3.5.20 \(\int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [420]

3.5.20.1 Optimal result
3.5.20.2 Mathematica [B] (verified)
3.5.20.3 Rubi [A] (verified)
3.5.20.4 Maple [A] (verified)
3.5.20.5 Fricas [A] (verification not implemented)
3.5.20.6 Sympy [B] (verification not implemented)
3.5.20.7 Maxima [B] (verification not implemented)
3.5.20.8 Giac [A] (verification not implemented)
3.5.20.9 Mupad [B] (verification not implemented)

3.5.20.1 Optimal result

Integrand size = 29, antiderivative size = 147 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 x}{8 a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {5 \cos ^3(c+d x)}{3 a^2 d}-\frac {4 \cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^7(c+d x)}{7 a^2 d}+\frac {5 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {5 \cos (c+d x) \sin ^3(c+d x)}{12 a^2 d}+\frac {\cos (c+d x) \sin ^5(c+d x)}{3 a^2 d} \]

output
-5/8*x/a^2-2*cos(d*x+c)/a^2/d+5/3*cos(d*x+c)^3/a^2/d-4/5*cos(d*x+c)^5/a^2/ 
d+1/7*cos(d*x+c)^7/a^2/d+5/8*cos(d*x+c)*sin(d*x+c)/a^2/d+5/12*cos(d*x+c)*s 
in(d*x+c)^3/a^2/d+1/3*cos(d*x+c)*sin(d*x+c)^5/a^2/d
 
3.5.20.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(418\) vs. \(2(147)=294\).

Time = 3.83 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-210 (1+40 d x) \cos \left (\frac {c}{2}\right )-7875 \cos \left (\frac {c}{2}+d x\right )-7875 \cos \left (\frac {3 c}{2}+d x\right )+3150 \cos \left (\frac {3 c}{2}+2 d x\right )-3150 \cos \left (\frac {5 c}{2}+2 d x\right )+1435 \cos \left (\frac {5 c}{2}+3 d x\right )+1435 \cos \left (\frac {7 c}{2}+3 d x\right )-630 \cos \left (\frac {7 c}{2}+4 d x\right )+630 \cos \left (\frac {9 c}{2}+4 d x\right )-231 \cos \left (\frac {9 c}{2}+5 d x\right )-231 \cos \left (\frac {11 c}{2}+5 d x\right )+70 \cos \left (\frac {11 c}{2}+6 d x\right )-70 \cos \left (\frac {13 c}{2}+6 d x\right )+15 \cos \left (\frac {13 c}{2}+7 d x\right )+15 \cos \left (\frac {15 c}{2}+7 d x\right )+210 \sin \left (\frac {c}{2}\right )-8400 d x \sin \left (\frac {c}{2}\right )+7875 \sin \left (\frac {c}{2}+d x\right )-7875 \sin \left (\frac {3 c}{2}+d x\right )+3150 \sin \left (\frac {3 c}{2}+2 d x\right )+3150 \sin \left (\frac {5 c}{2}+2 d x\right )-1435 \sin \left (\frac {5 c}{2}+3 d x\right )+1435 \sin \left (\frac {7 c}{2}+3 d x\right )-630 \sin \left (\frac {7 c}{2}+4 d x\right )-630 \sin \left (\frac {9 c}{2}+4 d x\right )+231 \sin \left (\frac {9 c}{2}+5 d x\right )-231 \sin \left (\frac {11 c}{2}+5 d x\right )+70 \sin \left (\frac {11 c}{2}+6 d x\right )+70 \sin \left (\frac {13 c}{2}+6 d x\right )-15 \sin \left (\frac {13 c}{2}+7 d x\right )+15 \sin \left (\frac {15 c}{2}+7 d x\right )}{13440 a^2 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

input
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]
 
output
(-210*(1 + 40*d*x)*Cos[c/2] - 7875*Cos[c/2 + d*x] - 7875*Cos[(3*c)/2 + d*x 
] + 3150*Cos[(3*c)/2 + 2*d*x] - 3150*Cos[(5*c)/2 + 2*d*x] + 1435*Cos[(5*c) 
/2 + 3*d*x] + 1435*Cos[(7*c)/2 + 3*d*x] - 630*Cos[(7*c)/2 + 4*d*x] + 630*C 
os[(9*c)/2 + 4*d*x] - 231*Cos[(9*c)/2 + 5*d*x] - 231*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] + 210*Sin[c/2] - 8400*d*x*Sin[c/2] + 78 
75*Sin[c/2 + d*x] - 7875*Sin[(3*c)/2 + d*x] + 3150*Sin[(3*c)/2 + 2*d*x] + 
3150*Sin[(5*c)/2 + 2*d*x] - 1435*Sin[(5*c)/2 + 3*d*x] + 1435*Sin[(7*c)/2 + 
 3*d*x] - 630*Sin[(7*c)/2 + 4*d*x] - 630*Sin[(9*c)/2 + 4*d*x] + 231*Sin[(9 
*c)/2 + 5*d*x] - 231*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])/(13440*a^2*d*(Cos[c/2] + Sin[c/2]))
 
3.5.20.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3348, 3042, 3236, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3348

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3236

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^2 \cos ^7(c+d x)}{7 d}-\frac {4 a^2 \cos ^5(c+d x)}{5 d}+\frac {5 a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \sin ^5(c+d x) \cos (c+d x)}{3 d}+\frac {5 a^2 \sin ^3(c+d x) \cos (c+d x)}{12 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {5 a^2 x}{8}}{a^4}\)

input
Int[(Cos[c + d*x]^4*Sin[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]
 
output
((-5*a^2*x)/8 - (2*a^2*Cos[c + d*x])/d + (5*a^2*Cos[c + d*x]^3)/(3*d) - (4 
*a^2*Cos[c + d*x]^5)/(5*d) + (a^2*Cos[c + d*x]^7)/(7*d) + (5*a^2*Cos[c + d 
*x]*Sin[c + d*x])/(8*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x]^3)/(12*d) + (a^ 
2*Cos[c + d*x]*Sin[c + d*x]^5)/(3*d))/a^4
 

3.5.20.3.1 Defintions of rubi rules used

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 3236
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + 
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt 
Q[m, 0] && RationalQ[n]
 

rule 3348
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) 
 + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m)   Int[(d* 
Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, 
 x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
 
3.5.20.4 Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61

method result size
parallelrisch \(\frac {-4200 d x +1435 \cos \left (3 d x +3 c \right )-7875 \cos \left (d x +c \right )+15 \cos \left (7 d x +7 c \right )+70 \sin \left (6 d x +6 c \right )-231 \cos \left (5 d x +5 c \right )-630 \sin \left (4 d x +4 c \right )+3150 \sin \left (2 d x +2 c \right )-6656}{6720 d \,a^{2}}\) \(89\)
risch \(-\frac {5 x}{8 a^{2}}-\frac {75 \cos \left (d x +c \right )}{64 a^{2} d}+\frac {\cos \left (7 d x +7 c \right )}{448 d \,a^{2}}+\frac {\sin \left (6 d x +6 c \right )}{96 d \,a^{2}}-\frac {11 \cos \left (5 d x +5 c \right )}{320 d \,a^{2}}-\frac {3 \sin \left (4 d x +4 c \right )}{32 d \,a^{2}}+\frac {41 \cos \left (3 d x +3 c \right )}{192 d \,a^{2}}+\frac {15 \sin \left (2 d x +2 c \right )}{32 d \,a^{2}}\) \(124\)
derivativedivides \(\frac {\frac {64 \left (-\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{256}-\frac {25 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192}-\frac {283 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{768}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {11 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}+\frac {283 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{768}-\frac {13 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20}+\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192}-\frac {13 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{256}-\frac {13}{420}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}-\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{2}}\) \(168\)
default \(\frac {\frac {64 \left (-\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{256}-\frac {25 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192}-\frac {283 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{768}-\frac {\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}-\frac {11 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}+\frac {283 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{768}-\frac {13 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20}+\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192}-\frac {13 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{256}-\frac {13}{420}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}-\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{2}}\) \(168\)

input
int(cos(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
1/6720*(-4200*d*x+1435*cos(3*d*x+3*c)-7875*cos(d*x+c)+15*cos(7*d*x+7*c)+70 
*sin(6*d*x+6*c)-231*cos(5*d*x+5*c)-630*sin(4*d*x+4*c)+3150*sin(2*d*x+2*c)- 
6656)/d/a^2
 
3.5.20.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {120 \, \cos \left (d x + c\right )^{7} - 672 \, \cos \left (d x + c\right )^{5} + 1400 \, \cos \left (d x + c\right )^{3} - 525 \, d x + 35 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 26 \, \cos \left (d x + c\right )^{3} + 33 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 1680 \, \cos \left (d x + c\right )}{840 \, a^{2} d} \]

input
integrate(cos(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 
output
1/840*(120*cos(d*x + c)^7 - 672*cos(d*x + c)^5 + 1400*cos(d*x + c)^3 - 525 
*d*x + 35*(8*cos(d*x + c)^5 - 26*cos(d*x + c)^3 + 33*cos(d*x + c))*sin(d*x 
 + c) - 1680*cos(d*x + c))/(a^2*d)
 
3.5.20.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2895 vs. \(2 (138) = 276\).

Time = 81.11 (sec) , antiderivative size = 2895, normalized size of antiderivative = 19.69 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \]

input
integrate(cos(d*x+c)**4*sin(d*x+c)**5/(a+a*sin(d*x+c))**2,x)
 
output
Piecewise((-525*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 
) - 3675*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) - 110 
25*d*x*tan(c/2 + d*x/2)**10/(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*t 
an(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) - 18375*d*x 
*tan(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) - 18375*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 
 + 5880*a**2*d*tan(c/2 + d*x/2)**2 + 840*a**2*d) - 11025*d*x*tan(c/2 + ...
 
3.5.20.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (133) = 266\).

Time = 0.31 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.69 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {525 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5824 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3500 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {17472 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {9905 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {24640 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {4480 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {9905 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {3500 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {525 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - 832}{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 {525 \, \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)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 
output
1/420*((525*sin(d*x + c)/(cos(d*x + c) + 1) - 5824*sin(d*x + c)^2/(cos(d*x 
 + c) + 1)^2 + 3500*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 17472*sin(d*x + 
c)^4/(cos(d*x + c) + 1)^4 + 9905*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 246 
40*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4480*sin(d*x + c)^8/(cos(d*x + c) 
 + 1)^8 - 9905*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 3500*sin(d*x + c)^11/ 
(cos(d*x + c) + 1)^11 - 525*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 832)/( 
a^2 + 7*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 21*a^2*sin(d*x + c)^4/(c 
os(d*x + c) + 1)^4 + 35*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 35*a^2*s 
in(d*x + c)^8/(cos(d*x + c) + 1)^8 + 21*a^2*sin(d*x + c)^10/(cos(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)^1 
4/(cos(d*x + c) + 1)^14) - 525*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 
)/d
 
3.5.20.8 Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {525 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 3500 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 9905 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 24640 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 9905 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 17472 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3500 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5824 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 832\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7} a^{2}}}{840 \, d} \]

input
integrate(cos(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 
output
-1/840*(525*(d*x + c)/a^2 + 2*(525*tan(1/2*d*x + 1/2*c)^13 + 3500*tan(1/2* 
d*x + 1/2*c)^11 + 9905*tan(1/2*d*x + 1/2*c)^9 + 4480*tan(1/2*d*x + 1/2*c)^ 
8 + 24640*tan(1/2*d*x + 1/2*c)^6 - 9905*tan(1/2*d*x + 1/2*c)^5 + 17472*tan 
(1/2*d*x + 1/2*c)^4 - 3500*tan(1/2*d*x + 1/2*c)^3 + 5824*tan(1/2*d*x + 1/2 
*c)^2 - 525*tan(1/2*d*x + 1/2*c) + 832)/((tan(1/2*d*x + 1/2*c)^2 + 1)^7*a^ 
2))/d
 
3.5.20.9 Mupad [B] (verification not implemented)

Time = 13.03 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^4(c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5\,x}{8\,a^2}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}+\frac {283\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {176\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-\frac {283\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {208\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}-\frac {25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {208\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {208}{105}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

input
int((cos(c + d*x)^4*sin(c + d*x)^5)/(a + a*sin(c + d*x))^2,x)
 
output
- (5*x)/(8*a^2) - ((208*tan(c/2 + (d*x)/2)^2)/15 - (5*tan(c/2 + (d*x)/2))/ 
4 - (25*tan(c/2 + (d*x)/2)^3)/3 + (208*tan(c/2 + (d*x)/2)^4)/5 - (283*tan( 
c/2 + (d*x)/2)^5)/12 + (176*tan(c/2 + (d*x)/2)^6)/3 + (32*tan(c/2 + (d*x)/ 
2)^8)/3 + (283*tan(c/2 + (d*x)/2)^9)/12 + (25*tan(c/2 + (d*x)/2)^11)/3 + ( 
5*tan(c/2 + (d*x)/2)^13)/4 + 208/105)/(a^2*d*(tan(c/2 + (d*x)/2)^2 + 1)^7)