\(\int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [625]

Optimal result

Integrand size = 29, antiderivative size = 141 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 x}{128 a}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {\cos ^7(c+d x)}{7 a d}-\frac {3 \cos (c+d x) \sin (c+d x)}{128 a d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{64 a d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^5(c+d x) \sin ^3(c+d x)}{8 a d} \] Output:

-3/128*x/a-1/5*cos(d*x+c)^5/a/d+1/7*cos(d*x+c)^7/a/d-3/128*cos(d*x+c)*sin( 
d*x+c)/a/d-1/64*cos(d*x+c)^3*sin(d*x+c)/a/d+1/16*cos(d*x+c)^5*sin(d*x+c)/a 
/d+1/8*cos(d*x+c)^5*sin(d*x+c)^3/a/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(375\) vs. \(2(141)=282\).

Time = 10.26 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.66 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {1680 (c-d x) \cos \left (\frac {c}{2}\right )-1680 \cos \left (\frac {c}{2}+d x\right )-1680 \cos \left (\frac {3 c}{2}+d x\right )-560 \cos \left (\frac {5 c}{2}+3 d x\right )-560 \cos \left (\frac {7 c}{2}+3 d x\right )+280 \cos \left (\frac {7 c}{2}+4 d x\right )-280 \cos \left (\frac {9 c}{2}+4 d x\right )+112 \cos \left (\frac {9 c}{2}+5 d x\right )+112 \cos \left (\frac {11 c}{2}+5 d x\right )+80 \cos \left (\frac {13 c}{2}+7 d x\right )+80 \cos \left (\frac {15 c}{2}+7 d x\right )-35 \cos \left (\frac {15 c}{2}+8 d x\right )+35 \cos \left (\frac {17 c}{2}+8 d x\right )-3360 \sin \left (\frac {c}{2}\right )+1680 c \sin \left (\frac {c}{2}\right )-1680 d x \sin \left (\frac {c}{2}\right )+1680 \sin \left (\frac {c}{2}+d x\right )-1680 \sin \left (\frac {3 c}{2}+d x\right )+560 \sin \left (\frac {5 c}{2}+3 d x\right )-560 \sin \left (\frac {7 c}{2}+3 d x\right )+280 \sin \left (\frac {7 c}{2}+4 d x\right )+280 \sin \left (\frac {9 c}{2}+4 d x\right )-112 \sin \left (\frac {9 c}{2}+5 d x\right )+112 \sin \left (\frac {11 c}{2}+5 d x\right )-80 \sin \left (\frac {13 c}{2}+7 d x\right )+80 \sin \left (\frac {15 c}{2}+7 d x\right )-35 \sin \left (\frac {15 c}{2}+8 d x\right )-35 \sin \left (\frac {17 c}{2}+8 d x\right )}{71680 a d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:

Integrate[(Cos[c + d*x]^6*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 

Output:

(1680*(c - d*x)*Cos[c/2] - 1680*Cos[c/2 + d*x] - 1680*Cos[(3*c)/2 + d*x] - 
 560*Cos[(5*c)/2 + 3*d*x] - 560*Cos[(7*c)/2 + 3*d*x] + 280*Cos[(7*c)/2 + 4 
*d*x] - 280*Cos[(9*c)/2 + 4*d*x] + 112*Cos[(9*c)/2 + 5*d*x] + 112*Cos[(11* 
c)/2 + 5*d*x] + 80*Cos[(13*c)/2 + 7*d*x] + 80*Cos[(15*c)/2 + 7*d*x] - 35*C 
os[(15*c)/2 + 8*d*x] + 35*Cos[(17*c)/2 + 8*d*x] - 3360*Sin[c/2] + 1680*c*S 
in[c/2] - 1680*d*x*Sin[c/2] + 1680*Sin[c/2 + d*x] - 1680*Sin[(3*c)/2 + d*x 
] + 560*Sin[(5*c)/2 + 3*d*x] - 560*Sin[(7*c)/2 + 3*d*x] + 280*Sin[(7*c)/2 
+ 4*d*x] + 280*Sin[(9*c)/2 + 4*d*x] - 112*Sin[(9*c)/2 + 5*d*x] + 112*Sin[( 
11*c)/2 + 5*d*x] - 80*Sin[(13*c)/2 + 7*d*x] + 80*Sin[(15*c)/2 + 7*d*x] - 3 
5*Sin[(15*c)/2 + 8*d*x] - 35*Sin[(17*c)/2 + 8*d*x])/(71680*a*d*(Cos[c/2] + 
 Sin[c/2]))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.483, Rules used = {3042, 3318, 3042, 3045, 244, 2009, 3048, 3042, 3048, 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.

\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^6(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^6}{a \sin (c+d x)+a}dx\)

\(\Big \downarrow \) 3318

\(\displaystyle \frac {\int \cos ^4(c+d x) \sin ^3(c+d x)dx}{a}-\frac {\int \cos ^4(c+d x) \sin ^4(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \cos (c+d x)^4 \sin (c+d x)^3dx}{a}-\frac {\int \cos (c+d x)^4 \sin (c+d x)^4dx}{a}\)

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\int \cos (c+d x)^4 \sin (c+d x)^4dx}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3048

\(\displaystyle -\frac {\frac {3}{8} \int \cos ^4(c+d x) \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{8} \int \cos (c+d x)^4 \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3048

\(\displaystyle -\frac {\frac {3}{8} \left (\frac {1}{6} \int \cos ^4(c+d x)dx-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{8} \left (\frac {1}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {\frac {3}{8} \left (\frac {1}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3}{8} \left (\frac {1}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {\frac {3}{8} \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}-\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)}{a d}-\frac {\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {\sin ^3(c+d x) \cos ^5(c+d x)}{8 d}}{a}\)

Input:

Int[(Cos[c + d*x]^6*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 

Output:

-((Cos[c + d*x]^5/5 - Cos[c + d*x]^7/7)/(a*d)) - (-1/8*(Cos[c + d*x]^5*Sin 
[c + d*x]^3)/d + (3*(-1/6*(Cos[c + d*x]^5*Sin[c + d*x])/d + ((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))/8)/a
 

Defintions of rubi rules used

rule 24

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

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3048

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 
1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Cos[e + f*x])^n 
*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

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 3318

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d)   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, 
d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.55

method	result	size
parallelrisch	\(\frac {-840 d x -35 \sin \left (8 d x +8 c \right )+80 \cos \left (7 d x +7 c \right )+280 \sin \left (4 d x +4 c \right )+112 \cos \left (5 d x +5 c \right )-560 \cos \left (3 d x +3 c \right )-1680 \cos \left (d x +c \right )-2048}{35840 d a}\)	\(78\)
risch	\(-\frac {3 x}{128 a}-\frac {3 \cos \left (d x +c \right )}{64 a d}-\frac {\sin \left (8 d x +8 c \right )}{1024 d a}+\frac {\cos \left (7 d x +7 c \right )}{448 a d}+\frac {\cos \left (5 d x +5 c \right )}{320 a d}+\frac {\sin \left (4 d x +4 c \right )}{128 d a}-\frac {\cos \left (3 d x +3 c \right )}{64 a d}\)	\(107\)
derivativedivides	\(\frac {\frac {16 \left (-\frac {1}{140}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1024}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{1024}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{20}-\frac {333 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{1024}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}+\frac {671 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{1024}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4}-\frac {671 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1024}+\frac {333 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{1024}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{1024}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{1024}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{d a}\)	\(207\)
default	\(\frac {\frac {16 \left (-\frac {1}{140}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1024}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{1024}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{20}-\frac {333 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{1024}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5}+\frac {671 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{1024}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4}-\frac {671 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1024}+\frac {333 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{1024}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{1024}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{1024}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{8}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}}{d a}\)	\(207\)

Input:

int(cos(d*x+c)^6*sin(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/35840*(-840*d*x-35*sin(8*d*x+8*c)+80*cos(7*d*x+7*c)+280*sin(4*d*x+4*c)+1 
12*cos(5*d*x+5*c)-560*cos(3*d*x+3*c)-1680*cos(d*x+c)-2048)/d/a
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {640 \, \cos \left (d x + c\right )^{7} - 896 \, \cos \left (d x + c\right )^{5} - 105 \, d x - 35 \, {\left (16 \, \cos \left (d x + c\right )^{7} - 24 \, \cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4480 \, a d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

1/4480*(640*cos(d*x + c)^7 - 896*cos(d*x + c)^5 - 105*d*x - 35*(16*cos(d*x 
 + c)^7 - 24*cos(d*x + c)^5 + 2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + 
 c))/(a*d)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3580 vs. \(2 (116) = 232\).

Time = 53.78 (sec) , antiderivative size = 3580, normalized size of antiderivative = 25.39 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)**6*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)
 

Output:

Piecewise((-105*d*x*tan(c/2 + d*x/2)**16/(4480*a*d*tan(c/2 + d*x/2)**16 + 
35840*a*d*tan(c/2 + d*x/2)**14 + 125440*a*d*tan(c/2 + d*x/2)**12 + 250880* 
a*d*tan(c/2 + d*x/2)**10 + 313600*a*d*tan(c/2 + d*x/2)**8 + 250880*a*d*tan 
(c/2 + d*x/2)**6 + 125440*a*d*tan(c/2 + d*x/2)**4 + 35840*a*d*tan(c/2 + d* 
x/2)**2 + 4480*a*d) - 840*d*x*tan(c/2 + d*x/2)**14/(4480*a*d*tan(c/2 + d*x 
/2)**16 + 35840*a*d*tan(c/2 + d*x/2)**14 + 125440*a*d*tan(c/2 + d*x/2)**12 
 + 250880*a*d*tan(c/2 + d*x/2)**10 + 313600*a*d*tan(c/2 + d*x/2)**8 + 2508 
80*a*d*tan(c/2 + d*x/2)**6 + 125440*a*d*tan(c/2 + d*x/2)**4 + 35840*a*d*ta 
n(c/2 + d*x/2)**2 + 4480*a*d) - 2940*d*x*tan(c/2 + d*x/2)**12/(4480*a*d*ta 
n(c/2 + d*x/2)**16 + 35840*a*d*tan(c/2 + d*x/2)**14 + 125440*a*d*tan(c/2 + 
 d*x/2)**12 + 250880*a*d*tan(c/2 + d*x/2)**10 + 313600*a*d*tan(c/2 + d*x/2 
)**8 + 250880*a*d*tan(c/2 + d*x/2)**6 + 125440*a*d*tan(c/2 + d*x/2)**4 + 3 
5840*a*d*tan(c/2 + d*x/2)**2 + 4480*a*d) - 5880*d*x*tan(c/2 + d*x/2)**10/( 
4480*a*d*tan(c/2 + d*x/2)**16 + 35840*a*d*tan(c/2 + d*x/2)**14 + 125440*a* 
d*tan(c/2 + d*x/2)**12 + 250880*a*d*tan(c/2 + d*x/2)**10 + 313600*a*d*tan( 
c/2 + d*x/2)**8 + 250880*a*d*tan(c/2 + d*x/2)**6 + 125440*a*d*tan(c/2 + d* 
x/2)**4 + 35840*a*d*tan(c/2 + d*x/2)**2 + 4480*a*d) - 7350*d*x*tan(c/2 + d 
*x/2)**8/(4480*a*d*tan(c/2 + d*x/2)**16 + 35840*a*d*tan(c/2 + d*x/2)**14 + 
 125440*a*d*tan(c/2 + d*x/2)**12 + 250880*a*d*tan(c/2 + d*x/2)**10 + 31360 
0*a*d*tan(c/2 + d*x/2)**8 + 250880*a*d*tan(c/2 + d*x/2)**6 + 125440*a*d...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (127) = 254\).

Time = 0.16 (sec) , antiderivative size = 461, normalized size of antiderivative = 3.27 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

1/2240*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 2048*sin(d*x + c)^2/(cos(d* 
x + c) + 1)^2 + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1792*sin(d*x + c 
)^4/(cos(d*x + c) + 1)^4 - 11655*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 143 
36*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 23485*sin(d*x + c)^7/(cos(d*x + c 
) + 1)^7 - 8960*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 23485*sin(d*x + c)^9 
/(cos(d*x + c) + 1)^9 + 11655*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 8960 
*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 805*sin(d*x + c)^13/(cos(d*x + c) 
 + 1)^13 - 105*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 - 256)/(a + 8*a*sin(d 
*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 
+ 56*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a*sin(d*x + c)^8/(cos(d*x 
+ c) + 1)^8 + 56*a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a*sin(d*x + 
c)^12/(cos(d*x + c) + 1)^12 + 8*a*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 
a*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) - 105*arctan(sin(d*x + c)/(cos(d* 
x + c) + 1))/a)/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.45 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {105 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 805 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 8960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 11655 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 23485 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 8960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 23485 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 14336 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 11655 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1792 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 805 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2048 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 256\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8} a}}{4480 \, d} \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

-1/4480*(105*(d*x + c)/a + 2*(105*tan(1/2*d*x + 1/2*c)^15 + 805*tan(1/2*d* 
x + 1/2*c)^13 + 8960*tan(1/2*d*x + 1/2*c)^12 - 11655*tan(1/2*d*x + 1/2*c)^ 
11 + 23485*tan(1/2*d*x + 1/2*c)^9 + 8960*tan(1/2*d*x + 1/2*c)^8 - 23485*ta 
n(1/2*d*x + 1/2*c)^7 + 14336*tan(1/2*d*x + 1/2*c)^6 + 11655*tan(1/2*d*x + 
1/2*c)^5 - 1792*tan(1/2*d*x + 1/2*c)^4 - 805*tan(1/2*d*x + 1/2*c)^3 + 2048 
*tan(1/2*d*x + 1/2*c)^2 - 105*tan(1/2*d*x + 1/2*c) + 256)/((tan(1/2*d*x + 
1/2*c)^2 + 1)^8*a))/d
 

Mupad [B] (verification not implemented)

Time = 36.54 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3\,x}{128\,a}-\frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {333\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{64}+\frac {671\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {671\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{64}+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {333\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{64}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}-\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{64}+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {4}{35}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \] Input:

int((cos(c + d*x)^6*sin(c + d*x)^3)/(a + a*sin(c + d*x)),x)
 

Output:

- (3*x)/(128*a) - ((32*tan(c/2 + (d*x)/2)^2)/35 - (3*tan(c/2 + (d*x)/2))/6 
4 - (23*tan(c/2 + (d*x)/2)^3)/64 - (4*tan(c/2 + (d*x)/2)^4)/5 + (333*tan(c 
/2 + (d*x)/2)^5)/64 + (32*tan(c/2 + (d*x)/2)^6)/5 - (671*tan(c/2 + (d*x)/2 
)^7)/64 + 4*tan(c/2 + (d*x)/2)^8 + (671*tan(c/2 + (d*x)/2)^9)/64 - (333*ta 
n(c/2 + (d*x)/2)^11)/64 + 4*tan(c/2 + (d*x)/2)^12 + (23*tan(c/2 + (d*x)/2) 
^13)/64 + (3*tan(c/2 + (d*x)/2)^15)/64 + 4/35)/(a*d*(tan(c/2 + (d*x)/2)^2 
+ 1)^8)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^6(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {560 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}-640 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+1024 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+105 \cos \left (d x +c \right ) \sin \left (d x +c \right )-256 \cos \left (d x +c \right )-105 d x +256}{4480 a d} \] Input:

int(cos(d*x+c)^6*sin(d*x+c)^3/(a+a*sin(d*x+c)),x)
 

Output:

(560*cos(c + d*x)*sin(c + d*x)**7 - 640*cos(c + d*x)*sin(c + d*x)**6 - 840 
*cos(c + d*x)*sin(c + d*x)**5 + 1024*cos(c + d*x)*sin(c + d*x)**4 + 70*cos 
(c + d*x)*sin(c + d*x)**3 - 128*cos(c + d*x)*sin(c + d*x)**2 + 105*cos(c + 
 d*x)*sin(c + d*x) - 256*cos(c + d*x) - 105*d*x + 256)/(4480*a*d)