Integrand size = 36, antiderivative size = 93 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=\frac {(A+B) \sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {(4 A-B) \tan (e+f x)}{5 a^2 c^3 f}+\frac {(4 A-B) \tan ^3(e+f x)}{15 a^2 c^3 f} \] Output:
1/5*(A+B)*sec(f*x+e)^3/a^2/f/(c^3-c^3*sin(f*x+e))+1/5*(4*A-B)*tan(f*x+e)/a ^2/c^3/f+1/15*(4*A-B)*tan(f*x+e)^3/a^2/c^3/f
Leaf count is larger than twice the leaf count of optimal. \(237\) vs. \(2(93)=186\).
Time = 4.98 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.55 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-240 B+54 (A+B) \cos (e+f x)-32 (4 A-B) \cos (2 (e+f x))+18 A \cos (3 (e+f x))+18 B \cos (3 (e+f x))-64 A \cos (4 (e+f x))+16 B \cos (4 (e+f x))-384 A \sin (e+f x)+96 B \sin (e+f x)-18 A \sin (2 (e+f x))-18 B \sin (2 (e+f x))-128 A \sin (3 (e+f x))+32 B \sin (3 (e+f x))-9 A \sin (4 (e+f x))-9 B \sin (4 (e+f x)))}{960 a^2 c^3 f (-1+\sin (e+f x))^3 (1+\sin (e+f x))^2} \] Input:
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x] )^3),x]
Output:
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(-240*B + 54*(A + B)*Cos[e + f*x] - 32*(4*A - B)*Cos[2*(e + f*x)] + 18* A*Cos[3*(e + f*x)] + 18*B*Cos[3*(e + f*x)] - 64*A*Cos[4*(e + f*x)] + 16*B* Cos[4*(e + f*x)] - 384*A*Sin[e + f*x] + 96*B*Sin[e + f*x] - 18*A*Sin[2*(e + f*x)] - 18*B*Sin[2*(e + f*x)] - 128*A*Sin[3*(e + f*x)] + 32*B*Sin[3*(e + f*x)] - 9*A*Sin[4*(e + f*x)] - 9*B*Sin[4*(e + f*x)]))/(960*a^2*c^3*f*(-1 + Sin[e + f*x])^3*(1 + Sin[e + f*x])^2)
Time = 0.56 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 3446, 3042, 3338, 3042, 4254, 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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \frac {\sec ^4(e+f x) (A+B \sin (e+f x))}{c-c \sin (e+f x)}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A+B \sin (e+f x)}{\cos (e+f x)^4 (c-c \sin (e+f x))}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle \frac {\frac {(4 A-B) \int \sec ^4(e+f x)dx}{5 c}+\frac {(A+B) \sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(4 A-B) \int \csc \left (e+f x+\frac {\pi }{2}\right )^4dx}{5 c}+\frac {(A+B) \sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}}{a^2 c^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {(A+B) \sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}-\frac {(4 A-B) \int \left (\tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{5 c f}}{a^2 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {(A+B) \sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}-\frac {(4 A-B) \left (-\frac {1}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{5 c f}}{a^2 c^2}\) |
Input:
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3),x ]
Output:
(((A + B)*Sec[e + f*x]^3)/(5*f*(c - c*Sin[e + f*x])) - ((4*A - B)*(-Tan[e + f*x] - Tan[e + f*x]^3/3))/(5*c*f))/(a^2*c^2)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) )), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 ]) && NeQ[2*m + p + 1, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Result contains complex when optimal does not.
Time = 1.01 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.46
| method | result | size |
| risch | \(-\frac {4 i \left (24 i A \,{\mathrm e}^{3 i \left (f x +e \right )}-6 i B \,{\mathrm e}^{3 i \left (f x +e \right )}+15 B \,{\mathrm e}^{4 i \left (f x +e \right )}+8 i A \,{\mathrm e}^{i \left (f x +e \right )}+8 A \,{\mathrm e}^{2 i \left (f x +e \right )}-2 i B \,{\mathrm e}^{i \left (f x +e \right )}-2 B \,{\mathrm e}^{2 i \left (f x +e \right )}+4 A -B \right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,a^{2} c^{3}}\) | \(136\) |
| parallelrisch | \(\frac {-30 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+\left (30 A -30 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (10 A +20 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (-50 A -10 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-26 A -16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (42 A -18 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-18 A +12 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-6 A -6 B}{15 f \,a^{2} c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(171\) |
| derivativedivides | \(\frac {-\frac {2 \left (A +B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}+2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-\frac {A}{4}+\frac {B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{4}-\frac {B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {5 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{2} c^{3}}\) | \(183\) |
| default | \(\frac {-\frac {2 \left (A +B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}+2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-\frac {A}{4}+\frac {B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{4}-\frac {B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {5 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{2} c^{3}}\) | \(183\) |
| norman | \(\frac {\frac {6 A -4 B}{10 a f c}-\frac {4 \left (4 A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 a f c}-\frac {A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a f c}+\frac {\left (14 A -16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{10 a f c}+\frac {\left (6 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 a f c}+\frac {\left (2 A -8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 a f c}-\frac {2 \left (8 A -2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 a f c}-\frac {\left (16 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 a f c}+\frac {2 \left (8 A -2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{15 a f c}-\frac {\left (38 A +28 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{15 a f c}}{a \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(321\) |
Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x,method=_RETUR NVERBOSE)
Output:
-4/15*I*(24*I*A*exp(3*I*(f*x+e))-6*I*B*exp(3*I*(f*x+e))+15*B*exp(4*I*(f*x+ e))+8*I*A*exp(I*(f*x+e))+8*A*exp(2*I*(f*x+e))-2*I*B*exp(I*(f*x+e))-2*B*exp (2*I*(f*x+e))+4*A-B)/(exp(I*(f*x+e))+I)^3/(exp(I*(f*x+e))-I)^5/f/a^2/c^3
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (4 \, A - B\right )} \cos \left (f x + e\right )^{4} - {\left (4 \, A - B\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, {\left (4 \, A - B\right )} \cos \left (f x + e\right )^{2} + 4 \, A - B\right )} \sin \left (f x + e\right ) - A + 4 \, B}{15 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algori thm="fricas")
Output:
-1/15*(2*(4*A - B)*cos(f*x + e)^4 - (4*A - B)*cos(f*x + e)^2 + (2*(4*A - B )*cos(f*x + e)^2 + 4*A - B)*sin(f*x + e) - A + 4*B)/(a^2*c^3*f*cos(f*x + e )^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3)
Leaf count of result is larger than twice the leaf count of optimal. 2674 vs. \(2 (82) = 164\).
Time = 9.52 (sec) , antiderivative size = 2674, normalized size of antiderivative = 28.75 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**3,x)
Output:
Piecewise((-30*A*tan(e/2 + f*x/2)**7/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15 *a**2*c**3*f) + 30*A*tan(e/2 + f*x/2)**6/(15*a**2*c**3*f*tan(e/2 + f*x/2)* *8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)* *6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)* *3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 10*A*tan(e/2 + f*x/2)**5/(15*a**2*c**3*f*tan(e/2 + f*x /2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x /2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x /2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x /2) - 15*a**2*c**3*f) - 50*A*tan(e/2 + f*x/2)**4/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 26*A*tan(e/2 + f*x/2)**3/(15*a**2*c**3*f*tan(e /2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e /2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e /2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*ta...
Leaf count of result is larger than twice the leaf count of optimal. 651 vs. \(2 (88) = 176\).
Time = 0.06 (sec) , antiderivative size = 651, normalized size of antiderivative = 7.00 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algori thm="maxima")
Output:
2/15*(A*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 25*sin(f*x + e)^4/(co s(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 15*sin(f*x + e )^6/(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 3)/(a^ 2*c^3 - 2*a^2*c^3*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^2*c^3*sin(f*x + e) ^2/(cos(f*x + e) + 1)^2 + 6*a^2*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 6*a^2*c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^2*c^3*sin(f*x + e)^6/( cos(f*x + e) + 1)^6 + 2*a^2*c^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^2* c^3*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) - B*(6*sin(f*x + e)/(cos(f*x + e) + 1) - 9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 8*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 10*sin(f*x + e)^5/(c os(f*x + e) + 1)^5 - 15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3)/(a^2*c^3 - 2*a^2*c^3*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^2*c^3*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 + 6*a^2*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 6*a^2* c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^2*c^3*sin(f*x + e)^6/(cos(f* x + e) + 1)^6 + 2*a^2*c^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^2*c^3*si n(f*x + e)^8/(cos(f*x + e) + 1)^8))/f
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (88) = 176\).
Time = 0.23 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.38 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=-\frac {\frac {5 \, {\left (15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 24 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, A - 7 \, B\right )}}{a^{2} c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {165 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 45 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 480 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 60 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 70 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 400 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 20 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113 \, A + 13 \, B}{a^{2} c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{120 \, f} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algori thm="giac")
Output:
-1/120*(5*(15*A*tan(1/2*f*x + 1/2*e)^2 - 9*B*tan(1/2*f*x + 1/2*e)^2 + 24*A *tan(1/2*f*x + 1/2*e) - 12*B*tan(1/2*f*x + 1/2*e) + 13*A - 7*B)/(a^2*c^3*( tan(1/2*f*x + 1/2*e) + 1)^3) + (165*A*tan(1/2*f*x + 1/2*e)^4 + 45*B*tan(1/ 2*f*x + 1/2*e)^4 - 480*A*tan(1/2*f*x + 1/2*e)^3 - 60*B*tan(1/2*f*x + 1/2*e )^3 + 650*A*tan(1/2*f*x + 1/2*e)^2 + 70*B*tan(1/2*f*x + 1/2*e)^2 - 400*A*t an(1/2*f*x + 1/2*e) - 20*B*tan(1/2*f*x + 1/2*e) + 113*A + 13*B)/(a^2*c^3*( tan(1/2*f*x + 1/2*e) - 1)^5))/f
Time = 36.99 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.97 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=\frac {\left (\frac {8\,A}{15}-\frac {2\,B}{15}-\frac {16\,A\,\sin \left (e+f\,x\right )}{15}+\frac {4\,B\,\sin \left (e+f\,x\right )}{15}\right )\,{\cos \left (e+f\,x\right )}^2+\frac {2\,A}{15}-\frac {8\,B}{15}-\frac {8\,A\,\sin \left (e+f\,x\right )}{15}+\frac {2\,B\,\sin \left (e+f\,x\right )}{15}}{a^2\,c^3\,f\,\left (2\,{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )-2\,{\cos \left (e+f\,x\right )}^3\right )}-\frac {\frac {2\,A}{5}+\frac {2\,B}{5}-\frac {2\,A\,\sin \left (e+f\,x\right )}{5}-\frac {2\,B\,\sin \left (e+f\,x\right )}{5}}{a^2\,c^3\,f\,\left (2\,\sin \left (e+f\,x\right )-2\right )}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {16\,A}{15}-\frac {4\,B}{15}\right )}{a^2\,c^3\,f\,\left (2\,\sin \left (e+f\,x\right )-2\right )} \] Input:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^3),x )
Output:
((2*A)/15 - (8*B)/15 - (8*A*sin(e + f*x))/15 + (2*B*sin(e + f*x))/15 + cos (e + f*x)^2*((8*A)/15 - (2*B)/15 - (16*A*sin(e + f*x))/15 + (4*B*sin(e + f *x))/15))/(a^2*c^3*f*(2*cos(e + f*x)^3*sin(e + f*x) - 2*cos(e + f*x)^3)) - ((2*A)/5 + (2*B)/5 - (2*A*sin(e + f*x))/5 - (2*B*sin(e + f*x))/5)/(a^2*c^ 3*f*(2*sin(e + f*x) - 2)) - (cos(e + f*x)*((16*A)/15 - (4*B)/15))/(a^2*c^3 *f*(2*sin(e + f*x) - 2))
Time = 0.19 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.75 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx=\frac {-12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a +3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a -3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +12 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -12 \cos \left (f x +e \right ) a +3 \cos \left (f x +e \right ) b +8 \sin \left (f x +e \right )^{4} a -2 \sin \left (f x +e \right )^{4} b -8 \sin \left (f x +e \right )^{3} a +2 \sin \left (f x +e \right )^{3} b -12 \sin \left (f x +e \right )^{2} a +3 \sin \left (f x +e \right )^{2} b +12 a \sin \left (f x +e \right )-3 \sin \left (f x +e \right ) b +3 a +3 b}{15 \cos \left (f x +e \right ) a^{2} c^{3} f \left (\sin \left (f x +e \right )^{3}-\sin \left (f x +e \right )^{2}-\sin \left (f x +e \right )+1\right )} \] Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x)
Output:
( - 12*cos(e + f*x)*sin(e + f*x)**3*a + 3*cos(e + f*x)*sin(e + f*x)**3*b + 12*cos(e + f*x)*sin(e + f*x)**2*a - 3*cos(e + f*x)*sin(e + f*x)**2*b + 12 *cos(e + f*x)*sin(e + f*x)*a - 3*cos(e + f*x)*sin(e + f*x)*b - 12*cos(e + f*x)*a + 3*cos(e + f*x)*b + 8*sin(e + f*x)**4*a - 2*sin(e + f*x)**4*b - 8* sin(e + f*x)**3*a + 2*sin(e + f*x)**3*b - 12*sin(e + f*x)**2*a + 3*sin(e + f*x)**2*b + 12*sin(e + f*x)*a - 3*sin(e + f*x)*b + 3*a + 3*b)/(15*cos(e + f*x)*a**2*c**3*f*(sin(e + f*x)**3 - sin(e + f*x)**2 - sin(e + f*x) + 1))