\(\int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [411]

Optimal result

Integrand size = 29, antiderivative size = 91 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{8 a}+\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {\cos (c+d x) \sin (c+d x)}{8 a d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \] Output:

1/8*x/a+1/3*cos(d*x+c)^3/a/d-1/5*cos(d*x+c)^5/a/d+1/8*cos(d*x+c)*sin(d*x+c 
)/a/d-1/4*cos(d*x+c)^3*sin(d*x+c)/a/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(258\) vs. \(2(91)=182\).

Time = 1.69 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {120 d x \cos \left (\frac {c}{2}\right )+60 \cos \left (\frac {c}{2}+d x\right )+60 \cos \left (\frac {3 c}{2}+d x\right )+10 \cos \left (\frac {5 c}{2}+3 d x\right )+10 \cos \left (\frac {7 c}{2}+3 d x\right )-15 \cos \left (\frac {7 c}{2}+4 d x\right )+15 \cos \left (\frac {9 c}{2}+4 d x\right )-6 \cos \left (\frac {9 c}{2}+5 d x\right )-6 \cos \left (\frac {11 c}{2}+5 d x\right )+120 \sin \left (\frac {c}{2}\right )+120 d x \sin \left (\frac {c}{2}\right )-60 \sin \left (\frac {c}{2}+d x\right )+60 \sin \left (\frac {3 c}{2}+d x\right )-10 \sin \left (\frac {5 c}{2}+3 d x\right )+10 \sin \left (\frac {7 c}{2}+3 d x\right )-15 \sin \left (\frac {7 c}{2}+4 d x\right )-15 \sin \left (\frac {9 c}{2}+4 d x\right )+6 \sin \left (\frac {9 c}{2}+5 d x\right )-6 \sin \left (\frac {11 c}{2}+5 d x\right )}{960 a d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
 

Output:

(120*d*x*Cos[c/2] + 60*Cos[c/2 + d*x] + 60*Cos[(3*c)/2 + d*x] + 10*Cos[(5* 
c)/2 + 3*d*x] + 10*Cos[(7*c)/2 + 3*d*x] - 15*Cos[(7*c)/2 + 4*d*x] + 15*Cos 
[(9*c)/2 + 4*d*x] - 6*Cos[(9*c)/2 + 5*d*x] - 6*Cos[(11*c)/2 + 5*d*x] + 120 
*Sin[c/2] + 120*d*x*Sin[c/2] - 60*Sin[c/2 + d*x] + 60*Sin[(3*c)/2 + d*x] - 
 10*Sin[(5*c)/2 + 3*d*x] + 10*Sin[(7*c)/2 + 3*d*x] - 15*Sin[(7*c)/2 + 4*d* 
x] - 15*Sin[(9*c)/2 + 4*d*x] + 6*Sin[(9*c)/2 + 5*d*x] - 6*Sin[(11*c)/2 + 5 
*d*x])/(960*a*d*(Cos[c/2] + Sin[c/2]))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3318, 3042, 3045, 244, 2009, 3048, 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.

\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^4(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3048

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

Input:

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
 

Output:

(Cos[c + d*x]^3/3 - Cos[c + d*x]^5/5)/(a*d) + (-1/4*(Cos[c + d*x]^3*Sin[c 
+ d*x])/d + (x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))/4)/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 = 0.52 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {60 d x +10 \cos \left (3 d x +3 c \right )+60 \cos \left (d x +c \right )-6 \cos \left (5 d x +5 c \right )-15 \sin \left (4 d x +4 c \right )+64}{480 d a}\)	\(56\)
risch	\(\frac {x}{8 a}+\frac {\cos \left (d x +c \right )}{8 a d}-\frac {\cos \left (5 d x +5 c \right )}{80 a d}-\frac {\sin \left (4 d x +4 c \right )}{32 d a}+\frac {\cos \left (3 d x +3 c \right )}{48 a d}\)	\(73\)
derivativedivides	\(\frac {\frac {8 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{32}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{6}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}+\frac {1}{30}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\)	\(129\)
default	\(\frac {\frac {8 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{32}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{6}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}+\frac {1}{30}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\)	\(129\)
norman	\(\frac {\frac {1}{60 a d}+\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8 a}+\frac {x}{8 a}+\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{20 d a}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{4 d a}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d a}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{6 d a}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d a}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d a}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{30 d a}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{20 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4 d a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a}+\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 a}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(490\)

Input:

int(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/480*(60*d*x+10*cos(3*d*x+3*c)+60*cos(d*x+c)-6*cos(5*d*x+5*c)-15*sin(4*d* 
x+4*c)+64)/d/a
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {24 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} - 15 \, d x + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, a d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

-1/120*(24*cos(d*x + c)^5 - 40*cos(d*x + c)^3 - 15*d*x + 15*(2*cos(d*x + c 
)^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. 1464 vs. \(2 (70) = 140\).

Time = 13.58 (sec) , antiderivative size = 1464, normalized size of antiderivative = 16.09 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)**2/(a+a*sin(d*x+c)),x)
 

Output:

Piecewise((15*d*x*tan(c/2 + d*x/2)**10/(120*a*d*tan(c/2 + d*x/2)**10 + 600 
*a*d*tan(c/2 + d*x/2)**8 + 1200*a*d*tan(c/2 + d*x/2)**6 + 1200*a*d*tan(c/2 
 + d*x/2)**4 + 600*a*d*tan(c/2 + d*x/2)**2 + 120*a*d) + 75*d*x*tan(c/2 + d 
*x/2)**8/(120*a*d*tan(c/2 + d*x/2)**10 + 600*a*d*tan(c/2 + d*x/2)**8 + 120 
0*a*d*tan(c/2 + d*x/2)**6 + 1200*a*d*tan(c/2 + d*x/2)**4 + 600*a*d*tan(c/2 
 + d*x/2)**2 + 120*a*d) + 150*d*x*tan(c/2 + d*x/2)**6/(120*a*d*tan(c/2 + d 
*x/2)**10 + 600*a*d*tan(c/2 + d*x/2)**8 + 1200*a*d*tan(c/2 + d*x/2)**6 + 1 
200*a*d*tan(c/2 + d*x/2)**4 + 600*a*d*tan(c/2 + d*x/2)**2 + 120*a*d) + 150 
*d*x*tan(c/2 + d*x/2)**4/(120*a*d*tan(c/2 + d*x/2)**10 + 600*a*d*tan(c/2 + 
 d*x/2)**8 + 1200*a*d*tan(c/2 + d*x/2)**6 + 1200*a*d*tan(c/2 + d*x/2)**4 + 
 600*a*d*tan(c/2 + d*x/2)**2 + 120*a*d) + 75*d*x*tan(c/2 + d*x/2)**2/(120* 
a*d*tan(c/2 + d*x/2)**10 + 600*a*d*tan(c/2 + d*x/2)**8 + 1200*a*d*tan(c/2 
+ d*x/2)**6 + 1200*a*d*tan(c/2 + d*x/2)**4 + 600*a*d*tan(c/2 + d*x/2)**2 + 
 120*a*d) + 15*d*x/(120*a*d*tan(c/2 + d*x/2)**10 + 600*a*d*tan(c/2 + d*x/2 
)**8 + 1200*a*d*tan(c/2 + d*x/2)**6 + 1200*a*d*tan(c/2 + d*x/2)**4 + 600*a 
*d*tan(c/2 + d*x/2)**2 + 120*a*d) + 30*tan(c/2 + d*x/2)**9/(120*a*d*tan(c/ 
2 + d*x/2)**10 + 600*a*d*tan(c/2 + d*x/2)**8 + 1200*a*d*tan(c/2 + d*x/2)** 
6 + 1200*a*d*tan(c/2 + d*x/2)**4 + 600*a*d*tan(c/2 + d*x/2)**2 + 120*a*d) 
- 180*tan(c/2 + d*x/2)**7/(120*a*d*tan(c/2 + d*x/2)**10 + 600*a*d*tan(c/2 
+ d*x/2)**8 + 1200*a*d*tan(c/2 + d*x/2)**6 + 1200*a*d*tan(c/2 + d*x/2)*...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (81) = 162\).

Time = 0.11 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.05 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {80 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {90 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {80 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {240 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {90 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 16}{a + \frac {5 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{60 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

-1/60*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 80*sin(d*x + c)^2/(cos(d*x + 
c) + 1)^2 - 90*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 80*sin(d*x + c)^4/(co 
s(d*x + c) + 1)^4 - 240*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 90*sin(d*x + 
 c)^7/(cos(d*x + c) + 1)^7 - 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 16)/ 
(a + 5*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a*sin(d*x + c)^4/(cos(d* 
x + c) + 1)^4 + 10*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a*sin(d*x + c 
)^8/(cos(d*x + c) + 1)^8 + a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) - 15*a 
rctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {15 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 90 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 90 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a}}{120 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/120*(15*(d*x + c)/a + 2*(15*tan(1/2*d*x + 1/2*c)^9 - 90*tan(1/2*d*x + 1/ 
2*c)^7 + 240*tan(1/2*d*x + 1/2*c)^6 - 80*tan(1/2*d*x + 1/2*c)^4 + 90*tan(1 
/2*d*x + 1/2*c)^3 + 80*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 
16)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a))/d
 

Mupad [B] (verification not implemented)

Time = 19.92 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{8\,a}+\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {4}{15}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \] Input:

int((cos(c + d*x)^4*sin(c + d*x)^2)/(a + a*sin(c + d*x)),x)
 

Output:

x/(8*a) + ((4*tan(c/2 + (d*x)/2)^2)/3 - tan(c/2 + (d*x)/2)/4 + (3*tan(c/2 
+ (d*x)/2)^3)/2 - (4*tan(c/2 + (d*x)/2)^4)/3 + 4*tan(c/2 + (d*x)/2)^6 - (3 
*tan(c/2 + (d*x)/2)^7)/2 + tan(c/2 + (d*x)/2)^9/4 + 4/15)/(a*d*(tan(c/2 + 
(d*x)/2)^2 + 1)^5)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )+16 \cos \left (d x +c \right )+15 d x -16}{120 a d} \] Input:

int(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)
 

Output:

( - 24*cos(c + d*x)*sin(c + d*x)**4 + 30*cos(c + d*x)*sin(c + d*x)**3 + 8* 
cos(c + d*x)*sin(c + d*x)**2 - 15*cos(c + d*x)*sin(c + d*x) + 16*cos(c + d 
*x) + 15*d*x - 16)/(120*a*d)