Integrand size = 36, antiderivative size = 112 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {a^2 B x}{c^3}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \] Output:
-a^2*B*x/c^3+1/5*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^5-2/3*a^2*B *cos(f*x+e)^3/f/(c-c*sin(f*x+e))^3+2*a^2*B*cos(f*x+e)/f/(c^3-c^3*sin(f*x+e ))
Leaf count is larger than twice the leaf count of optimal. \(278\) vs. \(2(112)=224\).
Time = 11.48 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.48 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-4 (3 A+8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-15 B (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-8 (3 A+8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+2 (3 A+43 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^3} \] Input:
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^3,x]
Output:
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 4*(3*A + 8*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 15*B*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 24*(A + B)*Sin [(e + f*x)/2] - 8*(3*A + 8*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[ (e + f*x)/2] + 2*(3*A + 43*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[ (e + f*x)/2])*(1 + Sin[e + f*x])^2)/(15*f*(Cos[(e + f*x)/2] + Sin[(e + f*x )/2])^4*(c - c*Sin[e + f*x])^3)
Time = 0.67 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04, 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, 3159, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {B \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^4}dx}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {B \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^4}dx}{c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {B \left (\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^2}dx}{c^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {B \left (\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^2}dx}{c^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {B \left (\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\frac {2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}-\frac {\int 1dx}{c^2}}{c^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {B \left (\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\frac {2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}-\frac {x}{c^2}}{c^2}\right )}{c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x ]
Output:
a^2*c^2*(((A + B)*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^5) - (B*((2*Co s[e + f*x]^3)/(3*c*f*(c - c*Sin[e + f*x])^3) - (-(x/c^2) + (2*Cos[e + f*x] )/(f*(c^2 - c^2*Sin[e + f*x])))/c^2))/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 _)])^(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 = 0.71 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12
| method | result | size |
| derivativedivides | \(\frac {2 a^{2} \left (-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\) | \(126\) |
| default | \(\frac {2 a^{2} \left (-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\) | \(126\) |
| parallelrisch | \(-\frac {2 \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} x f B}{2}+\left (-\frac {5}{2} f x B +A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+B \left (5 f x -4\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (-5 f x B +2 A +\frac {34}{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+B \left (\frac {5 f x}{2}-\frac {20}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {f x B}{2}+\frac {A}{5}+\frac {23 B}{15}\right ) a^{2}}{f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(133\) |
| risch | \(-\frac {a^{2} B x}{c^{3}}+\frac {-4 A \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 A \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-\frac {100 B \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{3}-24 i B \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}+\frac {56 i B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}}{3}+10 B \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 a^{2} A}{5}+\frac {86 a^{2} B}{15}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\) | \(139\) |
| norman | \(\frac {\frac {a^{2} x B}{c}+\frac {8 a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f c}+\frac {48 a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f c}-\frac {6 a^{2} A +46 a^{2} B}{15 f c}+\frac {40 a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f c}+\frac {64 a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f c}+\frac {112 a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 f c}-\frac {\left (2 a^{2} A +2 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{f c}-\frac {\left (30 a^{2} A +86 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 f c}-\frac {2 \left (38 a^{2} A +198 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 f c}-\frac {\left (78 a^{2} A +478 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 f c}-\frac {2 \left (138 a^{2} A +578 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 f c}-\frac {5 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}+\frac {13 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c}-\frac {25 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c}+\frac {38 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}-\frac {46 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}+\frac {46 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}-\frac {38 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}+\frac {25 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c}-\frac {13 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c}+\frac {5 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{c}-\frac {a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(566\) |
Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x,method=_RETUR NVERBOSE)
Output:
2/f*a^2/c^3*(-(A+B)/(tan(1/2*f*x+1/2*e)-1)-1/5*(16*A+16*B)/(tan(1/2*f*x+1/ 2*e)-1)^5-1/3*(24*A+16*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/4*(32*A+32*B)/(tan(1/ 2*f*x+1/2*e)-1)^4-4*A/(tan(1/2*f*x+1/2*e)-1)^2-B*arctan(tan(1/2*f*x+1/2*e) ))
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (111) = 222\).
Time = 0.09 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.47 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {60 \, B a^{2} f x - {\left (15 \, B a^{2} f x - {\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (A + B\right )} a^{2} - {\left (45 \, B a^{2} f x - {\left (9 \, A - 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B a^{2} f x - {\left (A + 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) - {\left (60 \, B a^{2} f x + 12 \, {\left (A + B\right )} a^{2} - {\left (15 \, B a^{2} f x + {\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B a^{2} f x + {\left (A - 9 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algori thm="fricas")
Output:
1/15*(60*B*a^2*f*x - (15*B*a^2*f*x - (3*A + 43*B)*a^2)*cos(f*x + e)^3 - 12 *(A + B)*a^2 - (45*B*a^2*f*x - (9*A - 11*B)*a^2)*cos(f*x + e)^2 + 6*(5*B*a ^2*f*x - (A + 11*B)*a^2)*cos(f*x + e) - (60*B*a^2*f*x + 12*(A + B)*a^2 - ( 15*B*a^2*f*x + (3*A + 43*B)*a^2)*cos(f*x + e)^2 + 6*(5*B*a^2*f*x + (A - 9* B)*a^2)*cos(f*x + e))*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. 1647 vs. \(2 (102) = 204\).
Time = 8.59 (sec) , antiderivative size = 1647, normalized size of antiderivative = 14.71 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)
Output:
Piecewise((-30*A*a**2*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) - 60*A*a* *2*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) - 6*A*a**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) - 15*B*a**2*f*x*tan(e/2 + f*x/2)**5/(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) + 75*B*a* *2*f*x*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) - 150*B*a**2*f*x*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) + 150*B*a**2*f*x*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) - 75*B*a**2*f*x*tan(e/2 + f*x/2)/(15*c**3*f*tan(e...
Leaf count of result is larger than twice the leaf count of optimal. 1139 vs. \(2 (111) = 222\).
Time = 0.15 (sec) , antiderivative size = 1139, normalized size of antiderivative = 10.17 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algori thm="maxima")
Output:
-2/15*(B*a^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) - 145*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 22)/(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/(c os(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) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1 ))/c^3) + A*a^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(c os(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/(c os(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) - 6*A*a^2*(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^2*(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*...
Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.35 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} B a^{2}}{c^{3}} + \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 170 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 100 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A a^{2} + 23 \, B a^{2}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{15 \, f} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algori thm="giac")
Output:
-1/15*(15*(f*x + e)*B*a^2/c^3 + 2*(15*A*a^2*tan(1/2*f*x + 1/2*e)^4 + 15*B* a^2*tan(1/2*f*x + 1/2*e)^4 - 60*B*a^2*tan(1/2*f*x + 1/2*e)^3 + 30*A*a^2*ta n(1/2*f*x + 1/2*e)^2 + 170*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 100*B*a^2*tan(1/ 2*f*x + 1/2*e) + 3*A*a^2 + 23*B*a^2)/(c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f
Time = 33.52 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.08 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=-\frac {B\,a^2\,x}{c^3}-\frac {\frac {a^2\,\left (6\,A+46\,B-15\,B\,\left (e+f\,x\right )\right )}{15}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {a^2\,\left (120\,B-150\,B\,\left (e+f\,x\right )\right )}{15}+10\,B\,a^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (30\,A+30\,B-75\,B\,\left (e+f\,x\right )\right )}{15}+5\,B\,a^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (60\,A+340\,B-150\,B\,\left (e+f\,x\right )\right )}{15}+10\,B\,a^2\,\left (e+f\,x\right )\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {a^2\,\left (200\,B-75\,B\,\left (e+f\,x\right )\right )}{15}+5\,B\,a^2\,\left (e+f\,x\right )\right )+B\,a^2\,\left (e+f\,x\right )}{c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^3,x )
Output:
- (B*a^2*x)/c^3 - ((a^2*(6*A + 46*B - 15*B*(e + f*x)))/15 - tan(e/2 + (f*x )/2)^3*((a^2*(120*B - 150*B*(e + f*x)))/15 + 10*B*a^2*(e + f*x)) + tan(e/2 + (f*x)/2)^4*((a^2*(30*A + 30*B - 75*B*(e + f*x)))/15 + 5*B*a^2*(e + f*x) ) + tan(e/2 + (f*x)/2)^2*((a^2*(60*A + 340*B - 150*B*(e + f*x)))/15 + 10*B *a^2*(e + f*x)) - tan(e/2 + (f*x)/2)*((a^2*(200*B - 75*B*(e + f*x)))/15 + 5*B*a^2*(e + f*x)) + B*a^2*(e + f*x))/(c^3*f*(tan(e/2 + (f*x)/2) - 1)^5)
Time = 0.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.29 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx=\frac {a^{2} \left (-6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a -15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b f x -6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b +75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} b f x -60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a -150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b f x +60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b +150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b f x -280 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b f x +170 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b +15 b f x -40 b \right )}{15 c^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x)
Output:
(a**2*( - 6*tan((e + f*x)/2)**5*a - 15*tan((e + f*x)/2)**5*b*f*x - 6*tan(( e + f*x)/2)**5*b + 75*tan((e + f*x)/2)**4*b*f*x - 60*tan((e + f*x)/2)**3*a - 150*tan((e + f*x)/2)**3*b*f*x + 60*tan((e + f*x)/2)**3*b + 150*tan((e + f*x)/2)**2*b*f*x - 280*tan((e + f*x)/2)**2*b - 30*tan((e + f*x)/2)*a - 75 *tan((e + f*x)/2)*b*f*x + 170*tan((e + f*x)/2)*b + 15*b*f*x - 40*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))