2)^13)/192 + (11*tan(c/2 + (d*x)/2)^15)/64 + 8/35)/(a^2*d*(tan( c/2 + (d*x)/2)^2 + 1)^8)
Time = 0.23 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^8(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-1680 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+3840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-6144 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+3710 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+768 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-1155 \cos \left (d x +c \right ) \sin \left (d x +c \right )+1536 \cos \left (d x +c \right )+1155 d x -1536}{13440 a^{2} d} \] Input:
int(cos(d*x+c)^8*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - 1680*cos(c + d*x)*sin(c + d*x)**7 + 3840*cos(c + d*x)*sin(c + d*x)**6 + 280*cos(c + d*x)*sin(c + d*x)**5 - 6144*cos(c + d*x)*sin(c + d*x)**4 + 3 710*cos(c + d*x)*sin(c + d*x)**3 + 768*cos(c + d*x)*sin(c + d*x)**2 - 1155 *cos(c + d*x)*sin(c + d*x) + 1536*cos(c + d*x) + 1155*d*x - 1536)/(13440*a **2*d)