3.5.10 \(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [410]

3.5.10.1 Optimal result
3.5.10.2 Mathematica [B] (verified)
3.5.10.3 Rubi [A] (verified)
3.5.10.4 Maple [A] (verified)
3.5.10.5 Fricas [A] (verification not implemented)
3.5.10.6 Sympy [B] (verification not implemented)
3.5.10.7 Maxima [B] (verification not implemented)
3.5.10.8 Giac [A] (verification not implemented)
3.5.10.9 Mupad [B] (verification not implemented)

3.5.10.1 Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{16 a}-\frac {\cos ^3(c+d x)}{3 a d}+\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cos (c+d x) \sin (c+d x)}{16 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d} \]

output
-1/16*x/a-1/3*cos(d*x+c)^3/a/d+1/5*cos(d*x+c)^5/a/d-1/16*cos(d*x+c)*sin(d* 
x+c)/a/d+1/8*cos(d*x+c)^3*sin(d*x+c)/a/d+1/6*cos(d*x+c)^3*sin(d*x+c)^3/a/d
 
3.5.10.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(377\) vs. \(2(117)=234\).

Time = 3.61 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.22 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 (3 c-4 d x) \cos \left (\frac {c}{2}\right )-120 \cos \left (\frac {c}{2}+d x\right )-120 \cos \left (\frac {3 c}{2}+d x\right )+15 \cos \left (\frac {3 c}{2}+2 d x\right )-15 \cos \left (\frac {5 c}{2}+2 d x\right )-20 \cos \left (\frac {5 c}{2}+3 d x\right )-20 \cos \left (\frac {7 c}{2}+3 d x\right )+15 \cos \left (\frac {7 c}{2}+4 d x\right )-15 \cos \left (\frac {9 c}{2}+4 d x\right )+12 \cos \left (\frac {9 c}{2}+5 d x\right )+12 \cos \left (\frac {11 c}{2}+5 d x\right )-5 \cos \left (\frac {11 c}{2}+6 d x\right )+5 \cos \left (\frac {13 c}{2}+6 d x\right )-180 \sin \left (\frac {c}{2}\right )+90 c \sin \left (\frac {c}{2}\right )-120 d x \sin \left (\frac {c}{2}\right )+120 \sin \left (\frac {c}{2}+d x\right )-120 \sin \left (\frac {3 c}{2}+d x\right )+15 \sin \left (\frac {3 c}{2}+2 d x\right )+15 \sin \left (\frac {5 c}{2}+2 d x\right )+20 \sin \left (\frac {5 c}{2}+3 d x\right )-20 \sin \left (\frac {7 c}{2}+3 d x\right )+15 \sin \left (\frac {7 c}{2}+4 d x\right )+15 \sin \left (\frac {9 c}{2}+4 d x\right )-12 \sin \left (\frac {9 c}{2}+5 d x\right )+12 \sin \left (\frac {11 c}{2}+5 d x\right )-5 \sin \left (\frac {11 c}{2}+6 d x\right )-5 \sin \left (\frac {13 c}{2}+6 d x\right )}{1920 a d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

input
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 
output
(30*(3*c - 4*d*x)*Cos[c/2] - 120*Cos[c/2 + d*x] - 120*Cos[(3*c)/2 + d*x] + 
 15*Cos[(3*c)/2 + 2*d*x] - 15*Cos[(5*c)/2 + 2*d*x] - 20*Cos[(5*c)/2 + 3*d* 
x] - 20*Cos[(7*c)/2 + 3*d*x] + 15*Cos[(7*c)/2 + 4*d*x] - 15*Cos[(9*c)/2 + 
4*d*x] + 12*Cos[(9*c)/2 + 5*d*x] + 12*Cos[(11*c)/2 + 5*d*x] - 5*Cos[(11*c) 
/2 + 6*d*x] + 5*Cos[(13*c)/2 + 6*d*x] - 180*Sin[c/2] + 90*c*Sin[c/2] - 120 
*d*x*Sin[c/2] + 120*Sin[c/2 + d*x] - 120*Sin[(3*c)/2 + d*x] + 15*Sin[(3*c) 
/2 + 2*d*x] + 15*Sin[(5*c)/2 + 2*d*x] + 20*Sin[(5*c)/2 + 3*d*x] - 20*Sin[( 
7*c)/2 + 3*d*x] + 15*Sin[(7*c)/2 + 4*d*x] + 15*Sin[(9*c)/2 + 4*d*x] - 12*S 
in[(9*c)/2 + 5*d*x] + 12*Sin[(11*c)/2 + 5*d*x] - 5*Sin[(11*c)/2 + 6*d*x] - 
 5*Sin[(13*c)/2 + 6*d*x])/(1920*a*d*(Cos[c/2] + Sin[c/2]))
 
3.5.10.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3318, 3042, 3045, 244, 2009, 3048, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3048

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3048

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

input
Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
 
output
-((Cos[c + d*x]^3/3 - Cos[c + d*x]^5/5)/(a*d)) - (-1/6*(Cos[c + d*x]^3*Sin 
[c + d*x]^3)/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)/2)/a
 

3.5.10.3.1 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]
 
3.5.10.4 Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67

method result size
parallelrisch \(\frac {-60 d x -20 \cos \left (3 d x +3 c \right )-5 \sin \left (6 d x +6 c \right )+12 \cos \left (5 d x +5 c \right )+15 \sin \left (4 d x +4 c \right )+15 \sin \left (2 d x +2 c \right )-120 \cos \left (d x +c \right )-128}{960 d a}\) \(78\)
risch \(-\frac {x}{16 a}-\frac {\cos \left (d x +c \right )}{8 a d}-\frac {\sin \left (6 d x +6 c \right )}{192 d a}+\frac {\cos \left (5 d x +5 c \right )}{80 a d}+\frac {\sin \left (4 d x +4 c \right )}{64 d a}-\frac {\cos \left (3 d x +3 c \right )}{48 a d}+\frac {\sin \left (2 d x +2 c \right )}{64 d a}\) \(107\)
derivativedivides \(\frac {\frac {16 \left (-\frac {\left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128}-\frac {17 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {19 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128}-\frac {1}{60}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{d a}\) \(155\)
default \(\frac {\frac {16 \left (-\frac {\left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128}-\frac {17 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {19 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64}+\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128}-\frac {1}{60}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{d a}\) \(155\)
norman \(\frac {-\frac {21 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {7 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {7 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {21 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {35 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {21 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {21 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {17}{120 a d}-\frac {7 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {7 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {x}{16 a}-\frac {53 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {13 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}-\frac {x \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a}+\frac {3 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 d a}+\frac {\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {41 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a}+\frac {7 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {19 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d a}-\frac {55 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {181 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d a}-\frac {7 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {223 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 d a}+\frac {8 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) \(562\)

input
int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 
output
1/960*(-60*d*x-20*cos(3*d*x+3*c)-5*sin(6*d*x+6*c)+12*cos(5*d*x+5*c)+15*sin 
(4*d*x+4*c)+15*sin(2*d*x+2*c)-120*cos(d*x+c)-128)/d/a
 
3.5.10.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {48 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 15 \, d x - 5 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 14 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, a d} \]

input
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 
output
1/240*(48*cos(d*x + c)^5 - 80*cos(d*x + c)^3 - 15*d*x - 5*(8*cos(d*x + c)^ 
5 - 14*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c))/(a*d)
 
3.5.10.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2067 vs. \(2 (92) = 184\).

Time = 19.66 (sec) , antiderivative size = 2067, normalized size of antiderivative = 17.67 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

input
integrate(cos(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)
 
output
Piecewise((-15*d*x*tan(c/2 + d*x/2)**12/(240*a*d*tan(c/2 + d*x/2)**12 + 14 
40*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d*tan( 
c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d*x/2) 
**2 + 240*a*d) - 90*d*x*tan(c/2 + d*x/2)**10/(240*a*d*tan(c/2 + d*x/2)**12 
 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 4800*a*d 
*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/2 + d 
*x/2)**2 + 240*a*d) - 225*d*x*tan(c/2 + d*x/2)**8/(240*a*d*tan(c/2 + d*x/2 
)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 480 
0*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*tan(c/ 
2 + d*x/2)**2 + 240*a*d) - 300*d*x*tan(c/2 + d*x/2)**6/(240*a*d*tan(c/2 + 
d*x/2)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 
+ 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440*a*d*t 
an(c/2 + d*x/2)**2 + 240*a*d) - 225*d*x*tan(c/2 + d*x/2)**4/(240*a*d*tan(c 
/2 + d*x/2)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2 
)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1440* 
a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 90*d*x*tan(c/2 + d*x/2)**2/(240*a*d*t 
an(c/2 + d*x/2)**12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d 
*x/2)**8 + 4800*a*d*tan(c/2 + d*x/2)**6 + 3600*a*d*tan(c/2 + d*x/2)**4 + 1 
440*a*d*tan(c/2 + d*x/2)**2 + 240*a*d) - 15*d*x/(240*a*d*tan(c/2 + d*x/2)* 
*12 + 1440*a*d*tan(c/2 + d*x/2)**10 + 3600*a*d*tan(c/2 + d*x/2)**8 + 48...
 
3.5.10.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (105) = 210\).

Time = 0.32 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.90 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {192 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {85 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {570 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {320 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {570 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {480 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {85 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {15 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - 32}{a + \frac {6 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{120 \, d} \]

input
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 
output
1/120*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 192*sin(d*x + c)^2/(cos(d*x + 
 c) + 1)^2 + 85*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 570*sin(d*x + c)^5/( 
cos(d*x + c) + 1)^5 - 320*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 570*sin(d* 
x + c)^7/(cos(d*x + c) + 1)^7 - 480*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 
85*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 15*sin(d*x + c)^11/(cos(d*x + c) 
+ 1)^11 - 32)/(a + 6*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a*sin(d*x 
+ c)^4/(cos(d*x + c) + 1)^4 + 20*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1 
5*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6*a*sin(d*x + c)^10/(cos(d*x + c 
) + 1)^10 + a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 15*arctan(sin(d*x + 
 c)/(cos(d*x + c) + 1))/a)/d
 
3.5.10.8 Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 570 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 570 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 85 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 192 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 32\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a}}{240 \, d} \]

input
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
 
output
-1/240*(15*(d*x + c)/a + 2*(15*tan(1/2*d*x + 1/2*c)^11 + 85*tan(1/2*d*x + 
1/2*c)^9 + 480*tan(1/2*d*x + 1/2*c)^8 - 570*tan(1/2*d*x + 1/2*c)^7 + 320*t 
an(1/2*d*x + 1/2*c)^6 + 570*tan(1/2*d*x + 1/2*c)^5 - 85*tan(1/2*d*x + 1/2* 
c)^3 + 192*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 32)/((tan(1/ 
2*d*x + 1/2*c)^2 + 1)^6*a))/d
 
3.5.10.9 Mupad [B] (verification not implemented)

Time = 12.16 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{16\,a}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {19\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {4}{15}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

input
int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + a*sin(c + d*x)),x)
 
output
- x/(16*a) - ((8*tan(c/2 + (d*x)/2)^2)/5 - tan(c/2 + (d*x)/2)/8 - (17*tan( 
c/2 + (d*x)/2)^3)/24 + (19*tan(c/2 + (d*x)/2)^5)/4 + (8*tan(c/2 + (d*x)/2) 
^6)/3 - (19*tan(c/2 + (d*x)/2)^7)/4 + 4*tan(c/2 + (d*x)/2)^8 + (17*tan(c/2 
 + (d*x)/2)^9)/24 + tan(c/2 + (d*x)/2)^11/8 + 4/15)/(a*d*(tan(c/2 + (d*x)/ 
2)^2 + 1)^6)