Integrand size = 36, antiderivative size = 109 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a^2 (A+4 B) x}{c^2}-\frac {a^2 (A+4 B) \cos (e+f x)}{c^2 f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {2 a^2 (A+4 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \] Output:
a^2*(A+4*B)*x/c^2-a^2*(A+4*B)*cos(f*x+e)/c^2/f+1/3*a^2*(A+B)*c^2*cos(f*x+e )^5/f/(c-c*sin(f*x+e))^4-2/3*a^2*(A+4*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^2
Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(109)=218\).
Time = 11.42 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.18 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (4 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+3 (A+4 B) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-3 B \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+8 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-8 (2 A+5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^2} \] Input:
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^2,x]
Output:
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) + 3*(A + 4*B)*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x )/2])^3 - 3*B*Cos[e + f*x]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 8*(A + B)*Sin[(e + f*x)/2] - 8*(2*A + 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2] )^2*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(3*f*(Cos[(e + f*x)/2] + Sin[( e + f*x)/2])^4*(c - c*Sin[e + f*x])^2)
Time = 0.67 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97, 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, 3159, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {(A+4 B) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^3}dx}{3 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {(A+4 B) \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^3}dx}{3 c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {(A+4 B) \left (\frac {2 \cos ^3(e+f x)}{c f (c-c \sin (e+f x))^2}-\frac {3 \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)}dx}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {(A+4 B) \left (\frac {2 \cos ^3(e+f x)}{c f (c-c \sin (e+f x))^2}-\frac {3 \int \frac {\cos (e+f x)^2}{c-c \sin (e+f x)}dx}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {(A+4 B) \left (\frac {2 \cos ^3(e+f x)}{c f (c-c \sin (e+f x))^2}-\frac {3 \left (\frac {\int 1dx}{c}-\frac {\cos (e+f x)}{c f}\right )}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {(A+4 B) \left (\frac {2 \cos ^3(e+f x)}{c f (c-c \sin (e+f x))^2}-\frac {3 \left (\frac {x}{c}-\frac {\cos (e+f x)}{c f}\right )}{c^2}\right )}{3 c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x ]
Output:
a^2*c^2*(((A + B)*Cos[e + f*x]^5)/(3*f*(c - c*Sin[e + f*x])^4) - ((A + 4*B )*((-3*(x/c - Cos[e + f*x]/(c*f)))/c^2 + (2*Cos[e + f*x]^3)/(c*f*(c - c*Si n[e + f*x])^2)))/(3*c))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[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 = 0.76 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {2 a^{2} \left (-\frac {B}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A +4 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8 A +8 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {4 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f \,c^{2}}\) | \(107\) |
| default | \(\frac {2 a^{2} \left (-\frac {B}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A +4 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8 A +8 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {4 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f \,c^{2}}\) | \(107\) |
| risch | \(\frac {a^{2} x A}{c^{2}}+\frac {4 a^{2} x B}{c^{2}}-\frac {B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}}{2 c^{2} f}-\frac {B \,a^{2} {\mathrm e}^{-i \left (f x +e \right )}}{2 c^{2} f}-\frac {8 \left (-3 i A \,a^{2} {\mathrm e}^{i \left (f x +e \right )}+3 A \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-9 i B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}+6 B \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-2 a^{2} A -5 a^{2} B \right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\) | \(160\) |
| parallelrisch | \(\frac {3 a^{2} \left (\left (4 \left (-f x +2\right ) B -f x A +\frac {4 A}{3}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {\left (\left (\frac {11}{6}+4 f x \right ) B +f x A +\frac {4 A}{3}\right ) \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{3}+\left (f x A +4 f x B -\frac {4}{3} A -\frac {14}{3} B \right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {\left (f x A +4 f x B -4 A -\frac {29}{2} B \right ) \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )}{3}-\frac {B \left (\cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )\right )}{6}\right )}{f \,c^{2} \left (-3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+3 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\) | \(186\) |
| norman | \(\frac {\frac {8 a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c f}+\frac {a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c}+\frac {8 a^{2} A +38 a^{2} B}{3 c f}-\frac {a^{2} \left (A +4 B \right ) x}{c}-\frac {2 \left (4 a^{2} A +13 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c f}+\frac {2 \left (4 a^{2} A +25 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}+\frac {2 \left (4 a^{2} A +35 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}+\frac {2 \left (4 a^{2} A +61 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f}-\frac {\left (8 a^{2} A +30 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {2 \left (12 a^{2} A +41 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}-\frac {2 \left (12 a^{2} A +43 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}+\frac {3 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {6 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c}+\frac {10 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c}-\frac {12 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}+\frac {12 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}-\frac {10 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}+\frac {6 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}-\frac {3 a^{2} \left (A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) | \(536\) |
Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x,method=_RETUR NVERBOSE)
Output:
2/f*a^2/c^2*(-B/(1+tan(1/2*f*x+1/2*e)^2)+(A+4*B)*arctan(tan(1/2*f*x+1/2*e) )-1/3*(8*A+8*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(8*A+8*B)/(tan(1/2*f*x+1/2*e) -1)^2+4*B/(tan(1/2*f*x+1/2*e)-1))
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (107) = 214\).
Time = 0.09 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.17 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=-\frac {3 \, B a^{2} \cos \left (f x + e\right )^{3} + 6 \, {\left (A + 4 \, B\right )} a^{2} f x + 4 \, {\left (A + B\right )} a^{2} - {\left (3 \, {\left (A + 4 \, B\right )} a^{2} f x + {\left (8 \, A + 23 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (3 \, {\left (A + 4 \, B\right )} a^{2} f x - 2 \, {\left (2 \, A + 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) - {\left (6 \, {\left (A + 4 \, B\right )} a^{2} f x - 3 \, B a^{2} \cos \left (f x + e\right )^{2} - 4 \, {\left (A + B\right )} a^{2} + {\left (3 \, {\left (A + 4 \, B\right )} a^{2} f x - 2 \, {\left (4 \, A + 13 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algori thm="fricas")
Output:
-1/3*(3*B*a^2*cos(f*x + e)^3 + 6*(A + 4*B)*a^2*f*x + 4*(A + B)*a^2 - (3*(A + 4*B)*a^2*f*x + (8*A + 23*B)*a^2)*cos(f*x + e)^2 + (3*(A + 4*B)*a^2*f*x - 2*(2*A + 11*B)*a^2)*cos(f*x + e) - (6*(A + 4*B)*a^2*f*x - 3*B*a^2*cos(f* x + e)^2 - 4*(A + B)*a^2 + (3*(A + 4*B)*a^2*f*x - 2*(4*A + 13*B)*a^2)*cos( f*x + e))*sin(f*x + e))/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2 *f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 2474 vs. \(2 (100) = 200\).
Time = 4.37 (sec) , antiderivative size = 2474, normalized size of antiderivative = 22.70 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)
Output:
Piecewise((3*A*a**2*f*x*tan(e/2 + f*x/2)**5/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f *tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 9*A*a**2*f* x*tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f *x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 12*A*a**2*f*x*tan(e/2 + f*x/2)**3 /(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f* tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f *x/2) - 3*c**2*f) - 12*A*a**2*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 9 *A*a**2*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan( e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/ 2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 3*A*a**2*f*x/(3*c**2*f*tan (e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/ 2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2 *f) - 24*A*a**2*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2 *f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 8*A*a**2*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 ...
Leaf count of result is larger than twice the leaf count of optimal. 839 vs. \(2 (107) = 214\).
Time = 0.14 (sec) , antiderivative size = 839, normalized size of antiderivative = 7.70 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algori thm="maxima")
Output:
2/3*(2*B*a^2*((12*sin(f*x + e)/(cos(f*x + e) + 1) - 11*sin(f*x + e)^2/(cos (f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 3*sin(f*x + e)^ 4/(cos(f*x + e) + 1)^4 - 5)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e )^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + A*a^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f *x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 2*B*a^2*((9*sin(f*x + e )/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3 *c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/ (cos(f*x + e) + 1))/c^2) - A*a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 3*si n(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3 /(cos(f*x + e) + 1)^3) + 2*A*a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/( c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f* x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + B*a^2*(3*sin(f* x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) ...
Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {\frac {3 \, {\left (A a^{2} + 4 \, B a^{2}\right )} {\left (f x + e\right )}}{c^{2}} - \frac {6 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} c^{2}} + \frac {8 \, {\left (3 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A a^{2} + 4 \, B a^{2}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algori thm="giac")
Output:
1/3*(3*(A*a^2 + 4*B*a^2)*(f*x + e)/c^2 - 6*B*a^2/((tan(1/2*f*x + 1/2*e)^2 + 1)*c^2) + 8*(3*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 3*A*a^2*tan(1/2*f*x + 1/2* e) - 9*B*a^2*tan(1/2*f*x + 1/2*e) + A*a^2 + 4*B*a^2)/(c^2*(tan(1/2*f*x + 1 /2*e) - 1)^3))/f
Time = 37.40 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.26 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {2\,a^2\,\mathrm {atan}\left (\frac {2\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A+4\,B\right )}{2\,A\,a^2+8\,B\,a^2}\right )\,\left (A+4\,B\right )}{c^2\,f}-\frac {\frac {8\,A\,a^2}{3}-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (8\,A\,a^2+30\,B\,a^2\right )+\frac {38\,B\,a^2}{3}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (8\,A\,a^2+26\,B\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {8\,A\,a^2}{3}+\frac {74\,B\,a^2}{3}\right )+8\,B\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{f\,\left (-c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+3\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c^2\right )} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^2,x )
Output:
(2*a^2*atan((2*a^2*tan(e/2 + (f*x)/2)*(A + 4*B))/(2*A*a^2 + 8*B*a^2))*(A + 4*B))/(c^2*f) - ((8*A*a^2)/3 - tan(e/2 + (f*x)/2)*(8*A*a^2 + 30*B*a^2) + (38*B*a^2)/3 - tan(e/2 + (f*x)/2)^3*(8*A*a^2 + 26*B*a^2) + tan(e/2 + (f*x) /2)^2*((8*A*a^2)/3 + (74*B*a^2)/3) + 8*B*a^2*tan(e/2 + (f*x)/2)^4)/(f*(4*c ^2*tan(e/2 + (f*x)/2)^2 - 4*c^2*tan(e/2 + (f*x)/2)^3 + 3*c^2*tan(e/2 + (f* x)/2)^4 - c^2*tan(e/2 + (f*x)/2)^5 + c^2 - 3*c^2*tan(e/2 + (f*x)/2)))
Time = 0.18 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.54 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a^{2} \left (-3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a f x +4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +12 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b f x +15 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -3 \cos \left (f x +e \right ) a f x -12 \cos \left (f x +e \right ) b f x -8 \cos \left (f x +e \right ) b +3 \sin \left (f x +e \right )^{3} b +3 \sin \left (f x +e \right )^{2} a f x -12 \sin \left (f x +e \right )^{2} a +12 \sin \left (f x +e \right )^{2} b f x -34 \sin \left (f x +e \right )^{2} b -6 \sin \left (f x +e \right ) a f x +4 a \sin \left (f x +e \right )-24 \sin \left (f x +e \right ) b f x +15 \sin \left (f x +e \right ) b +3 a f x +12 b f x +8 b \right )}{3 c^{2} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-\cos \left (f x +e \right )+\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1\right )} \] Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x)
Output:
(a**2*( - 3*cos(e + f*x)*sin(e + f*x)**2*b + 3*cos(e + f*x)*sin(e + f*x)*a *f*x + 4*cos(e + f*x)*sin(e + f*x)*a + 12*cos(e + f*x)*sin(e + f*x)*b*f*x + 15*cos(e + f*x)*sin(e + f*x)*b - 3*cos(e + f*x)*a*f*x - 12*cos(e + f*x)* b*f*x - 8*cos(e + f*x)*b + 3*sin(e + f*x)**3*b + 3*sin(e + f*x)**2*a*f*x - 12*sin(e + f*x)**2*a + 12*sin(e + f*x)**2*b*f*x - 34*sin(e + f*x)**2*b - 6*sin(e + f*x)*a*f*x + 4*sin(e + f*x)*a - 24*sin(e + f*x)*b*f*x + 15*sin(e + f*x)*b + 3*a*f*x + 12*b*f*x + 8*b))/(3*c**2*f*(cos(e + f*x)*sin(e + f*x ) - cos(e + f*x) + sin(e + f*x)**2 - 2*sin(e + f*x) + 1))