Integrand size = 36, antiderivative size = 117 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {3 a^2 (2 A+3 B) x}{2 c}+\frac {3 a^2 (2 A+3 B) \cos (e+f x)}{2 c f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {a^2 (2 A+3 B) \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))} \] Output:
-3/2*a^2*(2*A+3*B)*x/c+3/2*a^2*(2*A+3*B)*cos(f*x+e)/c/f+a^2*(A+B)*c^2*cos( f*x+e)^5/f/(c-c*sin(f*x+e))^3+1/2*a^2*(2*A+3*B)*cos(f*x+e)^3/f/(c-c*sin(f* x+e))
Time = 12.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.63 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2 \left (\cos \left (\frac {1}{2} (e+f x)\right ) (6 (2 A+3 B) (e+f x)-4 (A+3 B) \cos (e+f x)-B \sin (2 (e+f x)))-\sin \left (\frac {1}{2} (e+f x)\right ) (4 A (8+3 e+3 f x)+2 B (16+9 e+9 f x)-4 (A+3 B) \cos (e+f x)-B \sin (2 (e+f x)))\right )}{4 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (-1+\sin (e+f x))} \] Input:
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ]),x]
Output:
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2*(Cos[(e + f*x)/2]*(6*(2*A + 3*B)*(e + f*x) - 4*(A + 3*B)*Cos[e + f*x] - B*Sin[2*(e + f*x)]) - Sin[(e + f*x)/2]*(4*A*(8 + 3*e + 3*f*x) + 2*B*(16 + 9*e + 9*f*x) - 4*(A + 3*B)*Cos[e + f*x] - B*Sin[2*(e + f*x)])))/(4*c*f*(Cos[(e + f*x)/ 2] + Sin[(e + f*x)/2])^4*(-1 + Sin[e + f*x]))
Time = 0.66 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3446, 3042, 3338, 3042, 3158, 3042, 3161, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\frac {(2 A+3 B) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^2}dx}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\frac {(2 A+3 B) \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^2}dx}{c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\frac {(2 A+3 B) \left (\frac {3 \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)}dx}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\frac {(2 A+3 B) \left (\frac {3 \int \frac {\cos (e+f x)^2}{c-c \sin (e+f x)}dx}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\frac {(2 A+3 B) \left (\frac {3 \left (\frac {\int 1dx}{c}-\frac {\cos (e+f x)}{c f}\right )}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\frac {(2 A+3 B) \left (\frac {3 \left (\frac {x}{c}-\frac {\cos (e+f x)}{c f}\right )}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x]),x]
Output:
a^2*c^2*(((A + B)*Cos[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^3) - ((2*A + 3*B )*((3*(x/c - Cos[e + f*x]/(c*f)))/(2*c) - Cos[e + f*x]^3/(2*f*(c^2 - c^2*S in[e + f*x]))))/c)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si mp[g^2/a Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x ] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
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]))
Time = 0.77 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79
| method | result | size |
| parallelrisch | \(\frac {4 a^{2} \left (\frac {\left (A +3 B \right ) \cos \left (2 f x +2 e \right )}{8}+\frac {B \sin \left (3 f x +3 e \right )}{32}+\frac {\left (-3 f x A -\frac {9}{2} f x B +5 A +7 B \right ) \cos \left (f x +e \right )}{4}+\left (A +\frac {33 B}{32}\right ) \sin \left (f x +e \right )+\frac {9 A}{8}+\frac {11 B}{8}\right )}{c f \cos \left (f x +e \right )}\) | \(92\) |
| derivativedivides | \(\frac {2 a^{2} \left (-\frac {4 A +4 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (-A -3 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-A -3 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {3 \left (2 A +3 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f c}\) | \(123\) |
| default | \(\frac {2 a^{2} \left (-\frac {4 A +4 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (-A -3 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-A -3 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {3 \left (2 A +3 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f c}\) | \(123\) |
| risch | \(-\frac {3 a^{2} x A}{c}-\frac {9 a^{2} x B}{2 c}+\frac {a^{2} {\mathrm e}^{i \left (f x +e \right )} A}{2 c f}+\frac {3 a^{2} {\mathrm e}^{i \left (f x +e \right )} B}{2 c f}+\frac {a^{2} {\mathrm e}^{-i \left (f x +e \right )} A}{2 c f}+\frac {3 a^{2} {\mathrm e}^{-i \left (f x +e \right )} B}{2 c f}+\frac {8 a^{2} A}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}+\frac {8 a^{2} B}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}+\frac {a^{2} B \sin \left (2 f x +2 e \right )}{4 c f}\) | \(179\) |
| norman | \(\frac {-\frac {2 a^{2} A +5 a^{2} B}{c f}+\frac {3 a^{2} \left (2 A +3 B \right ) x}{2 c}-\frac {\left (2 a^{2} A +3 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}-\frac {\left (4 a^{2} A +8 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}-\frac {\left (6 a^{2} A +4 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {\left (8 a^{2} A +9 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c f}-\frac {\left (20 a^{2} A +15 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}-\frac {\left (22 a^{2} A +20 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}-\frac {3 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c}+\frac {9 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 c}-\frac {9 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 c}+\frac {9 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2 c}-\frac {9 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 c}+\frac {3 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{2 c}-\frac {3 a^{2} \left (2 A +3 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{2 c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) | \(445\) |
Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x,method=_RETURNV ERBOSE)
Output:
4*a^2*(1/8*(A+3*B)*cos(2*f*x+2*e)+1/32*B*sin(3*f*x+3*e)+1/4*(-3*f*x*A-9/2* f*x*B+5*A+7*B)*cos(f*x+e)+(A+33/32*B)*sin(f*x+e)+9/8*A+11/8*B)/c/f/cos(f*x +e)
Time = 0.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.51 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {B a^{2} \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, A + 3 \, B\right )} a^{2} f x + 2 \, {\left (A + 3 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 8 \, {\left (A + B\right )} a^{2} - {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{2} f x - {\left (10 \, A + 13 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) + {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{2} f x + B a^{2} \cos \left (f x + e\right )^{2} - {\left (2 \, A + 5 \, B\right )} a^{2} \cos \left (f x + e\right ) + 8 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorith m="fricas")
Output:
1/2*(B*a^2*cos(f*x + e)^3 - 3*(2*A + 3*B)*a^2*f*x + 2*(A + 3*B)*a^2*cos(f* x + e)^2 + 8*(A + B)*a^2 - (3*(2*A + 3*B)*a^2*f*x - (10*A + 13*B)*a^2)*cos (f*x + e) + (3*(2*A + 3*B)*a^2*f*x + B*a^2*cos(f*x + e)^2 - (2*A + 5*B)*a^ 2*cos(f*x + e) + 8*(A + B)*a^2)*sin(f*x + e))/(c*f*cos(f*x + e) - c*f*sin( f*x + e) + c*f)
Leaf count of result is larger than twice the leaf count of optimal. 2365 vs. \(2 (104) = 208\).
Time = 2.16 (sec) , antiderivative size = 2365, normalized size of antiderivative = 20.21 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x)
Output:
Piecewise((-6*A*a**2*f*x*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f* x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 6*A*a**2*f*x*tan(e/2 + f*x/2)* *4/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 12*A*a**2*f*x*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan (e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 12*A*a**2*f*x*tan(e/2 + f*x/2)**2/(2*c* f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2) **3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 6*A*a* *2*f*x*tan(e/2 + f*x/2)/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2 )**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e /2 + f*x/2) - 2*c*f) + 6*A*a**2*f*x/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan (e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 16*A*a**2*tan(e/2 + f*x/2)**4/(2*c*f*ta n(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 4*A*a**2*t an(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 36*A*a**2*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)...
Leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (114) = 228\).
Time = 0.13 (sec) , antiderivative size = 624, normalized size of antiderivative = 5.33 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorith m="maxima")
Output:
-(2*A*a^2*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f* x + e)/(cos(f*x + e) + 1))/c) + 4*B*a^2*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos(f* x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c) + B*a^2*((sin( f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*si n(f*x + e)^3/(cos(f*x + e) + 1)^3 - 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + 2*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c*sin(f*x + e)^4/( cos(f*x + e) + 1)^4 - c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(si n(f*x + e)/(cos(f*x + e) + 1))/c) + 4*A*a^2*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(f*x + e)/(cos(f*x + e) + 1))) + 2*B*a^2*(arcta n(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(f*x + e)/(cos(f*x + e) + 1))) - 2*A*a^2/(c - c*sin(f*x + e)/(cos(f*x + e) + 1)))/f
Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.33 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {\frac {3 \, {\left (2 \, A a^{2} + 3 \, B a^{2}\right )} {\left (f x + e\right )}}{c} + \frac {16 \, {\left (A a^{2} + B a^{2}\right )}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A a^{2} - 6 \, B a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} c}}{2 \, f} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorith m="giac")
Output:
-1/2*(3*(2*A*a^2 + 3*B*a^2)*(f*x + e)/c + 16*(A*a^2 + B*a^2)/(c*(tan(1/2*f *x + 1/2*e) - 1)) + 2*(B*a^2*tan(1/2*f*x + 1/2*e)^3 - 2*A*a^2*tan(1/2*f*x + 1/2*e)^2 - 6*B*a^2*tan(1/2*f*x + 1/2*e)^2 - B*a^2*tan(1/2*f*x + 1/2*e) - 2*A*a^2 - 6*B*a^2)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*c))/f
Time = 37.23 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.09 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {10\,A\,a^2-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A\,a^2+5\,B\,a^2\right )+14\,B\,a^2-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,A\,a^2+7\,B\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (8\,A\,a^2+9\,B\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (18\,A\,a^2+21\,B\,a^2\right )}{f\,\left (-c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )}-\frac {3\,a^2\,\mathrm {atan}\left (\frac {3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A+3\,B\right )}{6\,A\,a^2+9\,B\,a^2}\right )\,\left (2\,A+3\,B\right )}{c\,f} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x)),x)
Output:
(10*A*a^2 - tan(e/2 + (f*x)/2)*(2*A*a^2 + 5*B*a^2) + 14*B*a^2 - tan(e/2 + (f*x)/2)^3*(2*A*a^2 + 7*B*a^2) + tan(e/2 + (f*x)/2)^4*(8*A*a^2 + 9*B*a^2) + tan(e/2 + (f*x)/2)^2*(18*A*a^2 + 21*B*a^2))/(f*(c - c*tan(e/2 + (f*x)/2) + 2*c*tan(e/2 + (f*x)/2)^2 - 2*c*tan(e/2 + (f*x)/2)^3 + c*tan(e/2 + (f*x) /2)^4 - c*tan(e/2 + (f*x)/2)^5)) - (3*a^2*atan((3*a^2*tan(e/2 + (f*x)/2)*( 2*A + 3*B))/(6*A*a^2 + 9*B*a^2))*(2*A + 3*B))/(c*f)
Time = 0.18 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.74 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {a^{2} \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -6 \cos \left (f x +e \right ) a f x -16 \cos \left (f x +e \right ) a -9 \cos \left (f x +e \right ) b f x -18 \cos \left (f x +e \right ) b -\sin \left (f x +e \right )^{3} b -2 \sin \left (f x +e \right )^{2} a -6 \sin \left (f x +e \right )^{2} b -6 \sin \left (f x +e \right ) a f x +2 a \sin \left (f x +e \right )-9 \sin \left (f x +e \right ) b f x +5 \sin \left (f x +e \right ) b +6 a f x +16 a +9 b f x +18 b \right )}{2 c f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right )} \] Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x)
Output:
(a**2*(cos(e + f*x)*sin(e + f*x)**2*b + 2*cos(e + f*x)*sin(e + f*x)*a + 5* cos(e + f*x)*sin(e + f*x)*b - 6*cos(e + f*x)*a*f*x - 16*cos(e + f*x)*a - 9 *cos(e + f*x)*b*f*x - 18*cos(e + f*x)*b - sin(e + f*x)**3*b - 2*sin(e + f* x)**2*a - 6*sin(e + f*x)**2*b - 6*sin(e + f*x)*a*f*x + 2*sin(e + f*x)*a - 9*sin(e + f*x)*b*f*x + 5*sin(e + f*x)*b + 6*a*f*x + 16*a + 9*b*f*x + 18*b) )/(2*c*f*(cos(e + f*x) + sin(e + f*x) - 1))