Integrand size = 32, antiderivative size = 103 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {4 A x}{a^3}+\frac {A \cos (c+d x)}{a^3 d}+\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {31 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac {104 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))} \] Output:
4*A*x/a^3+A*cos(d*x+c)/a^3/d+2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3-31/15 *A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2+104/15*A*cos(d*x+c)/a^3/d/(1+sin(d*x+ c))
Time = 4.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.50 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A \sec (c+d x) \left (-120 \arcsin \left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \sqrt {1-\sin (c+d x)} \sqrt {a (1+\sin (c+d x))}+\sqrt {a} \left (-94-128 \sin (c+d x)+73 \sin ^2(c+d x)+134 \sin ^3(c+d x)+15 \sin ^4(c+d x)\right )\right )}{15 a^{7/2} d (1+\sin (c+d x))^2} \] Input:
Integrate[(Sin[c + d*x]^3*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
-1/15*(A*Sec[c + d*x]*(-120*ArcSin[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqr t[a])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Sqrt[1 - Sin[c + d*x]]*Sqrt [a*(1 + Sin[c + d*x])] + Sqrt[a]*(-94 - 128*Sin[c + d*x] + 73*Sin[c + d*x] ^2 + 134*Sin[c + d*x]^3 + 15*Sin[c + d*x]^4)))/(a^(7/2)*d*(1 + Sin[c + d*x ])^2)
Time = 0.41 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3042, 3445, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3445 |
| \(\displaystyle \int \left (-\frac {A \sin (c+d x)}{a^3}-\frac {9 A}{a^3 (\sin (c+d x)+1)}+\frac {7 A}{a^3 (\sin (c+d x)+1)^2}-\frac {2 A}{a^3 (\sin (c+d x)+1)^3}+\frac {4 A}{a^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {A \cos (c+d x)}{a^3 d}+\frac {104 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}-\frac {31 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac {4 A x}{a^3}\) |
Input:
Int[(Sin[c + d*x]^3*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
(4*A*x)/a^3 + (A*Cos[c + d*x])/(a^3*d) + (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (31*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) + ( 104*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x]))
Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[si n[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; FreeQ[{ a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ [m] && IntegerQ[n]
Time = 0.68 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.12
| method | result | size |
| derivativedivides | \(\frac {16 A \left (\frac {1}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2}+\frac {1}{8+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{d \,a^{3}}\) | \(115\) |
| default | \(\frac {16 A \left (\frac {1}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2}+\frac {1}{8+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{d \,a^{3}}\) | \(115\) |
| risch | \(\frac {4 A x}{a^{3}}+\frac {A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}+\frac {A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}+\frac {2 A \left (435 i {\mathrm e}^{3 i \left (d x +c \right )}+135 \,{\mathrm e}^{4 i \left (d x +c \right )}-385 i {\mathrm e}^{i \left (d x +c \right )}-605 \,{\mathrm e}^{2 i \left (d x +c \right )}+104\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}\) | \(116\) |
| parallelrisch | \(\frac {4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} x d +\left (5 d x +2\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (11 d x +10\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (15 d x +\frac {64}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (15 d x +\frac {77}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (11 d x +\frac {367}{15}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (5 d x +\frac {41}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+d x +\frac {47}{15}\right ) A}{d \,a^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(163\) |
| norman | \(\frac {\frac {20 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}+\frac {56 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {120 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a}+\frac {204 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {284 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}+\frac {336 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {336 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a}+\frac {284 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a}+\frac {204 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a}+\frac {120 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a}+\frac {56 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a}+\frac {20 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{a}+\frac {4 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{a}+\frac {4 A x}{a}+\frac {164 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a d}+\frac {2032 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 a d}+\frac {800 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {6248 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{15 a d}+\frac {512 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {8552 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 a d}+\frac {1448 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}+\frac {5668 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{15 a d}+\frac {668 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 a d}+\frac {328 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 a d}+\frac {40 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a d}+\frac {8 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{a d}+\frac {188 A}{15 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(520\) |
Input:
int(sin(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
16/d*A/a^3*(1/5/(tan(1/2*d*x+1/2*c)+1)^5-1/2/(tan(1/2*d*x+1/2*c)+1)^4+1/12 /(tan(1/2*d*x+1/2*c)+1)^3+3/8/(tan(1/2*d*x+1/2*c)+1)^2+1/2/(tan(1/2*d*x+1/ 2*c)+1)+1/8/(1+tan(1/2*d*x+1/2*c)^2)+1/2*arctan(tan(1/2*d*x+1/2*c)))
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (97) = 194\).
Time = 0.09 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.18 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {15 \, A \cos \left (d x + c\right )^{4} + {\left (60 \, A d x + 149 \, A\right )} \cos \left (d x + c\right )^{3} - 240 \, A d x + {\left (180 \, A d x - 103 \, A\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (40 \, A d x + 81 \, A\right )} \cos \left (d x + c\right ) + {\left (15 \, A \cos \left (d x + c\right )^{3} - 240 \, A d x + 2 \, {\left (30 \, A d x - 67 \, A\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (40 \, A d x + 79 \, A\right )} \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right ) - 6 \, A}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="f ricas")
Output:
1/15*(15*A*cos(d*x + c)^4 + (60*A*d*x + 149*A)*cos(d*x + c)^3 - 240*A*d*x + (180*A*d*x - 103*A)*cos(d*x + c)^2 - 3*(40*A*d*x + 81*A)*cos(d*x + c) + (15*A*cos(d*x + c)^3 - 240*A*d*x + 2*(30*A*d*x - 67*A)*cos(d*x + c)^2 - 3* (40*A*d*x + 79*A)*cos(d*x + c) + 6*A)*sin(d*x + c) - 6*A)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*c os(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 2290 vs. \(2 (100) = 200\).
Time = 14.48 (sec) , antiderivative size = 2290, normalized size of antiderivative = 22.23 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)**3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((60*A*d*x*tan(c/2 + d*x/2)**7/(15*a**3*d*tan(c/2 + d*x/2)**7 + 7 5*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d *tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 300*A*d*x*tan(c/2 + d*x/2)**6/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)** 6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225* a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*ta n(c/2 + d*x/2) + 15*a**3*d) + 660*A*d*x*tan(c/2 + d*x/2)**5/(15*a**3*d*tan (c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d* x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 900*A*d*x*tan(c/2 + d*x/2)**4/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3* d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/ 2 + d*x/2)**4 + 225*a**3*d*tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/ 2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 900*A*d*x*tan(c/2 + d*x/ 2)**3/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165 *a**3*d*tan(c/2 + d*x/2)**5 + 225*a**3*d*tan(c/2 + d*x/2)**4 + 225*a**3*d* tan(c/2 + d*x/2)**3 + 165*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) + 15*a**3*d) + 660*A*d*x*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/2 + d*x/2)**7 + 75*a**3*d*tan(c/2 + d*x/2)**6 + 165*a**3*d*tan(c/2 + d*x/2)...
Leaf count of result is larger than twice the leaf count of optimal. 543 vs. \(2 (97) = 194\).
Time = 0.12 (sec) , antiderivative size = 543, normalized size of antiderivative = 5.27 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="m axima")
Output:
2/15*(3*A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 189*sin(d*x + c)^2/(cos( d*x + c) + 1)^2 + 200*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 160*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 75*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*si n(d*x + c)^6/(cos(d*x + c) + 1)^6 + 24)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 11*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) + 15*arctan(sin(d* x + c)/(cos(d*x + c) + 1))/a^3) + A*((95*sin(d*x + c)/(cos(d*x + c) + 1) + 145*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 75*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 22)/(a^3 + 5*a^3*sin(d* x + c)/(cos(d*x + c) + 1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1 0*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {30 \, {\left (d x + c\right )} A}{a^{3}} + \frac {15 \, A}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} + \frac {60 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 285 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 505 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 335 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 79 \, A}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}\right )}}{15 \, d} \] Input:
integrate(sin(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="g iac")
Output:
2/15*(30*(d*x + c)*A/a^3 + 15*A/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) + (60*A *tan(1/2*d*x + 1/2*c)^4 + 285*A*tan(1/2*d*x + 1/2*c)^3 + 505*A*tan(1/2*d*x + 1/2*c)^2 + 335*A*tan(1/2*d*x + 1/2*c) + 79*A)/(a^3*(tan(1/2*d*x + 1/2*c ) + 1)^5))/d
Time = 38.48 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.53 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {4\,A\,x}{a^3}-\frac {\left (20\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (75\,c+75\,d\,x+30\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (44\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (165\,c+165\,d\,x+150\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (60\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (225\,c+225\,d\,x+320\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (60\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (225\,c+225\,d\,x+385\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (44\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (165\,c+165\,d\,x+367\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (20\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (75\,c+75\,d\,x+205\right )}{15}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,A\,\left (c+d\,x\right )-\frac {4\,A\,\left (15\,c+15\,d\,x+47\right )}{15}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((sin(c + d*x)^3*(A - A*sin(c + d*x)))/(a + a*sin(c + d*x))^3,x)
Output:
(4*A*x)/a^3 - (tan(c/2 + (d*x)/2)*(20*A*(c + d*x) - (4*A*(75*c + 75*d*x + 205))/15) + tan(c/2 + (d*x)/2)^6*(20*A*(c + d*x) - (4*A*(75*c + 75*d*x + 3 0))/15) + tan(c/2 + (d*x)/2)^5*(44*A*(c + d*x) - (4*A*(165*c + 165*d*x + 1 50))/15) + tan(c/2 + (d*x)/2)^2*(44*A*(c + d*x) - (4*A*(165*c + 165*d*x + 367))/15) + tan(c/2 + (d*x)/2)^4*(60*A*(c + d*x) - (4*A*(225*c + 225*d*x + 320))/15) + tan(c/2 + (d*x)/2)^3*(60*A*(c + d*x) - (4*A*(225*c + 225*d*x + 385))/15) + 4*A*(c + d*x) - (4*A*(15*c + 15*d*x + 47))/15)/(a^3*d*(tan(c /2 + (d*x)/2) + 1)^5*(tan(c/2 + (d*x)/2)^2 + 1))
Time = 0.17 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.44 \[ \int \frac {\sin ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x -79 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -82 \cos \left (d x +c \right ) \sin \left (d x +c \right )+60 \cos \left (d x +c \right ) d x -24 \cos \left (d x +c \right )-15 \sin \left (d x +c \right )^{4}-60 \sin \left (d x +c \right )^{3} d x -204 \sin \left (d x +c \right )^{3}-180 \sin \left (d x +c \right )^{2} d x -283 \sin \left (d x +c \right )^{2}-180 \sin \left (d x +c \right ) d x -82 \sin \left (d x +c \right )-60 d x +24}{15 a^{2} d \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+\cos \left (d x +c \right )-\sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )^{2}-3 \sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)
Output:
( - 15*cos(c + d*x)*sin(c + d*x)**3 + 60*cos(c + d*x)*sin(c + d*x)**2*d*x - 79*cos(c + d*x)*sin(c + d*x)**2 + 120*cos(c + d*x)*sin(c + d*x)*d*x - 82 *cos(c + d*x)*sin(c + d*x) + 60*cos(c + d*x)*d*x - 24*cos(c + d*x) - 15*si n(c + d*x)**4 - 60*sin(c + d*x)**3*d*x - 204*sin(c + d*x)**3 - 180*sin(c + d*x)**2*d*x - 283*sin(c + d*x)**2 - 180*sin(c + d*x)*d*x - 82*sin(c + d*x ) - 60*d*x + 24)/(15*a**2*d*(cos(c + d*x)*sin(c + d*x)**2 + 2*cos(c + d*x) *sin(c + d*x) + cos(c + d*x) - sin(c + d*x)**3 - 3*sin(c + d*x)**2 - 3*sin (c + d*x) - 1))