Integrand size = 32, antiderivative size = 89 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \] Output:
-A*x/a^3-2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3+7/5*A*cos(d*x+c)/a^3/d/(1 +sin(d*x+c))^2-13/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 4.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A \sec (c+d x) \sqrt {1-\sin (c+d x)} \left (20 \sqrt {2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1+\sin (c+d x))+\sqrt {1-\sin (c+d x)} \left (-4+3 \sin (c+d x)+\sin ^2(c+d x)\right )\right )}{15 a^3 d (1+\sin (c+d x))^2} \] Input:
Integrate[(Sin[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
(A*Sec[c + d*x]*Sqrt[1 - Sin[c + d*x]]*(20*Sqrt[2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 + Sin[c + d*x])/2]*(1 + Sin[c + d*x]) + Sqrt[1 - Sin[c + d *x]]*(-4 + 3*Sin[c + d*x] + Sin[c + d*x]^2)))/(15*a^3*d*(1 + Sin[c + d*x]) ^2)
Time = 0.40 (sec) , antiderivative size = 89, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3445 |
| \(\displaystyle \int \left (\frac {4 A}{a^3 (\sin (c+d x)+1)}-\frac {5 A}{a^3 (\sin (c+d x)+1)^2}+\frac {2 A}{a^3 (\sin (c+d x)+1)^3}-\frac {A}{a^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac {7 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac {A x}{a^3}\) |
Input:
Int[(Sin[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]
Output:
-((A*x)/a^3) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (7*A*Co s[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^2) - (13*A*Cos[c + d*x])/(5*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]
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(-\frac {A x}{a^{3}}-\frac {2 \left (-75 A \,{\mathrm e}^{2 i \left (d x +c \right )}+55 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 i A \,{\mathrm e}^{i \left (d x +c \right )}+13 A \right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}\) | \(85\) |
| derivativedivides | \(\frac {8 A \left (-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}\right )}{d \,a^{3}}\) | \(96\) |
| default | \(\frac {8 A \left (-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}\right )}{d \,a^{3}}\) | \(96\) |
| parallelrisch | \(-\frac {A \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} d x +\left (5 d x +2\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (10 d x +10\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (10 d x +22\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (5 d x +14\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+d x +\frac {16}{5}\right )}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(112\) |
| norman | \(\frac {-\frac {A x}{a}-\frac {16 A}{5 a d}-\frac {5 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {13 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {25 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a}-\frac {38 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {46 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}-\frac {46 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {38 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a}-\frac {25 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a}-\frac {13 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a}-\frac {5 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a}-\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a}-\frac {376 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5 a d}-\frac {388 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 a d}-\frac {158 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 a d}-\frac {28 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a d}-\frac {52 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {72 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {44 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {10 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}-\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a d}-\frac {14 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(444\) |
Input:
int(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
-A*x/a^3-2/5*(-75*A*exp(2*I*(d*x+c))+55*I*A*exp(3*I*(d*x+c))+20*A*exp(4*I* (d*x+c))-45*I*A*exp(I*(d*x+c))+13*A)/d/a^3/(exp(I*(d*x+c))+I)^5
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (83) = 166\).
Time = 0.08 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.29 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\left (5 \, A d x + 13 \, A\right )} \cos \left (d x + c\right )^{3} - 20 \, A d x + 3 \, {\left (5 \, A d x - 2 \, A\right )} \cos \left (d x + c\right )^{2} - {\left (10 \, A d x + 21 \, A\right )} \cos \left (d x + c\right ) - {\left (20 \, A d x - {\left (5 \, A d x - 13 \, A\right )} \cos \left (d x + c\right )^{2} + {\left (10 \, A d x + 19 \, A\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right ) - 2 \, A}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="f ricas")
Output:
-1/5*((5*A*d*x + 13*A)*cos(d*x + c)^3 - 20*A*d*x + 3*(5*A*d*x - 2*A)*cos(d *x + c)^2 - (10*A*d*x + 21*A)*cos(d*x + c) - (20*A*d*x - (5*A*d*x - 13*A)* cos(d*x + c)^2 + (10*A*d*x + 19*A)*cos(d*x + c) - 2*A)*sin(d*x + c) - 2*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*cos(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. 1268 vs. \(2 (85) = 170\).
Time = 8.27 (sec) , antiderivative size = 1268, normalized size of antiderivative = 14.25 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)**2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((-5*A*d*x*tan(c/2 + d*x/2)**5/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25 *a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*ta n(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 25*A*d*x*tan( c/2 + d*x/2)**4/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2) **4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a **3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 50*A*d*x*tan(c/2 + d*x/2)**3/(5*a**3* d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 50*A*d*x*tan(c/2 + d*x/2)**2/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d* tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 25*A*d*x*ta n(c/2 + d*x/2)/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)* *4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a* *3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 5*A*d*x/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3* d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 10*A*tan( c/2 + d*x/2)**4/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2) **4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a **3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 50*A*tan(c/2 + d*x/2)**3/(5*a**3*d*ta n(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 +...
Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (83) = 166\).
Time = 0.12 (sec) , antiderivative size = 392, normalized size of antiderivative = 4.40 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (A {\left (\frac {\frac {95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 22}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {2 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}\right )}}{15 \, d} \] Input:
integrate(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="m axima")
Output:
-2/15*(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 + 10*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) + 2*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/(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 + 10*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))/d
Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {5 \, {\left (d x + c\right )} A}{a^{3}} + \frac {2 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 55 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{5 \, d} \] Input:
integrate(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="g iac")
Output:
-1/5*(5*(d*x + c)*A/a^3 + 2*(5*A*tan(1/2*d*x + 1/2*c)^4 + 25*A*tan(1/2*d*x + 1/2*c)^3 + 55*A*tan(1/2*d*x + 1/2*c)^2 + 35*A*tan(1/2*d*x + 1/2*c) + 8* A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 36.36 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.00 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {\left (5\,A\,\left (c+d\,x\right )-\frac {A\,\left (25\,c+25\,d\,x+10\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (10\,A\,\left (c+d\,x\right )-\frac {A\,\left (50\,c+50\,d\,x+50\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (10\,A\,\left (c+d\,x\right )-\frac {A\,\left (50\,c+50\,d\,x+110\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (5\,A\,\left (c+d\,x\right )-\frac {A\,\left (25\,c+25\,d\,x+70\right )}{5}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,\left (c+d\,x\right )-\frac {A\,\left (5\,c+5\,d\,x+16\right )}{5}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5}-\frac {A\,x}{a^3} \] Input:
int((sin(c + d*x)^2*(A - A*sin(c + d*x)))/(a + a*sin(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)*(5*A*(c + d*x) - (A*(25*c + 25*d*x + 70))/5) + tan(c/2 + (d*x)/2)^4*(5*A*(c + d*x) - (A*(25*c + 25*d*x + 10))/5) + tan(c/2 + (d* x)/2)^3*(10*A*(c + d*x) - (A*(50*c + 50*d*x + 50))/5) + tan(c/2 + (d*x)/2) ^2*(10*A*(c + d*x) - (A*(50*c + 50*d*x + 110))/5) + A*(c + d*x) - (A*(5*c + 5*d*x + 16))/5)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^5) - (A*x)/a^3
Time = 0.18 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.27 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} d x +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} d x -50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} d x -30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} d x -90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x -60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-5 d x -14}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \] Input:
int(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)
Output:
( - 5*tan((c + d*x)/2)**5*d*x + 2*tan((c + d*x)/2)**5 - 25*tan((c + d*x)/2 )**4*d*x - 50*tan((c + d*x)/2)**3*d*x - 30*tan((c + d*x)/2)**3 - 50*tan((c + d*x)/2)**2*d*x - 90*tan((c + d*x)/2)**2 - 25*tan((c + d*x)/2)*d*x - 60* tan((c + d*x)/2) - 5*d*x - 14)/(5*a**2*d*(tan((c + d*x)/2)**5 + 5*tan((c + d*x)/2)**4 + 10*tan((c + d*x)/2)**3 + 10*tan((c + d*x)/2)**2 + 5*tan((c + d*x)/2) + 1))