\(\int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [726]

Optimal result
Mathematica [B] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 124 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {x}{8 a^2}-\frac {2 \cos ^7(c+d x)}{35 a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{8 a^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{12 a^2 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{15 a^2 d}-\frac {\cos ^9(c+d x)}{5 d (a+a \sin (c+d x))^2} \] Output:

-1/8*x/a^2-2/35*cos(d*x+c)^7/a^2/d-1/8*cos(d*x+c)*sin(d*x+c)/a^2/d-1/12*co 
s(d*x+c)^3*sin(d*x+c)/a^2/d-1/15*cos(d*x+c)^5*sin(d*x+c)/a^2/d-1/5*cos(d*x 
+c)^9/d/(a+a*sin(d*x+c))^2
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(418\) vs. \(2(124)=248\).

Time = 5.99 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.37 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {70 (7+24 d x) \cos \left (\frac {c}{2}\right )+1155 \cos \left (\frac {c}{2}+d x\right )+1155 \cos \left (\frac {3 c}{2}+d x\right )+210 \cos \left (\frac {3 c}{2}+2 d x\right )-210 \cos \left (\frac {5 c}{2}+2 d x\right )+525 \cos \left (\frac {5 c}{2}+3 d x\right )+525 \cos \left (\frac {7 c}{2}+3 d x\right )-210 \cos \left (\frac {7 c}{2}+4 d x\right )+210 \cos \left (\frac {9 c}{2}+4 d x\right )+63 \cos \left (\frac {9 c}{2}+5 d x\right )+63 \cos \left (\frac {11 c}{2}+5 d x\right )-70 \cos \left (\frac {11 c}{2}+6 d x\right )+70 \cos \left (\frac {13 c}{2}+6 d x\right )-15 \cos \left (\frac {13 c}{2}+7 d x\right )-15 \cos \left (\frac {15 c}{2}+7 d x\right )-490 \sin \left (\frac {c}{2}\right )+1680 d x \sin \left (\frac {c}{2}\right )-1155 \sin \left (\frac {c}{2}+d x\right )+1155 \sin \left (\frac {3 c}{2}+d x\right )+210 \sin \left (\frac {3 c}{2}+2 d x\right )+210 \sin \left (\frac {5 c}{2}+2 d x\right )-525 \sin \left (\frac {5 c}{2}+3 d x\right )+525 \sin \left (\frac {7 c}{2}+3 d x\right )-210 \sin \left (\frac {7 c}{2}+4 d x\right )-210 \sin \left (\frac {9 c}{2}+4 d x\right )-63 \sin \left (\frac {9 c}{2}+5 d x\right )+63 \sin \left (\frac {11 c}{2}+5 d x\right )-70 \sin \left (\frac {11 c}{2}+6 d x\right )-70 \sin \left (\frac {13 c}{2}+6 d x\right )+15 \sin \left (\frac {13 c}{2}+7 d x\right )-15 \sin \left (\frac {15 c}{2}+7 d x\right )}{13440 a^2 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:

Integrate[(Cos[c + d*x]^8*Sin[c + d*x])/(a + a*Sin[c + d*x])^2,x]
 

Output:

-1/13440*(70*(7 + 24*d*x)*Cos[c/2] + 1155*Cos[c/2 + d*x] + 1155*Cos[(3*c)/ 
2 + d*x] + 210*Cos[(3*c)/2 + 2*d*x] - 210*Cos[(5*c)/2 + 2*d*x] + 525*Cos[( 
5*c)/2 + 3*d*x] + 525*Cos[(7*c)/2 + 3*d*x] - 210*Cos[(7*c)/2 + 4*d*x] + 21 
0*Cos[(9*c)/2 + 4*d*x] + 63*Cos[(9*c)/2 + 5*d*x] + 63*Cos[(11*c)/2 + 5*d*x 
] - 70*Cos[(11*c)/2 + 6*d*x] + 70*Cos[(13*c)/2 + 6*d*x] - 15*Cos[(13*c)/2 
+ 7*d*x] - 15*Cos[(15*c)/2 + 7*d*x] - 490*Sin[c/2] + 1680*d*x*Sin[c/2] - 1 
155*Sin[c/2 + d*x] + 1155*Sin[(3*c)/2 + d*x] + 210*Sin[(3*c)/2 + 2*d*x] + 
210*Sin[(5*c)/2 + 2*d*x] - 525*Sin[(5*c)/2 + 3*d*x] + 525*Sin[(7*c)/2 + 3* 
d*x] - 210*Sin[(7*c)/2 + 4*d*x] - 210*Sin[(9*c)/2 + 4*d*x] - 63*Sin[(9*c)/ 
2 + 5*d*x] + 63*Sin[(11*c)/2 + 5*d*x] - 70*Sin[(11*c)/2 + 6*d*x] - 70*Sin[ 
(13*c)/2 + 6*d*x] + 15*Sin[(13*c)/2 + 7*d*x] - 15*Sin[(15*c)/2 + 7*d*x])/( 
a^2*d*(Cos[c/2] + Sin[c/2]))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3338, 3042, 3161, 3042, 3115, 3042, 3115, 3042, 3115, 24}

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.

\(\displaystyle \int \frac {\sin (c+d x) \cos ^8(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^8}{(a \sin (c+d x)+a)^2}dx\)

\(\Big \downarrow \) 3338

\(\displaystyle -\frac {2 \int \frac {\cos ^8(c+d x)}{\sin (c+d x) a+a}dx}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 \int \frac {\cos (c+d x)^8}{\sin (c+d x) a+a}dx}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3161

\(\displaystyle -\frac {2 \left (\frac {\int \cos ^6(c+d x)dx}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {2 \left (\frac {\frac {5}{6} \int \cos ^4(c+d x)dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 \left (\frac {\frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {2 \left (\frac {\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 \left (\frac {\frac {5}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {2 \left (\frac {\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a}+\frac {\cos ^7(c+d x)}{7 a d}\right )}{5 a}-\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {\cos ^9(c+d x)}{5 d (a \sin (c+d x)+a)^2}-\frac {2 \left (\frac {\cos ^7(c+d x)}{7 a d}+\frac {\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a}\right )}{5 a}\)

Input:

Int[(Cos[c + d*x]^8*Sin[c + d*x])/(a + a*Sin[c + d*x])^2,x]
 

Output:

-1/5*Cos[c + d*x]^9/(d*(a + a*Sin[c + d*x])^2) - (2*(Cos[c + d*x]^7/(7*a*d 
) + ((Cos[c + d*x]^5*Sin[c + d*x])/(6*d) + (5*((Cos[c + d*x]^3*Sin[c + d*x 
])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6)/a))/(5*a)
 

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3161
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 

rule 3338
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - 
 a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) 
)), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1))   Int[(g*Cos[e 
+ f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 
]) && NeQ[2*m + p + 1, 0]
 
Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.72

method result size
parallelrisch \(\frac {-840 d x +70 \sin \left (6 d x +6 c \right )+210 \sin \left (4 d x +4 c \right )-210 \sin \left (2 d x +2 c \right )+15 \cos \left (7 d x +7 c \right )-63 \cos \left (5 d x +5 c \right )-525 \cos \left (3 d x +3 c \right )-1155 \cos \left (d x +c \right )-1728}{6720 d \,a^{2}}\) \(89\)
risch \(-\frac {x}{8 a^{2}}-\frac {11 \cos \left (d x +c \right )}{64 a^{2} d}+\frac {\cos \left (7 d x +7 c \right )}{448 d \,a^{2}}+\frac {\sin \left (6 d x +6 c \right )}{96 d \,a^{2}}-\frac {3 \cos \left (5 d x +5 c \right )}{320 a^{2} d}+\frac {\sin \left (4 d x +4 c \right )}{32 d \,a^{2}}-\frac {5 \cos \left (3 d x +3 c \right )}{64 d \,a^{2}}-\frac {\sin \left (2 d x +2 c \right )}{32 d \,a^{2}}\) \(124\)
derivativedivides \(\frac {\frac {4 \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{16}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{48}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{48}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{10}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {9}{70}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{2}}\) \(194\)
default \(\frac {\frac {4 \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{16}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{48}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{48}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{10}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {9}{70}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{2}}\) \(194\)

Input:

int(cos(d*x+c)^8*sin(d*x+c)/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/6720*(-840*d*x+70*sin(6*d*x+6*c)+210*sin(4*d*x+4*c)-210*sin(2*d*x+2*c)+1 
5*cos(7*d*x+7*c)-63*cos(5*d*x+5*c)-525*cos(3*d*x+3*c)-1155*cos(d*x+c)-1728 
)/d/a^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {120 \, \cos \left (d x + c\right )^{7} - 336 \, \cos \left (d x + c\right )^{5} - 105 \, d x + 35 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, a^{2} d} \] Input:

integrate(cos(d*x+c)^8*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

1/840*(120*cos(d*x + c)^7 - 336*cos(d*x + c)^5 - 105*d*x + 35*(8*cos(d*x + 
 c)^5 - 2*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x + c))/(a^2*d)
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3196 vs. \(2 (112) = 224\).

Time = 93.66 (sec) , antiderivative size = 3196, normalized size of antiderivative = 25.77 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)**8*sin(d*x+c)/(a+a*sin(d*x+c))**2,x)
 

Output:

Piecewise((-105*d*x*tan(c/2 + d*x/2)**14/(840*a**2*d*tan(c/2 + d*x/2)**14 
+ 5880*a**2*d*tan(c/2 + d*x/2)**12 + 17640*a**2*d*tan(c/2 + d*x/2)**10 + 2 
9400*a**2*d*tan(c/2 + d*x/2)**8 + 29400*a**2*d*tan(c/2 + d*x/2)**6 + 17640 
*a**2*d*tan(c/2 + d*x/2)**4 + 5880*a**2*d*tan(c/2 + d*x/2)**2 + 840*a**2*d 
) - 735*d*x*tan(c/2 + d*x/2)**12/(840*a**2*d*tan(c/2 + d*x/2)**14 + 5880*a 
**2*d*tan(c/2 + d*x/2)**12 + 17640*a**2*d*tan(c/2 + d*x/2)**10 + 29400*a** 
2*d*tan(c/2 + d*x/2)**8 + 29400*a**2*d*tan(c/2 + d*x/2)**6 + 17640*a**2*d* 
tan(c/2 + d*x/2)**4 + 5880*a**2*d*tan(c/2 + d*x/2)**2 + 840*a**2*d) - 2205 
*d*x*tan(c/2 + d*x/2)**10/(840*a**2*d*tan(c/2 + d*x/2)**14 + 5880*a**2*d*t 
an(c/2 + d*x/2)**12 + 17640*a**2*d*tan(c/2 + d*x/2)**10 + 29400*a**2*d*tan 
(c/2 + d*x/2)**8 + 29400*a**2*d*tan(c/2 + d*x/2)**6 + 17640*a**2*d*tan(c/2 
 + d*x/2)**4 + 5880*a**2*d*tan(c/2 + d*x/2)**2 + 840*a**2*d) - 3675*d*x*ta 
n(c/2 + d*x/2)**8/(840*a**2*d*tan(c/2 + d*x/2)**14 + 5880*a**2*d*tan(c/2 + 
 d*x/2)**12 + 17640*a**2*d*tan(c/2 + d*x/2)**10 + 29400*a**2*d*tan(c/2 + d 
*x/2)**8 + 29400*a**2*d*tan(c/2 + d*x/2)**6 + 17640*a**2*d*tan(c/2 + d*x/2 
)**4 + 5880*a**2*d*tan(c/2 + d*x/2)**2 + 840*a**2*d) - 3675*d*x*tan(c/2 + 
d*x/2)**6/(840*a**2*d*tan(c/2 + d*x/2)**14 + 5880*a**2*d*tan(c/2 + d*x/2)* 
*12 + 17640*a**2*d*tan(c/2 + d*x/2)**10 + 29400*a**2*d*tan(c/2 + d*x/2)**8 
 + 29400*a**2*d*tan(c/2 + d*x/2)**6 + 17640*a**2*d*tan(c/2 + d*x/2)**4 + 5 
880*a**2*d*tan(c/2 + d*x/2)**2 + 840*a**2*d) - 2205*d*x*tan(c/2 + d*x/2...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 436 vs. \(2 (112) = 224\).

Time = 0.11 (sec) , antiderivative size = 436, normalized size of antiderivative = 3.52 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {672 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1540 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {1176 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1085 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {6720 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {840 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {1085 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {3360 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {1540 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {840 \, \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {105 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - 216}{a^{2} + \frac {7 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {21 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {35 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {35 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {21 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {7 \, a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {a^{2} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}}} - \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{420 \, d} \] Input:

integrate(cos(d*x+c)^8*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima" 
)
 

Output:

1/420*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 672*sin(d*x + c)^2/(cos(d*x 
+ c) + 1)^2 - 1540*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1176*sin(d*x + c) 
^4/(cos(d*x + c) + 1)^4 + 1085*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 6720* 
sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 840*sin(d*x + c)^8/(cos(d*x + c) + 1 
)^8 - 1085*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 3360*sin(d*x + c)^10/(cos 
(d*x + c) + 1)^10 + 1540*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 840*sin(d 
*x + c)^12/(cos(d*x + c) + 1)^12 - 105*sin(d*x + c)^13/(cos(d*x + c) + 1)^ 
13 - 216)/(a^2 + 7*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 21*a^2*sin(d* 
x + c)^4/(cos(d*x + c) + 1)^4 + 35*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 
 + 35*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 21*a^2*sin(d*x + c)^10/(co 
s(d*x + c) + 1)^10 + 7*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + a^2*sin 
(d*x + c)^14/(cos(d*x + c) + 1)^14) - 105*arctan(sin(d*x + c)/(cos(d*x + c 
) + 1))/a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.55 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {105 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 1540 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 1085 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 6720 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1085 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1176 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1540 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 672 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7} a^{2}}}{840 \, d} \] Input:

integrate(cos(d*x+c)^8*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/840*(105*(d*x + c)/a^2 + 2*(105*tan(1/2*d*x + 1/2*c)^13 + 840*tan(1/2*d 
*x + 1/2*c)^12 - 1540*tan(1/2*d*x + 1/2*c)^11 + 3360*tan(1/2*d*x + 1/2*c)^ 
10 + 1085*tan(1/2*d*x + 1/2*c)^9 + 840*tan(1/2*d*x + 1/2*c)^8 + 6720*tan(1 
/2*d*x + 1/2*c)^6 - 1085*tan(1/2*d*x + 1/2*c)^5 + 1176*tan(1/2*d*x + 1/2*c 
)^4 + 1540*tan(1/2*d*x + 1/2*c)^3 + 672*tan(1/2*d*x + 1/2*c)^2 - 105*tan(1 
/2*d*x + 1/2*c) + 216)/((tan(1/2*d*x + 1/2*c)^2 + 1)^7*a^2))/d
 

Mupad [B] (verification not implemented)

Time = 35.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {x}{8\,a^2}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {18}{35}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \] Input:

int((cos(c + d*x)^8*sin(c + d*x))/(a + a*sin(c + d*x))^2,x)
 

Output:

- x/(8*a^2) - ((8*tan(c/2 + (d*x)/2)^2)/5 - tan(c/2 + (d*x)/2)/4 + (11*tan 
(c/2 + (d*x)/2)^3)/3 + (14*tan(c/2 + (d*x)/2)^4)/5 - (31*tan(c/2 + (d*x)/2 
)^5)/12 + 16*tan(c/2 + (d*x)/2)^6 + 2*tan(c/2 + (d*x)/2)^8 + (31*tan(c/2 + 
 (d*x)/2)^9)/12 + 8*tan(c/2 + (d*x)/2)^10 - (11*tan(c/2 + (d*x)/2)^11)/3 + 
 2*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^13/4 + 18/35)/(a^2*d*(tan(c/ 
2 + (d*x)/2)^2 + 1)^7)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^8(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-490 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+312 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+105 \cos \left (d x +c \right ) \sin \left (d x +c \right )-216 \cos \left (d x +c \right )-105 d x +216}{840 a^{2} d} \] Input:

int(cos(d*x+c)^8*sin(d*x+c)/(a+a*sin(d*x+c))^2,x)
 

Output:

( - 120*cos(c + d*x)*sin(c + d*x)**6 + 280*cos(c + d*x)*sin(c + d*x)**5 + 
24*cos(c + d*x)*sin(c + d*x)**4 - 490*cos(c + d*x)*sin(c + d*x)**3 + 312*c 
os(c + d*x)*sin(c + d*x)**2 + 105*cos(c + d*x)*sin(c + d*x) - 216*cos(c + 
d*x) - 105*d*x + 216)/(840*a**2*d)