Integrand size = 29, antiderivative size = 109 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {51 x}{8 a^3}+\frac {7 \cos (c+d x)}{a^3 d}-\frac {\cos ^3(c+d x)}{a^3 d}-\frac {19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac {4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))} \] Output:
51/8*x/a^3+7*cos(d*x+c)/a^3/d-cos(d*x+c)^3/a^3/d-19/8*cos(d*x+c)*sin(d*x+c )/a^3/d-1/4*cos(d*x+c)*sin(d*x+c)^3/a^3/d+4*cos(d*x+c)/a^3/d/(1+sin(d*x+c) )
Time = 1.18 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2040 d x \cos \left (\frac {d x}{2}\right )+997 \cos \left (c+\frac {d x}{2}\right )+800 \cos \left (c+\frac {3 d x}{2}\right )+160 \cos \left (3 c+\frac {5 d x}{2}\right )-35 \cos \left (3 c+\frac {7 d x}{2}\right )-5 \cos \left (5 c+\frac {9 d x}{2}\right )-3563 \sin \left (\frac {d x}{2}\right )+2040 d x \sin \left (c+\frac {d x}{2}\right )+800 \sin \left (2 c+\frac {3 d x}{2}\right )-160 \sin \left (2 c+\frac {5 d x}{2}\right )-35 \sin \left (4 c+\frac {7 d x}{2}\right )+5 \sin \left (4 c+\frac {9 d x}{2}\right )}{320 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
(2040*d*x*Cos[(d*x)/2] + 997*Cos[c + (d*x)/2] + 800*Cos[c + (3*d*x)/2] + 1 60*Cos[3*c + (5*d*x)/2] - 35*Cos[3*c + (7*d*x)/2] - 5*Cos[5*c + (9*d*x)/2] - 3563*Sin[(d*x)/2] + 2040*d*x*Sin[c + (d*x)/2] + 800*Sin[2*c + (3*d*x)/2 ] - 160*Sin[2*c + (5*d*x)/2] - 35*Sin[4*c + (7*d*x)/2] + 5*Sin[4*c + (9*d* x)/2])/(320*a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2]))
Time = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3351, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^4(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^4}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sin (c+d x) (a-a \sin (c+d x))^3 \tan ^2(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^3 (a-a \sin (c+d x))^3}{\cos (c+d x)^2}dx}{a^6}\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (a \sin ^4(c+d x)-3 a \sin ^3(c+d x)+4 a \sin ^2(c+d x)-4 a \sin (c+d x)+4 a-\frac {4 a}{\sin (c+d x)+1}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a \cos ^3(c+d x)}{d}+\frac {7 a \cos (c+d x)}{d}-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {19 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {4 a \cos (c+d x)}{d (\sin (c+d x)+1)}+\frac {51 a x}{8}}{a^4}\) |
Input:
Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]
Output:
((51*a*x)/8 + (7*a*Cos[c + d*x])/d - (a*Cos[c + d*x]^3)/d - (19*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d) + (4*a*Co s[c + d*x])/(d*(1 + Sin[c + d*x])))/a^4
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06
| method | result | size |
| risch | \(\frac {51 x}{8 a^{3}}+\frac {25 \,{\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {25 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {8}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {\sin \left (4 d x +4 c \right )}{32 d \,a^{3}}-\frac {\cos \left (3 d x +3 c \right )}{4 d \,a^{3}}-\frac {5 \sin \left (2 d x +2 c \right )}{4 d \,a^{3}}\) | \(115\) |
| derivativedivides | \(\frac {\frac {8 \left (\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{32}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{32}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{32}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}+\frac {3}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {51 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {16}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2}}{a^{3} d}\) | \(143\) |
| default | \(\frac {\frac {8 \left (\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{32}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{32}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{32}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}+\frac {3}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {51 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {16}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2}}{a^{3} d}\) | \(143\) |
| parallelrisch | \(\frac {\left (408 d x +280\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (408 d x -632\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+160 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-32 \sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-7 \sin \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )+\sin \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )+160 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+32 \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-7 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )-\cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )}{64 d \,a^{3} \left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )+\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(149\) |
Input:
int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
51/8*x/a^3+25/8/d/a^3*exp(I*(d*x+c))+25/8/d/a^3*exp(-I*(d*x+c))+8/d/a^3/(e xp(I*(d*x+c))+I)+1/32/d/a^3*sin(4*d*x+4*c)-1/4/d/a^3*cos(3*d*x+3*c)-5/4/d/ a^3*sin(2*d*x+2*c)
Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{4} - 15 \, \cos \left (d x + c\right )^{3} - 51 \, d x - {\left (51 \, d x + 67\right )} \cos \left (d x + c\right ) - 56 \, \cos \left (d x + c\right )^{2} - {\left (2 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{3} + 51 \, d x - 21 \, \cos \left (d x + c\right )^{2} + 35 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right ) - 32}{8 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/8*(2*cos(d*x + c)^5 + 8*cos(d*x + c)^4 - 15*cos(d*x + c)^3 - 51*d*x - ( 51*d*x + 67)*cos(d*x + c) - 56*cos(d*x + c)^2 - (2*cos(d*x + c)^4 - 6*cos( d*x + c)^3 + 51*d*x - 21*cos(d*x + c)^2 + 35*cos(d*x + c) - 32)*sin(d*x + c) - 32)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 3578 vs. \(2 (100) = 200\).
Time = 65.13 (sec) , antiderivative size = 3578, normalized size of antiderivative = 32.83 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((51*d*x*tan(c/2 + d*x/2)**9/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a* *3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c /2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2 )**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a **3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 51*d*x*tan(c/2 + d*x/2)**8/(8*a**3*d* tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d *x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d *tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 204*d*x*tan (c/2 + d*x/2)**7/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2) **8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a **3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan( c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2 ) + 8*a**3*d) + 204*d*x*tan(c/2 + d*x/2)**6/(8*a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**7 + 32*a**3*d *tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**5 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**3 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d*tan(c/2 + d*x/2) + 8*a**3*d) + 306*d*x*tan(c/2 + d*x/2)**5/(8* a**3*d*tan(c/2 + d*x/2)**9 + 8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan( c/2 + d*x/2)**7 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d...
Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (103) = 206\).
Time = 0.12 (sec) , antiderivative size = 398, normalized size of antiderivative = 3.65 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {29 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {269 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {133 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {309 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {171 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {187 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {51 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {51 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 80}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac {51 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/4*((29*sin(d*x + c)/(cos(d*x + c) + 1) + 269*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 133*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 309*sin(d*x + c)^4/(c os(d*x + c) + 1)^4 + 171*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 187*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 51*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 51 *sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 80)/(a^3 + a^3*sin(d*x + c)/(cos(d* x + c) + 1) + 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6* a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^3*sin(d*x + c)^8 /(cos(d*x + c) + 1)^8 + a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 51*arct an(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d
Time = 0.17 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {51 \, {\left (d x + c\right )}}{a^{3}} + \frac {64}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} + \frac {2 \, {\left (19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 32 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 144 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \] Input:
integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/8*(51*(d*x + c)/a^3 + 64/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(19*tan(1/ 2*d*x + 1/2*c)^7 + 32*tan(1/2*d*x + 1/2*c)^6 + 27*tan(1/2*d*x + 1/2*c)^5 + 144*tan(1/2*d*x + 1/2*c)^4 - 27*tan(1/2*d*x + 1/2*c)^3 + 160*tan(1/2*d*x + 1/2*c)^2 - 19*tan(1/2*d*x + 1/2*c) + 48)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4 *a^3))/d
Time = 20.12 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {51\,x}{8\,a^3}+\frac {\frac {51\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+\frac {51\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {187\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+\frac {171\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {309\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {133\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {269\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {29\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+20}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \] Input:
int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + a*sin(c + d*x))^3,x)
Output:
(51*x)/(8*a^3) + ((29*tan(c/2 + (d*x)/2))/4 + (269*tan(c/2 + (d*x)/2)^2)/4 + (133*tan(c/2 + (d*x)/2)^3)/4 + (309*tan(c/2 + (d*x)/2)^4)/4 + (171*tan( c/2 + (d*x)/2)^5)/4 + (187*tan(c/2 + (d*x)/2)^6)/4 + (51*tan(c/2 + (d*x)/2 )^7)/4 + (51*tan(c/2 + (d*x)/2)^8)/4 + 20)/(a^3*d*(tan(c/2 + (d*x)/2) + 1) *(tan(c/2 + (d*x)/2)^2 + 1)^4)
Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+11 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-29 \cos \left (d x +c \right ) \sin \left (d x +c \right )+51 \cos \left (d x +c \right ) d x -102 \cos \left (d x +c \right )+2 \sin \left (d x +c \right )^{5}-8 \sin \left (d x +c \right )^{4}+17 \sin \left (d x +c \right )^{3}-40 \sin \left (d x +c \right )^{2}-51 \sin \left (d x +c \right ) d x -29 \sin \left (d x +c \right )-51 d x +102}{8 a^{3} d \left (\cos \left (d x +c \right )-\sin \left (d x +c \right )-1\right )} \] Input:
int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)
Output:
(2*cos(c + d*x)*sin(c + d*x)**4 - 6*cos(c + d*x)*sin(c + d*x)**3 + 11*cos( c + d*x)*sin(c + d*x)**2 - 29*cos(c + d*x)*sin(c + d*x) + 51*cos(c + d*x)* d*x - 102*cos(c + d*x) + 2*sin(c + d*x)**5 - 8*sin(c + d*x)**4 + 17*sin(c + d*x)**3 - 40*sin(c + d*x)**2 - 51*sin(c + d*x)*d*x - 29*sin(c + d*x) - 5 1*d*x + 102)/(8*a**3*d*(cos(c + d*x) - sin(c + d*x) - 1))