Integrand size = 34, antiderivative size = 104 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}-\frac {a (A+11 B) c \cos (e+f x)}{15 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac {a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )} \] Output:
2/5*a*(A+B)*cos(f*x+e)/f/(c-c*sin(f*x+e))^3-1/15*a*(A+11*B)*c*cos(f*x+e)/f /(c^2-c^2*sin(f*x+e))^2-1/15*a*(A-4*B)*cos(f*x+e)/f/(c^3-c^3*sin(f*x+e))
Time = 6.66 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {a \left (15 (A-B) \cos \left (e+\frac {f x}{2}\right )-5 (A-B) \cos \left (e+\frac {3 f x}{2}\right )+5 A \sin \left (\frac {f x}{2}\right )+25 B \sin \left (\frac {f x}{2}\right )+15 B \sin \left (2 e+\frac {3 f x}{2}\right )+A \sin \left (2 e+\frac {5 f x}{2}\right )-4 B \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 c^3 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:
Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x]) ^3,x]
Output:
(a*(15*(A - B)*Cos[e + (f*x)/2] - 5*(A - B)*Cos[e + (3*f*x)/2] + 5*A*Sin[( f*x)/2] + 25*B*Sin[(f*x)/2] + 15*B*Sin[2*e + (3*f*x)/2] + A*Sin[2*e + (5*f *x)/2] - 4*B*Sin[2*e + (5*f*x)/2]))/(30*c^3*f*(Cos[e/2] - Sin[e/2])*(Cos[( e + f*x)/2] - Sin[(e + f*x)/2])^5)
Time = 0.62 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3446, 3042, 3336, 25, 3042, 3229, 3042, 3127}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a c \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \frac {\cos (e+f x)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3336 |
| \(\displaystyle a c \left (\frac {\int -\frac {(A+6 B) c+5 B \sin (e+f x) c}{(c-c \sin (e+f x))^2}dx}{5 c^3}+\frac {2 (A+B) \cos (e+f x)}{5 c f (c-c \sin (e+f x))^3}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{5 c f (c-c \sin (e+f x))^3}-\frac {\int \frac {(A+6 B) c+5 B \sin (e+f x) c}{(c-c \sin (e+f x))^2}dx}{5 c^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{5 c f (c-c \sin (e+f x))^3}-\frac {\int \frac {(A+6 B) c+5 B \sin (e+f x) c}{(c-c \sin (e+f x))^2}dx}{5 c^3}\right )\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{5 c f (c-c \sin (e+f x))^3}-\frac {\frac {1}{3} (A-4 B) \int \frac {1}{c-c \sin (e+f x)}dx+\frac {c (A+11 B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^2}}{5 c^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{5 c f (c-c \sin (e+f x))^3}-\frac {\frac {1}{3} (A-4 B) \int \frac {1}{c-c \sin (e+f x)}dx+\frac {c (A+11 B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^2}}{5 c^3}\right )\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{5 c f (c-c \sin (e+f x))^3}-\frac {\frac {(A-4 B) \cos (e+f x)}{3 f (c-c \sin (e+f x))}+\frac {c (A+11 B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^2}}{5 c^3}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]
Output:
a*c*((2*(A + B)*Cos[e + f*x])/(5*c*f*(c - c*Sin[e + f*x])^3) - (((A + 11*B )*c*Cos[e + f*x])/(3*f*(c - c*Sin[e + f*x])^2) + ((A - 4*B)*Cos[e + f*x])/ (3*f*(c - c*Sin[e + f*x])))/(5*c^3))
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*( (c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[2*(b*c - a*d)*Cos [e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(2*m + 3))), x] + Simp[1/(b^ 3*(2*m + 3)) Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d* (2*m + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]
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]))
Time = 0.82 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90
| method | result | size |
| parallelrisch | \(-\frac {2 \left (A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\frac {\left (5 A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}+\frac {\left (-A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}+\frac {4 A}{15}-\frac {B}{15}\right ) a}{f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(94\) |
| derivativedivides | \(\frac {2 a \left (-\frac {14 A +10 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {6 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {8 A +8 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {16 A +16 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{3}}\) | \(115\) |
| default | \(\frac {2 a \left (-\frac {14 A +10 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {6 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {8 A +8 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {16 A +16 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{3}}\) | \(115\) |
| risch | \(\frac {-\frac {2 A a \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}-\frac {2 i A a \,{\mathrm e}^{i \left (f x +e \right )}}{3}-\frac {10 B a \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}-2 i B a \,{\mathrm e}^{3 i \left (f x +e \right )}+\frac {2 i B a \,{\mathrm e}^{i \left (f x +e \right )}}{3}+2 B a \,{\mathrm e}^{4 i \left (f x +e \right )}-\frac {2 A a}{15}+\frac {8 B a}{15}+2 i A a \,{\mathrm e}^{3 i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\) | \(127\) |
| norman | \(\frac {-\frac {8 A a -2 B a}{15 f c}-\frac {2 A a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{f c}+\frac {10 \left (A a -B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f c}+\frac {2 \left (A a -B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f c}+\frac {\left (2 A a -2 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f c}-\frac {2 \left (11 A a +B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 f c}-\frac {2 \left (11 A a +B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 f c}+\frac {2 \left (7 A a -7 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 f c}-\frac {2 \left (23 A a +3 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 f c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(262\) |
Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x,method=_RETURNV ERBOSE)
Output:
-2*(A*tan(1/2*f*x+1/2*e)^4+(-A+B)*tan(1/2*f*x+1/2*e)^3+1/3*(5*A+B)*tan(1/2 *f*x+1/2*e)^2+1/3*(-A+B)*tan(1/2*f*x+1/2*e)+4/15*A-1/15*B)*a/f/c^3/(tan(1/ 2*f*x+1/2*e)-1)^5
Time = 0.07 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.76 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {{\left (A - 4 \, B\right )} a \cos \left (f x + e\right )^{3} - {\left (2 \, A + 7 \, B\right )} a \cos \left (f x + e\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (f x + e\right ) + 6 \, {\left (A + B\right )} a + {\left ({\left (A - 4 \, B\right )} a \cos \left (f x + e\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (f x + e\right ) + 6 \, {\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorith m="fricas")
Output:
-1/15*((A - 4*B)*a*cos(f*x + e)^3 - (2*A + 7*B)*a*cos(f*x + e)^2 + 3*(A + B)*a*cos(f*x + e) + 6*(A + B)*a + ((A - 4*B)*a*cos(f*x + e)^2 + 3*(A + B)* a*cos(f*x + e) + 6*(A + B)*a)*sin(f*x + e))/(c^3*f*cos(f*x + e)^3 + 3*c^3* f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 1035 vs. \(2 (92) = 184\).
Time = 4.74 (sec) , antiderivative size = 1035, normalized size of antiderivative = 9.95 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)
Output:
Piecewise((-30*A*a*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75 *c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f* tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 30*A*a*tan (e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/ 2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 50*A*a*tan(e/2 + f*x/2)**2/(15*c **3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan (e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f* x/2) - 15*c**3*f) + 10*A*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c* *3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 8*A*a /(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3 *f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/ 2 + f*x/2) - 15*c**3*f) - 30*B*a*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f ) - 10*B*a*tan(e/2 + f*x/2)**2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f* tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 10*B*a*tan(e/2 + f *x/2)/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + ...
Leaf count of result is larger than twice the leaf count of optimal. 737 vs. \(2 (101) = 202\).
Time = 0.06 (sec) , antiderivative size = 737, normalized size of antiderivative = 7.09 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorith m="maxima")
Output:
-2/15*(A*a*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f* x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f* x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*A*a*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 5* sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^ 3 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^ 2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^ 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*a*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^2/(cos( f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3* sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1) ^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(co s(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2*B*a*(5*si n(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1 )/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin (f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 ))/f
Time = 0.25 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.26 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, A a - B a\right )}}{15 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorith m="giac")
Output:
-2/15*(15*A*a*tan(1/2*f*x + 1/2*e)^4 - 15*A*a*tan(1/2*f*x + 1/2*e)^3 + 15* B*a*tan(1/2*f*x + 1/2*e)^3 + 25*A*a*tan(1/2*f*x + 1/2*e)^2 + 5*B*a*tan(1/2 *f*x + 1/2*e)^2 - 5*A*a*tan(1/2*f*x + 1/2*e) + 5*B*a*tan(1/2*f*x + 1/2*e) + 4*A*a - B*a)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)
Time = 35.43 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.65 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {11\,A\,a\,\cos \left (e+f\,x\right )}{2}-\frac {B\,a}{4}-\frac {41\,A\,a}{4}+\frac {B\,a\,\cos \left (e+f\,x\right )}{2}+5\,A\,a\,\sin \left (e+f\,x\right )-5\,B\,a\,\sin \left (e+f\,x\right )+\frac {3\,A\,a\,\cos \left (2\,e+2\,f\,x\right )}{4}+\frac {3\,B\,a\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,A\,a\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,a\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,c^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}-\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}+\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x)))/(c - c*sin(e + f*x))^3,x)
Output:
(2*cos(e/2 + (f*x)/2)*((11*A*a*cos(e + f*x))/2 - (B*a)/4 - (41*A*a)/4 + (B *a*cos(e + f*x))/2 + 5*A*a*sin(e + f*x) - 5*B*a*sin(e + f*x) + (3*A*a*cos( 2*e + 2*f*x))/4 + (3*B*a*cos(2*e + 2*f*x))/4 - (5*A*a*sin(2*e + 2*f*x))/4 + (5*B*a*sin(2*e + 2*f*x))/4))/(15*c^3*f*((5*2^(1/2)*cos((3*e)/2 - pi/4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2 + pi/4 + (f*x)/2))/2 + (2^(1/2)*cos((5* e)/2 + pi/4 + (5*f*x)/2))/4))
Time = 0.18 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.66 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {2 a \left (-3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a -15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a -15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b +5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b -a +b \right )}{15 c^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x)
Output:
(2*a*( - 3*tan((e + f*x)/2)**5*a - 15*tan((e + f*x)/2)**3*a - 15*tan((e + f*x)/2)**3*b + 5*tan((e + f*x)/2)**2*a - 5*tan((e + f*x)/2)**2*b - 10*tan( (e + f*x)/2)*a - 5*tan((e + f*x)/2)*b - a + b))/(15*c**3*f*(tan((e + f*x)/ 2)**5 - 5*tan((e + f*x)/2)**4 + 10*tan((e + f*x)/2)**3 - 10*tan((e + f*x)/ 2)**2 + 5*tan((e + f*x)/2) - 1))