Integrand size = 36, antiderivative size = 102 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f} \] Output:
-1/5*(A-B)*sec(f*x+e)/a/c/f/(a+a*sin(f*x+e))^2-1/15*(3*A+2*B)*sec(f*x+e)/c /f/(a^3+a^3*sin(f*x+e))+2/15*(3*A+2*B)*tan(f*x+e)/a^3/c/f
Time = 3.93 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.53 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {\cos (e+f x) (-80 B-5 (9 A+B) \cos (e+f x)+32 (3 A+2 B) \cos (2 (e+f x))+9 A \cos (3 (e+f x))+B \cos (3 (e+f x))-120 A \sin (e+f x)-80 B \sin (e+f x)-36 A \sin (2 (e+f x))-4 B \sin (2 (e+f x))+24 A \sin (3 (e+f x))+16 B \sin (3 (e+f x)))}{240 a^3 c f (-1+\sin (e+f x)) (1+\sin (e+f x))^3} \] Input:
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x] )),x]
Output:
(Cos[e + f*x]*(-80*B - 5*(9*A + B)*Cos[e + f*x] + 32*(3*A + 2*B)*Cos[2*(e + f*x)] + 9*A*Cos[3*(e + f*x)] + B*Cos[3*(e + f*x)] - 120*A*Sin[e + f*x] - 80*B*Sin[e + f*x] - 36*A*Sin[2*(e + f*x)] - 4*B*Sin[2*(e + f*x)] + 24*A*S in[3*(e + f*x)] + 16*B*Sin[3*(e + f*x)]))/(240*a^3*c*f*(-1 + Sin[e + f*x]) *(1 + Sin[e + f*x])^3)
Time = 0.62 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3446, 3042, 3338, 3042, 3151, 3042, 4254, 24}
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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^2}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A+B \sin (e+f x)}{\cos (e+f x)^2 (\sin (e+f x) a+a)^2}dx}{a c}\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle \frac {\frac {(3 A+2 B) \int \frac {\sec ^2(e+f x)}{\sin (e+f x) a+a}dx}{5 a}-\frac {(A-B) \sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(3 A+2 B) \int \frac {1}{\cos (e+f x)^2 (\sin (e+f x) a+a)}dx}{5 a}-\frac {(A-B) \sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {\frac {(3 A+2 B) \left (\frac {2 \int \sec ^2(e+f x)dx}{3 a}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {(A-B) \sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(3 A+2 B) \left (\frac {2 \int \csc \left (e+f x+\frac {\pi }{2}\right )^2dx}{3 a}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {(A-B) \sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {(3 A+2 B) \left (-\frac {2 \int 1d(-\tan (e+f x))}{3 a f}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {(A-B) \sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {(3 A+2 B) \left (\frac {2 \tan (e+f x)}{3 a f}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {(A-B) \sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
Input:
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])),x]
Output:
(-1/5*((A - B)*Sec[e + f*x])/(f*(a + a*Sin[e + f*x])^2) + ((3*A + 2*B)*(-1 /3*Sec[e + f*x]/(f*(a + a*Sin[e + f*x])) + (2*Tan[e + f*x])/(3*a*f)))/(5*a ))/(a*c)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
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 = 0.62 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09
| method | result | size |
| risch | \(-\frac {4 \left (15 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-12 A \,{\mathrm e}^{i \left (f x +e \right )}+10 B \,{\mathrm e}^{3 i \left (f x +e \right )}-8 B \,{\mathrm e}^{i \left (f x +e \right )}-2 i B +10 i B \,{\mathrm e}^{2 i \left (f x +e \right )}-3 i A \right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) f \,a^{3} c}\) | \(111\) |
| parallelrisch | \(\frac {-30 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (-60 A -30 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-60 A -40 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-40 B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (18 A -8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+12 A -2 B}{15 f \,a^{3} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(128\) |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (2 A -2 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {9 A}{2}-\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{f \,a^{3} c}\) | \(145\) |
| default | \(\frac {-\frac {2 \left (\frac {A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (2 A -2 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {9 A}{2}-\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{f \,a^{3} c}\) | \(145\) |
| norman | \(\frac {\frac {12 A -2 B}{15 c f a}-\frac {2 \left (6 A +7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 c f a}-\frac {2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f c}+\frac {2 \left (9 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 c f a}+\frac {2 \left (2 A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 c f a}-\frac {2 \left (9 A +4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 c f a}-\frac {2 \left (2 A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c f a}-\frac {2 \left (8 B +7 A \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{5 c f a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) | \(258\) |
Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x,method=_RETURNV ERBOSE)
Output:
-4/15*(15*I*A*exp(2*I*(f*x+e))-12*A*exp(I*(f*x+e))+10*B*exp(3*I*(f*x+e))-8 *B*exp(I*(f*x+e))-2*I*B+10*I*B*exp(2*I*(f*x+e))-3*I*A)/(exp(I*(f*x+e))+I)^ 5/(exp(I*(f*x+e))-I)/f/a^3/c
Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {4 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right )^{2} - 9 \, A - 6 \, B\right )} \sin \left (f x + e\right ) - 6 \, A - 9 \, B}{15 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorith m="fricas")
Output:
1/15*(4*(3*A + 2*B)*cos(f*x + e)^2 + (2*(3*A + 2*B)*cos(f*x + e)^2 - 9*A - 6*B)*sin(f*x + e) - 6*A - 9*B)/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*cos(f* x + e)*sin(f*x + e) - 2*a^3*c*f*cos(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 1236 vs. \(2 (85) = 170\).
Time = 4.95 (sec) , antiderivative size = 1236, normalized size of antiderivative = 12.12 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e)),x)
Output:
Piecewise((-30*A*tan(e/2 + f*x/2)**5/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60 *a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3* c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 60 *A*tan(e/2 + f*x/2)**4/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan( e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 60*A*tan(e/2 + f *x/2)**3/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)** 5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60 *a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) + 18*A*tan(e/2 + f*x/2)/(15*a**3 *c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*t an(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) + 12*A/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3 *c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*t an(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 30*B*ta n(e/2 + f*x/2)**4/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2 )**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 40*B*tan(e/2 + f*x/2) **3/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 7 5*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3 *c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 40*B*tan(e/2 + f*x/2)**2/(15*a**...
Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (96) = 192\).
Time = 0.05 (sec) , antiderivative size = 423, normalized size of antiderivative = 4.15 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {B {\left (\frac {4 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 1\right )}}{a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}} - \frac {3 \, A {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + 2\right )}}{a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{15 \, f} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorith m="maxima")
Output:
2/15*(B*(4*sin(f*x + e)/(cos(f*x + e) + 1) + 20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(co s(f*x + e) + 1)^4 + 1)/(a^3*c + 4*a^3*c*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^3*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a^3*c*sin(f*x + e)^4/(cos( f*x + e) + 1)^4 - 4*a^3*c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - a^3*c*sin( f*x + e)^6/(cos(f*x + e) + 1)^6) - 3*A*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 10*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2)/(a^3*c + 4*a^3*c*sin( f*x + e)/(cos(f*x + e) + 1) + 5*a^3*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a^3*c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4*a^3*c*sin(f*x + e)^5/(co s(f*x + e) + 1)^5 - a^3*c*sin(f*x + e)^6/(cos(f*x + e) + 1)^6))/f
Time = 0.21 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {\frac {15 \, {\left (A + B\right )}}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {105 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 270 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 360 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 40 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 210 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 50 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 63 \, A + 7 \, B}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{60 \, f} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorith m="giac")
Output:
-1/60*(15*(A + B)/(a^3*c*(tan(1/2*f*x + 1/2*e) - 1)) + (105*A*tan(1/2*f*x + 1/2*e)^4 - 15*B*tan(1/2*f*x + 1/2*e)^4 + 270*A*tan(1/2*f*x + 1/2*e)^3 + 30*B*tan(1/2*f*x + 1/2*e)^3 + 360*A*tan(1/2*f*x + 1/2*e)^2 + 40*B*tan(1/2* f*x + 1/2*e)^2 + 210*A*tan(1/2*f*x + 1/2*e) + 50*B*tan(1/2*f*x + 1/2*e) + 63*A + 7*B)/(a^3*c*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
Time = 36.65 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.75 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {2\,\left (\frac {15\,A\,\cos \left (e+f\,x\right )}{4}-\frac {5\,B}{2}-\frac {5\,B\,\cos \left (e+f\,x\right )}{8}-\frac {15\,A\,\sin \left (e+f\,x\right )}{4}-\frac {5\,B\,\sin \left (e+f\,x\right )}{2}+3\,A\,\cos \left (2\,e+2\,f\,x\right )-\frac {3\,A\,\cos \left (3\,e+3\,f\,x\right )}{4}+2\,B\,\cos \left (2\,e+2\,f\,x\right )+\frac {B\,\cos \left (3\,e+3\,f\,x\right )}{8}+3\,A\,\sin \left (2\,e+2\,f\,x\right )+\frac {3\,A\,\sin \left (3\,e+3\,f\,x\right )}{4}-\frac {B\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {B\,\sin \left (3\,e+3\,f\,x\right )}{2}\right )}{15\,a^3\,c\,f\,\left (\frac {5\,\cos \left (e+f\,x\right )}{4}-\frac {\cos \left (3\,e+3\,f\,x\right )}{4}+\sin \left (2\,e+2\,f\,x\right )\right )} \] Input:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))),x)
Output:
-(2*((15*A*cos(e + f*x))/4 - (5*B)/2 - (5*B*cos(e + f*x))/8 - (15*A*sin(e + f*x))/4 - (5*B*sin(e + f*x))/2 + 3*A*cos(2*e + 2*f*x) - (3*A*cos(3*e + 3 *f*x))/4 + 2*B*cos(2*e + 2*f*x) + (B*cos(3*e + 3*f*x))/8 + 3*A*sin(2*e + 2 *f*x) + (3*A*sin(3*e + 3*f*x))/4 - (B*sin(2*e + 2*f*x))/2 + (B*sin(3*e + 3 *f*x))/2))/(15*a^3*c*f*((5*cos(e + f*x))/4 - cos(3*e + 3*f*x)/4 + sin(2*e + 2*f*x)))
Time = 0.18 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.86 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +6 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +3 \cos \left (f x +e \right ) a +2 \cos \left (f x +e \right ) b +12 \sin \left (f x +e \right )^{3} a +8 \sin \left (f x +e \right )^{3} b +24 \sin \left (f x +e \right )^{2} a +16 \sin \left (f x +e \right )^{2} b +6 a \sin \left (f x +e \right )+4 \sin \left (f x +e \right ) b -12 a +2 b}{30 \cos \left (f x +e \right ) a^{3} c f \left (\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1\right )} \] Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x)
Output:
(3*cos(e + f*x)*sin(e + f*x)**2*a + 2*cos(e + f*x)*sin(e + f*x)**2*b + 6*c os(e + f*x)*sin(e + f*x)*a + 4*cos(e + f*x)*sin(e + f*x)*b + 3*cos(e + f*x )*a + 2*cos(e + f*x)*b + 12*sin(e + f*x)**3*a + 8*sin(e + f*x)**3*b + 24*s in(e + f*x)**2*a + 16*sin(e + f*x)**2*b + 6*sin(e + f*x)*a + 4*sin(e + f*x )*b - 12*a + 2*b)/(30*cos(e + f*x)*a**3*c*f*(sin(e + f*x)**2 + 2*sin(e + f *x) + 1))