Integrand size = 36, antiderivative size = 90 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}+\frac {(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f} \] Output:
-1/5*(A-B)*sec(f*x+e)^3/c^2/f/(a^3+a^3*sin(f*x+e))+1/5*(4*A+B)*tan(f*x+e)/ a^3/c^2/f+1/15*(4*A+B)*tan(f*x+e)^3/a^3/c^2/f
Leaf count is larger than twice the leaf count of optimal. \(237\) vs. \(2(90)=180\).
Time = 4.79 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.63 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (240 B+54 (A-B) \cos (e+f x)-32 (4 A+B) \cos (2 (e+f x))+18 A \cos (3 (e+f x))-18 B \cos (3 (e+f x))-64 A \cos (4 (e+f x))-16 B \cos (4 (e+f x))+384 A \sin (e+f x)+96 B \sin (e+f x)+18 A \sin (2 (e+f x))-18 B \sin (2 (e+f x))+128 A \sin (3 (e+f x))+32 B \sin (3 (e+f x))+9 A \sin (4 (e+f x))-9 B \sin (4 (e+f x)))}{960 a^3 c^2 f (-1+\sin (e+f x))^2 (1+\sin (e+f x))^3} \] Input:
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x] )^2),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*C os[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^3*c^2*f*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^3)
Time = 0.52 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84, 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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \frac {\sec ^4(e+f x) (A+B \sin (e+f x))}{\sin (e+f x) a+a}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A+B \sin (e+f x)}{\cos (e+f x)^4 (\sin (e+f x) a+a)}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle \frac {\frac {(4 A+B) \int \sec ^4(e+f x)dx}{5 a}-\frac {(A-B) \sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(4 A+B) \int \csc \left (e+f x+\frac {\pi }{2}\right )^4dx}{5 a}-\frac {(A-B) \sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {(4 A+B) \int \left (\tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{5 a f}-\frac {(A-B) \sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {(4 A+B) \left (-\frac {1}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{5 a f}-\frac {(A-B) \sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
Input:
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^2),x ]
Output:
(-1/5*((A - B)*Sec[e + f*x]^3)/(f*(a + a*Sin[e + f*x])) - ((4*A + B)*(-Tan [e + f*x] - Tan[e + f*x]^3/3))/(5*a*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 = 1.00 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.51
| method | result | size |
| risch | \(\frac {4 i \left (24 i A \,{\mathrm e}^{3 i \left (f x +e \right )}+6 i B \,{\mathrm e}^{3 i \left (f x +e \right )}+15 B \,{\mathrm e}^{4 i \left (f x +e \right )}+8 i A \,{\mathrm e}^{i \left (f x +e \right )}-8 A \,{\mathrm e}^{2 i \left (f x +e \right )}+2 i B \,{\mathrm e}^{i \left (f x +e \right )}-2 B \,{\mathrm e}^{2 i \left (f x +e \right )}-4 A -B \right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,a^{3} c^{2}}\) | \(136\) |
| parallelrisch | \(\frac {-30 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+\left (-30 A -30 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (10 A -20 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (50 A -10 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-26 A +16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (-42 A -18 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-18 A -12 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+6 A -6 B}{15 f \,a^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(171\) |
| derivativedivides | \(\frac {-\frac {-2 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (A -B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}-2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {A}{4}+\frac {B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {\frac {A}{4}+\frac {B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}}{f \,a^{3} c^{2}}\) | \(185\) |
| default | \(\frac {-\frac {-2 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (A -B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}-2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {A}{4}+\frac {B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {\frac {A}{4}+\frac {B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}}{f \,a^{3} c^{2}}\) | \(185\) |
| norman | \(\frac {-\frac {6 A +4 B}{10 c f a}-\frac {4 \left (4 A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 c f a}+\frac {A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a f c}-\frac {\left (14 A +16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{10 c f a}-\frac {\left (6 A +4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 c f a}-\frac {\left (2 A +8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f a}-\frac {2 \left (8 A +2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 c f a}-\frac {\left (16 A +4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 c f a}+\frac {2 \left (8 A +2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{15 c f a}+\frac {\left (38 A -28 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{15 c f a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) | \(318\) |
Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,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)^5/(exp(I*(f*x+e))-I)^3/f/a^3/c^2
Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (4 \, A + B\right )} \cos \left (f x + e\right )^{4} - {\left (4 \, A + B\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, {\left (4 \, A + B\right )} \cos \left (f x + e\right )^{2} + 4 \, A + B\right )} \sin \left (f x + e\right ) - A - 4 \, B}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,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^3*c^2*f*cos(f*x + e )^3*sin(f*x + e) + a^3*c^2*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 = 9.82 (sec) , antiderivative size = 2674, normalized size of antiderivative = 29.71 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**2,x)
Output:
Piecewise((-30*A*tan(e/2 + f*x/2)**7/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15 *a**3*c**2*f) - 30*A*tan(e/2 + f*x/2)**6/(15*a**3*c**2*f*tan(e/2 + f*x/2)* *8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)* *6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)* *3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) + 10*A*tan(e/2 + f*x/2)**5/(15*a**3*c**2*f*tan(e/2 + f*x /2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x /2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x /2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x /2) - 15*a**3*c**2*f) + 50*A*tan(e/2 + f*x/2)**4/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 26*A*tan(e/2 + f*x/2)**3/(15*a**3*c**2*f*tan(e /2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e /2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e /2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*ta...
Leaf count of result is larger than twice the leaf count of optimal. 650 vs. \(2 (84) = 168\).
Time = 0.06 (sec) , antiderivative size = 650, normalized size of antiderivative = 7.22 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,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^ 3*c^2 + 2*a^3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e) ^2/(cos(f*x + e) + 1)^2 - 6*a^3*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6*a^3*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)^6/( cos(f*x + e) + 1)^6 - 2*a^3*c^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3* c^2*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^3*c^2 + 2*a^3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 - 6*a^3*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6*a^3* c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)^6/(cos(f* x + e) + 1)^6 - 2*a^3*c^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3*c^2*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. \(2 (84) = 168\).
Time = 0.27 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.46 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\frac {5 \, {\left (15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 24 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, A + 7 \, B\right )}}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}} + \frac {165 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 480 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 60 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 70 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 400 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 20 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113 \, A - 13 \, B}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{120 \, f} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,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^3*c^2*( 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^3*c^2*( tan(1/2*f*x + 1/2*e) + 1)^5))/f
Time = 36.71 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.03 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {\left (\frac {8\,A}{15}+\frac {2\,B}{15}+\frac {16\,A\,\sin \left (e+f\,x\right )}{15}+\frac {4\,B\,\sin \left (e+f\,x\right )}{15}\right )\,{\cos \left (e+f\,x\right )}^2+\frac {2\,A}{15}+\frac {8\,B}{15}+\frac {8\,A\,\sin \left (e+f\,x\right )}{15}+\frac {2\,B\,\sin \left (e+f\,x\right )}{15}}{a^3\,c^2\,f\,\left (2\,{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )+2\,{\cos \left (e+f\,x\right )}^3\right )}-\frac {\frac {2\,A}{5}-\frac {2\,B}{5}+\frac {2\,A\,\sin \left (e+f\,x\right )}{5}-\frac {2\,B\,\sin \left (e+f\,x\right )}{5}}{a^3\,c^2\,f\,\left (2\,\sin \left (e+f\,x\right )+2\right )}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {16\,A}{15}+\frac {4\,B}{15}\right )}{a^3\,c^2\,f\,\left (2\,\sin \left (e+f\,x\right )+2\right )} \] Input:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^2),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^3*c^2*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^3*c^ 2*f*(2*sin(e + f*x) + 2)) - (cos(e + f*x)*((16*A)/15 + (4*B)/15))/(a^3*c^2 *f*(2*sin(e + f*x) + 2))
Time = 0.18 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.82 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a +3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b -12 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -12 \cos \left (f x +e \right ) a -3 \cos \left (f x +e \right ) b +8 \sin \left (f x +e \right )^{4} a +2 \sin \left (f x +e \right )^{4} b +8 \sin \left (f x +e \right )^{3} a +2 \sin \left (f x +e \right )^{3} b -12 \sin \left (f x +e \right )^{2} a -3 \sin \left (f x +e \right )^{2} b -12 a \sin \left (f x +e \right )-3 \sin \left (f x +e \right ) b +3 a -3 b}{15 \cos \left (f x +e \right ) a^{3} c^{2} f \left (\sin \left (f x +e \right )^{3}+\sin \left (f x +e \right )^{2}-\sin \left (f x +e \right )-1\right )} \] Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,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*co s(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**3*c**2*f*(sin(e + f*x)**3 + sin(e + f*x)**2 - sin(e + f*x) - 1))