Integrand size = 29, antiderivative size = 185 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {9 x}{256 a^2}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {4 \cos ^7(c+d x)}{7 a^2 d}+\frac {2 \cos ^9(c+d x)}{9 a^2 d}+\frac {9 \cos (c+d x) \sin (c+d x)}{256 a^2 d}+\frac {3 \cos ^3(c+d x) \sin (c+d x)}{128 a^2 d}-\frac {3 \cos ^5(c+d x) \sin (c+d x)}{32 a^2 d}-\frac {3 \cos ^5(c+d x) \sin ^3(c+d x)}{16 a^2 d}-\frac {\cos ^5(c+d x) \sin ^5(c+d x)}{10 a^2 d} \] Output:
9/256*x/a^2+2/5*cos(d*x+c)^5/a^2/d-4/7*cos(d*x+c)^7/a^2/d+2/9*cos(d*x+c)^9 /a^2/d+9/256*cos(d*x+c)*sin(d*x+c)/a^2/d+3/128*cos(d*x+c)^3*sin(d*x+c)/a^2 /d-3/32*cos(d*x+c)^5*sin(d*x+c)/a^2/d-3/16*cos(d*x+c)^5*sin(d*x+c)^3/a^2/d -1/10*cos(d*x+c)^5*sin(d*x+c)^5/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(585\) vs. \(2(185)=370\).
Time = 10.24 (sec) , antiderivative size = 585, normalized size of antiderivative = 3.16 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-2520 (187 c-18 d x) \cos \left (\frac {c}{2}\right )+30240 \cos \left (\frac {c}{2}+d x\right )+30240 \cos \left (\frac {3 c}{2}+d x\right )-1260 \cos \left (\frac {3 c}{2}+2 d x\right )+1260 \cos \left (\frac {5 c}{2}+2 d x\right )+6720 \cos \left (\frac {5 c}{2}+3 d x\right )+6720 \cos \left (\frac {7 c}{2}+3 d x\right )-7560 \cos \left (\frac {7 c}{2}+4 d x\right )+7560 \cos \left (\frac {9 c}{2}+4 d x\right )-4032 \cos \left (\frac {9 c}{2}+5 d x\right )-4032 \cos \left (\frac {11 c}{2}+5 d x\right )+630 \cos \left (\frac {11 c}{2}+6 d x\right )-630 \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 )+945 \cos \left (\frac {15 c}{2}+8 d x\right )-945 \cos \left (\frac {17 c}{2}+8 d x\right )+560 \cos \left (\frac {17 c}{2}+9 d x\right )+560 \cos \left (\frac {19 c}{2}+9 d x\right )-126 \cos \left (\frac {19 c}{2}+10 d x\right )+126 \cos \left (\frac {21 c}{2}+10 d x\right )+327180 \sin \left (\frac {c}{2}\right )-471240 c \sin \left (\frac {c}{2}\right )+45360 d x \sin \left (\frac {c}{2}\right )-30240 \sin \left (\frac {c}{2}+d x\right )+30240 \sin \left (\frac {3 c}{2}+d x\right )-1260 \sin \left (\frac {3 c}{2}+2 d x\right )-1260 \sin \left (\frac {5 c}{2}+2 d x\right )-6720 \sin \left (\frac {5 c}{2}+3 d x\right )+6720 \sin \left (\frac {7 c}{2}+3 d x\right )-7560 \sin \left (\frac {7 c}{2}+4 d x\right )-7560 \sin \left (\frac {9 c}{2}+4 d x\right )+4032 \sin \left (\frac {9 c}{2}+5 d x\right )-4032 \sin \left (\frac {11 c}{2}+5 d x\right )+630 \sin \left (\frac {11 c}{2}+6 d x\right )+630 \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 )+945 \sin \left (\frac {15 c}{2}+8 d x\right )+945 \sin \left (\frac {17 c}{2}+8 d x\right )-560 \sin \left (\frac {17 c}{2}+9 d x\right )+560 \sin \left (\frac {19 c}{2}+9 d x\right )-126 \sin \left (\frac {19 c}{2}+10 d x\right )-126 \sin \left (\frac {21 c}{2}+10 d x\right )}{1290240 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]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
(-2520*(187*c - 18*d*x)*Cos[c/2] + 30240*Cos[c/2 + d*x] + 30240*Cos[(3*c)/ 2 + d*x] - 1260*Cos[(3*c)/2 + 2*d*x] + 1260*Cos[(5*c)/2 + 2*d*x] + 6720*Co s[(5*c)/2 + 3*d*x] + 6720*Cos[(7*c)/2 + 3*d*x] - 7560*Cos[(7*c)/2 + 4*d*x] + 7560*Cos[(9*c)/2 + 4*d*x] - 4032*Cos[(9*c)/2 + 5*d*x] - 4032*Cos[(11*c) /2 + 5*d*x] + 630*Cos[(11*c)/2 + 6*d*x] - 630*Cos[(13*c)/2 + 6*d*x] - 720* Cos[(13*c)/2 + 7*d*x] - 720*Cos[(15*c)/2 + 7*d*x] + 945*Cos[(15*c)/2 + 8*d *x] - 945*Cos[(17*c)/2 + 8*d*x] + 560*Cos[(17*c)/2 + 9*d*x] + 560*Cos[(19* c)/2 + 9*d*x] - 126*Cos[(19*c)/2 + 10*d*x] + 126*Cos[(21*c)/2 + 10*d*x] + 327180*Sin[c/2] - 471240*c*Sin[c/2] + 45360*d*x*Sin[c/2] - 30240*Sin[c/2 + d*x] + 30240*Sin[(3*c)/2 + d*x] - 1260*Sin[(3*c)/2 + 2*d*x] - 1260*Sin[(5 *c)/2 + 2*d*x] - 6720*Sin[(5*c)/2 + 3*d*x] + 6720*Sin[(7*c)/2 + 3*d*x] - 7 560*Sin[(7*c)/2 + 4*d*x] - 7560*Sin[(9*c)/2 + 4*d*x] + 4032*Sin[(9*c)/2 + 5*d*x] - 4032*Sin[(11*c)/2 + 5*d*x] + 630*Sin[(11*c)/2 + 6*d*x] + 630*Sin[ (13*c)/2 + 6*d*x] + 720*Sin[(13*c)/2 + 7*d*x] - 720*Sin[(15*c)/2 + 7*d*x] + 945*Sin[(15*c)/2 + 8*d*x] + 945*Sin[(17*c)/2 + 8*d*x] - 560*Sin[(17*c)/2 + 9*d*x] + 560*Sin[(19*c)/2 + 9*d*x] - 126*Sin[(19*c)/2 + 10*d*x] - 126*S in[(21*c)/2 + 10*d*x])/(1290240*a^2*d*(Cos[c/2] + Sin[c/2]))
Time = 0.70 (sec) , antiderivative size = 189, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x) \cos ^8(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4 \cos (c+d x)^8}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {\int \cos ^4(c+d x) \sin ^4(c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^4 \sin (c+d x)^4 (a-a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {\int \left (a^2 \cos ^4(c+d x) \sin ^6(c+d x)-2 a^2 \cos ^4(c+d x) \sin ^5(c+d x)+a^2 \cos ^4(c+d x) \sin ^4(c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 a^2 \cos ^9(c+d x)}{9 d}-\frac {4 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \sin ^5(c+d x) \cos ^5(c+d x)}{10 d}-\frac {3 a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{16 d}-\frac {3 a^2 \sin (c+d x) \cos ^5(c+d x)}{32 d}+\frac {3 a^2 \sin (c+d x) \cos ^3(c+d x)}{128 d}+\frac {9 a^2 \sin (c+d x) \cos (c+d x)}{256 d}+\frac {9 a^2 x}{256}}{a^4}\) |
Input:
Int[(Cos[c + d*x]^8*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
((9*a^2*x)/256 + (2*a^2*Cos[c + d*x]^5)/(5*d) - (4*a^2*Cos[c + d*x]^7)/(7* d) + (2*a^2*Cos[c + d*x]^9)/(9*d) + (9*a^2*Cos[c + d*x]*Sin[c + d*x])/(256 *d) + (3*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(128*d) - (3*a^2*Cos[c + d*x]^5* Sin[c + d*x])/(32*d) - (3*a^2*Cos[c + d*x]^5*Sin[c + d*x]^3)/(16*d) - (a^2 *Cos[c + d*x]^5*Sin[c + d*x]^5)/(10*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 = 1.99 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.66
method | result | size |
parallelrisch | \(\frac {32768+630 \sin \left (6 d x +6 c \right )-1260 \sin \left (2 d x +2 c \right )+6720 \cos \left (3 d x +3 c \right )-7560 \sin \left (4 d x +4 c \right )+945 \sin \left (8 d x +8 c \right )+22680 d x +30240 \cos \left (d x +c \right )-4032 \cos \left (5 d x +5 c \right )-720 \cos \left (7 d x +7 c \right )+560 \cos \left (9 d x +9 c \right )-126 \sin \left (10 d x +10 c \right )}{645120 d \,a^{2}}\) | \(122\) |
risch | \(\frac {9 x}{256 a^{2}}+\frac {3 \cos \left (d x +c \right )}{64 a^{2} d}-\frac {\sin \left (10 d x +10 c \right )}{5120 d \,a^{2}}+\frac {\cos \left (9 d x +9 c \right )}{1152 d \,a^{2}}+\frac {3 \sin \left (8 d x +8 c \right )}{2048 d \,a^{2}}-\frac {\cos \left (7 d x +7 c \right )}{896 d \,a^{2}}+\frac {\sin \left (6 d x +6 c \right )}{1024 d \,a^{2}}-\frac {\cos \left (5 d x +5 c \right )}{160 a^{2} d}-\frac {3 \sin \left (4 d x +4 c \right )}{256 d \,a^{2}}+\frac {\cos \left (3 d x +3 c \right )}{96 d \,a^{2}}-\frac {\sin \left (2 d x +2 c \right )}{512 d \,a^{2}}\) | \(175\) |
derivativedivides | \(\frac {\frac {32 \left (\frac {1}{315}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4096}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{63}-\frac {87 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4096}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{7}+\frac {553 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5120}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{7}+\frac {491 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{1024}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {2555 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2048}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{5}+\frac {2555 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2048}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3}-\frac {491 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{1024}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{3}-\frac {553 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{5120}+\frac {87 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{17}}{4096}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{19}}{4096}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{10}}+\frac {9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128}}{d \,a^{2}}\) | \(257\) |
default | \(\frac {\frac {32 \left (\frac {1}{315}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4096}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{63}-\frac {87 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4096}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{7}+\frac {553 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5120}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{7}+\frac {491 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{1024}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {2555 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2048}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{5}+\frac {2555 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2048}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3}-\frac {491 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{1024}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{3}-\frac {553 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{5120}+\frac {87 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{17}}{4096}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{19}}{4096}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{10}}+\frac {9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128}}{d \,a^{2}}\) | \(257\) |
Input:
int(cos(d*x+c)^8*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/645120*(32768+630*sin(6*d*x+6*c)-1260*sin(2*d*x+2*c)+6720*cos(3*d*x+3*c) -7560*sin(4*d*x+4*c)+945*sin(8*d*x+8*c)+22680*d*x+30240*cos(d*x+c)-4032*co s(5*d*x+5*c)-720*cos(7*d*x+7*c)+560*cos(9*d*x+9*c)-126*sin(10*d*x+10*c))/d /a^2
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.54 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {17920 \, \cos \left (d x + c\right )^{9} - 46080 \, \cos \left (d x + c\right )^{7} + 32256 \, \cos \left (d x + c\right )^{5} + 2835 \, d x - 63 \, {\left (128 \, \cos \left (d x + c\right )^{9} - 496 \, \cos \left (d x + c\right )^{7} + 488 \, \cos \left (d x + c\right )^{5} - 30 \, \cos \left (d x + c\right )^{3} - 45 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80640 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/80640*(17920*cos(d*x + c)^9 - 46080*cos(d*x + c)^7 + 32256*cos(d*x + c)^ 5 + 2835*d*x - 63*(128*cos(d*x + c)^9 - 496*cos(d*x + c)^7 + 488*cos(d*x + c)^5 - 30*cos(d*x + c)^3 - 45*cos(d*x + c))*sin(d*x + c))/(a^2*d)
Timed out. \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**8*sin(d*x+c)**4/(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 605 vs. \(2 (167) = 334\).
Time = 0.15 (sec) , antiderivative size = 605, normalized size of antiderivative = 3.27 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/40320*((2835*sin(d*x + c)/(cos(d*x + c) + 1) - 40960*sin(d*x + c)^2/(co s(d*x + c) + 1)^2 + 27405*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 184320*sin (d*x + c)^4/(cos(d*x + c) + 1)^4 - 139356*sin(d*x + c)^5/(cos(d*x + c) + 1 )^5 + 368640*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 618660*sin(d*x + c)^7/( cos(d*x + c) + 1)^7 - 1290240*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 160965 0*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 516096*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 1609650*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 430080*sin(d* x + c)^12/(cos(d*x + c) + 1)^12 + 618660*sin(d*x + c)^13/(cos(d*x + c) + 1 )^13 - 860160*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 139356*sin(d*x + c)^ 15/(cos(d*x + c) + 1)^15 - 27405*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 - 2 835*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 - 4096)/(a^2 + 10*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 210*a^2*sin(d*x + c)^8/(cos(d *x + c) + 1)^8 + 252*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 210*a^2*s in(d*x + c)^12/(cos(d*x + c) + 1)^12 + 120*a^2*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 45*a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 + 10*a^2*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 + a^2*sin(d*x + c)^20/(cos(d*x + c) + 1)^20 ) - 2835*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.14 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2835 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (2835 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{19} + 27405 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} - 139356 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 860160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} - 618660 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 430080 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 1609650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 516096 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 1609650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1290240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 618660 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 368640 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 139356 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 184320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 27405 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2835 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4096\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{10} a^{2}}}{80640 \, d} \] Input:
integrate(cos(d*x+c)^8*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/80640*(2835*(d*x + c)/a^2 + 2*(2835*tan(1/2*d*x + 1/2*c)^19 + 27405*tan( 1/2*d*x + 1/2*c)^17 - 139356*tan(1/2*d*x + 1/2*c)^15 + 860160*tan(1/2*d*x + 1/2*c)^14 - 618660*tan(1/2*d*x + 1/2*c)^13 - 430080*tan(1/2*d*x + 1/2*c) ^12 + 1609650*tan(1/2*d*x + 1/2*c)^11 + 516096*tan(1/2*d*x + 1/2*c)^10 - 1 609650*tan(1/2*d*x + 1/2*c)^9 + 1290240*tan(1/2*d*x + 1/2*c)^8 + 618660*ta n(1/2*d*x + 1/2*c)^7 - 368640*tan(1/2*d*x + 1/2*c)^6 + 139356*tan(1/2*d*x + 1/2*c)^5 + 184320*tan(1/2*d*x + 1/2*c)^4 - 27405*tan(1/2*d*x + 1/2*c)^3 + 40960*tan(1/2*d*x + 1/2*c)^2 - 2835*tan(1/2*d*x + 1/2*c) + 4096)/((tan(1 /2*d*x + 1/2*c)^2 + 1)^10*a^2))/d
Time = 35.33 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {9\,x}{256\,a^2}+\frac {\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{19}}{128}+\frac {87\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}-\frac {553\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{160}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{3}-\frac {491\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{32}-\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3}+\frac {2555\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{64}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{5}-\frac {2555\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {491\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{32}-\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{7}+\frac {553\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160}+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}-\frac {87\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{128}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{63}-\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}+\frac {32}{315}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^{10}} \] Input:
int((cos(c + d*x)^8*sin(c + d*x)^4)/(a + a*sin(c + d*x))^2,x)
Output:
(9*x)/(256*a^2) + ((64*tan(c/2 + (d*x)/2)^2)/63 - (9*tan(c/2 + (d*x)/2))/1 28 - (87*tan(c/2 + (d*x)/2)^3)/128 + (32*tan(c/2 + (d*x)/2)^4)/7 + (553*ta n(c/2 + (d*x)/2)^5)/160 - (64*tan(c/2 + (d*x)/2)^6)/7 + (491*tan(c/2 + (d* x)/2)^7)/32 + 32*tan(c/2 + (d*x)/2)^8 - (2555*tan(c/2 + (d*x)/2)^9)/64 + ( 64*tan(c/2 + (d*x)/2)^10)/5 + (2555*tan(c/2 + (d*x)/2)^11)/64 - (32*tan(c/ 2 + (d*x)/2)^12)/3 - (491*tan(c/2 + (d*x)/2)^13)/32 + (64*tan(c/2 + (d*x)/ 2)^14)/3 - (553*tan(c/2 + (d*x)/2)^15)/160 + (87*tan(c/2 + (d*x)/2)^17)/12 8 + (9*tan(c/2 + (d*x)/2)^19)/128 + 32/315)/(a^2*d*(tan(c/2 + (d*x)/2)^2 + 1)^10)
Time = 0.58 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^8(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-8064 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9}+17920 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}+1008 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}-25600 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+14616 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+1536 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-1890 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+2048 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-2835 \cos \left (d x +c \right ) \sin \left (d x +c \right )+4096 \cos \left (d x +c \right )+2835 d x -4096}{80640 a^{2} d} \] Input:
int(cos(d*x+c)^8*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x)
Output:
( - 8064*cos(c + d*x)*sin(c + d*x)**9 + 17920*cos(c + d*x)*sin(c + d*x)**8 + 1008*cos(c + d*x)*sin(c + d*x)**7 - 25600*cos(c + d*x)*sin(c + d*x)**6 + 14616*cos(c + d*x)*sin(c + d*x)**5 + 1536*cos(c + d*x)*sin(c + d*x)**4 - 1890*cos(c + d*x)*sin(c + d*x)**3 + 2048*cos(c + d*x)*sin(c + d*x)**2 - 2 835*cos(c + d*x)*sin(c + d*x) + 4096*cos(c + d*x) + 2835*d*x - 4096)/(8064 0*a**2*d)