Integrand size = 36, antiderivative size = 162 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {5 (2 A-5 B) c^3 x}{2 a^2}+\frac {5 (2 A-5 B) c^3 \cos (e+f x)}{2 a^2 f}-\frac {a^3 (A-B) c^3 \cos ^7(e+f x)}{3 f (a+a \sin (e+f x))^5}+\frac {2 a (2 A-5 B) c^3 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {5 (2 A-5 B) c^3 \cos ^3(e+f x)}{6 f \left (a^2+a^2 \sin (e+f x)\right )} \] Output:
5/2*(2*A-5*B)*c^3*x/a^2+5/2*(2*A-5*B)*c^3*cos(f*x+e)/a^2/f-1/3*a^3*(A-B)*c ^3*cos(f*x+e)^7/f/(a+a*sin(f*x+e))^5+2/3*a*(2*A-5*B)*c^3*cos(f*x+e)^5/f/(a +a*sin(f*x+e))^3+5/6*(2*A-5*B)*c^3*cos(f*x+e)^3/f/(a^2+a^2*sin(f*x+e))
Time = 11.53 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.69 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c-c \sin (e+f x))^3 \left (64 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )-32 (A-B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-32 (7 A-13 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+30 (2 A-5 B) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+12 (A-5 B) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+3 B \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sin (2 (e+f x))\right )}{12 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (1+\sin (e+f x))^2} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^3)/(a + a*Sin[e + f*x ])^2,x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^3*(64*(A - B)* Sin[(e + f*x)/2] - 32*(A - B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 32*( 7*A - 13*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 30* (2*A - 5*B)*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 12*(A - 5* B)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 3*B*(Cos[(e + f* x)/2] + Sin[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(1 + Sin[e + f*x])^2)
Time = 0.81 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {3042, 3446, 3042, 3338, 3042, 3159, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3 (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3 (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^2}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^5}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^4}dx}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^4}dx}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \left (-\frac {5 \int \frac {\cos ^4(e+f x)}{(\sin (e+f x) a+a)^2}dx}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \left (-\frac {5 \int \frac {\cos (e+f x)^4}{(\sin (e+f x) a+a)^2}dx}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \left (-\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{\sin (e+f x) a+a}dx}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \left (-\frac {5 \left (\frac {3 \int \frac {\cos (e+f x)^2}{\sin (e+f x) a+a}dx}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \left (-\frac {5 \left (\frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (e+f x)}{a f}\right )}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (-\frac {(2 A-5 B) \left (-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac {3 \left (\frac {\cos (e+f x)}{a f}+\frac {x}{a}\right )}{2 a}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a}-\frac {(A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}\right )\) |
Input:
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^3)/(a + a*Sin[e + f*x])^2,x ]
Output:
a^3*c^3*(-1/3*((A - B)*Cos[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^5) - ((2*A - 5*B)*((-2*Cos[e + f*x]^5)/(a*f*(a + a*Sin[e + f*x])^3) - (5*((3*(x/a + C os[e + f*x]/(a*f)))/(2*a) + Cos[e + f*x]^3/(2*f*(a^2 + a^2*Sin[e + f*x]))) )/a^2))/(3*a))
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 _)])^(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 = 1.95 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {2 c^{3} \left (-\frac {-16 A +16 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-4 A +12 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {16 A -16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A -5 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A -5 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {5 \left (2 A -5 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f \,a^{2}}\) | \(162\) |
| default | \(\frac {2 c^{3} \left (-\frac {-16 A +16 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-4 A +12 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {16 A -16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A -5 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A -5 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {5 \left (2 A -5 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f \,a^{2}}\) | \(162\) |
| parallelrisch | \(-\frac {c^{3} \left (\left (-30 f x A +75 f x B -56 A +\frac {607}{4} B \right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (10 f x A -25 f x B -\frac {19}{3} A +\frac {115}{12} B \right ) \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (-30 f x A +75 f x B -36 A +\frac {337}{4} B \right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (-10 f x A +25 f x B -37 A +\frac {353}{4} B \right ) \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (A -\frac {17 B}{4}\right ) \cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\left (-A +\frac {17 B}{4}\right ) \sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\frac {B \left (\cos \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )+\sin \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )\right )}{4}\right )}{2 f \,a^{2} \left (\sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+3 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )\right )}\) | \(220\) |
| risch | \(\frac {5 c^{3} x A}{a^{2}}-\frac {25 c^{3} x B}{2 a^{2}}-\frac {i B \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{2} f}+\frac {c^{3} {\mathrm e}^{i \left (f x +e \right )} A}{2 a^{2} f}-\frac {5 c^{3} {\mathrm e}^{i \left (f x +e \right )} B}{2 a^{2} f}+\frac {c^{3} {\mathrm e}^{-i \left (f x +e \right )} A}{2 a^{2} f}-\frac {5 c^{3} {\mathrm e}^{-i \left (f x +e \right )} B}{2 a^{2} f}+\frac {i B \,c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f}+\frac {32 i A \,c^{3} {\mathrm e}^{i \left (f x +e \right )}+24 A \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}-64 i B \,c^{3} {\mathrm e}^{i \left (f x +e \right )}-40 B \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}-\frac {56 A \,c^{3}}{3}+\frac {104 B \,c^{3}}{3}}{f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\) | \(247\) |
| norman | \(\frac {\frac {\left (8 A \,c^{3}-25 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a f}+\frac {\left (34 A \,c^{3}-77 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a f}+\frac {\left (38 A \,c^{3}-93 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {\left (148 A \,c^{3}-352 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}+\frac {46 A \,c^{3}-118 B \,c^{3}}{3 a f}+\frac {5 c^{3} \left (2 A -5 B \right ) x}{2 a}+\frac {2 \left (68 A \,c^{3}-194 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f}+\frac {2 \left (70 A \,c^{3}-160 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}+\frac {2 \left (108 A \,c^{3}-251 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}+\frac {\left (154 A \,c^{3}-478 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 a f}+\frac {2 \left (182 A \,c^{3}-545 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 a f}+\frac {\left (220 A \,c^{3}-595 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 a f}+\frac {15 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}+\frac {35 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 a}+\frac {65 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 a}+\frac {45 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}+\frac {55 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}+\frac {55 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}+\frac {45 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}+\frac {65 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 a}+\frac {35 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 a}+\frac {15 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{2 a}+\frac {5 c^{3} \left (2 A -5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{2 a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\) | \(680\) |
Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x,method=_RETUR NVERBOSE)
Output:
2/f*c^3/a^2*(-1/2*(-16*A+16*B)/(tan(1/2*f*x+1/2*e)+1)^2-(-4*A+12*B)/(tan(1 /2*f*x+1/2*e)+1)-1/3*(16*A-16*B)/(tan(1/2*f*x+1/2*e)+1)^3+(-1/2*B*tan(1/2* f*x+1/2*e)^3+(A-5*B)*tan(1/2*f*x+1/2*e)^2+1/2*B*tan(1/2*f*x+1/2*e)+A-5*B)/ (1+tan(1/2*f*x+1/2*e)^2)^2+5/2*(2*A-5*B)*arctan(tan(1/2*f*x+1/2*e)))
Time = 0.09 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {3 \, B c^{3} \cos \left (f x + e\right )^{4} + 6 \, {\left (A - 4 \, B\right )} c^{3} \cos \left (f x + e\right )^{3} - 30 \, {\left (2 \, A - 5 \, B\right )} c^{3} f x + 16 \, {\left (A - B\right )} c^{3} + {\left (15 \, {\left (2 \, A - 5 \, B\right )} c^{3} f x - {\left (62 \, A - 131 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, {\left (2 \, A - 5 \, B\right )} c^{3} f x + 2 \, {\left (26 \, A - 71 \, B\right )} c^{3}\right )} \cos \left (f x + e\right ) + {\left (3 \, B c^{3} \cos \left (f x + e\right )^{3} - 30 \, {\left (2 \, A - 5 \, B\right )} c^{3} f x - 3 \, {\left (2 \, A - 9 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 16 \, {\left (A - B\right )} c^{3} - {\left (15 \, {\left (2 \, A - 5 \, B\right )} c^{3} f x + 2 \, {\left (34 \, A - 79 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algori thm="fricas")
Output:
1/6*(3*B*c^3*cos(f*x + e)^4 + 6*(A - 4*B)*c^3*cos(f*x + e)^3 - 30*(2*A - 5 *B)*c^3*f*x + 16*(A - B)*c^3 + (15*(2*A - 5*B)*c^3*f*x - (62*A - 131*B)*c^ 3)*cos(f*x + e)^2 - (15*(2*A - 5*B)*c^3*f*x + 2*(26*A - 71*B)*c^3)*cos(f*x + e) + (3*B*c^3*cos(f*x + e)^3 - 30*(2*A - 5*B)*c^3*f*x - 3*(2*A - 9*B)*c ^3*cos(f*x + e)^2 - 16*(A - B)*c^3 - (15*(2*A - 5*B)*c^3*f*x + 2*(34*A - 7 9*B)*c^3)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f* x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 4665 vs. \(2 (148) = 296\).
Time = 7.78 (sec) , antiderivative size = 4665, normalized size of antiderivative = 28.80 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**3/(a+a*sin(f*x+e))**2,x)
Output:
Piecewise((30*A*c**3*f*x*tan(e/2 + f*x/2)**7/(6*a**2*f*tan(e/2 + f*x/2)**7 + 18*a**2*f*tan(e/2 + f*x/2)**6 + 30*a**2*f*tan(e/2 + f*x/2)**5 + 42*a**2 *f*tan(e/2 + f*x/2)**4 + 42*a**2*f*tan(e/2 + f*x/2)**3 + 30*a**2*f*tan(e/2 + f*x/2)**2 + 18*a**2*f*tan(e/2 + f*x/2) + 6*a**2*f) + 90*A*c**3*f*x*tan( e/2 + f*x/2)**6/(6*a**2*f*tan(e/2 + f*x/2)**7 + 18*a**2*f*tan(e/2 + f*x/2) **6 + 30*a**2*f*tan(e/2 + f*x/2)**5 + 42*a**2*f*tan(e/2 + f*x/2)**4 + 42*a **2*f*tan(e/2 + f*x/2)**3 + 30*a**2*f*tan(e/2 + f*x/2)**2 + 18*a**2*f*tan( e/2 + f*x/2) + 6*a**2*f) + 150*A*c**3*f*x*tan(e/2 + f*x/2)**5/(6*a**2*f*ta n(e/2 + f*x/2)**7 + 18*a**2*f*tan(e/2 + f*x/2)**6 + 30*a**2*f*tan(e/2 + f* x/2)**5 + 42*a**2*f*tan(e/2 + f*x/2)**4 + 42*a**2*f*tan(e/2 + f*x/2)**3 + 30*a**2*f*tan(e/2 + f*x/2)**2 + 18*a**2*f*tan(e/2 + f*x/2) + 6*a**2*f) + 2 10*A*c**3*f*x*tan(e/2 + f*x/2)**4/(6*a**2*f*tan(e/2 + f*x/2)**7 + 18*a**2* f*tan(e/2 + f*x/2)**6 + 30*a**2*f*tan(e/2 + f*x/2)**5 + 42*a**2*f*tan(e/2 + f*x/2)**4 + 42*a**2*f*tan(e/2 + f*x/2)**3 + 30*a**2*f*tan(e/2 + f*x/2)** 2 + 18*a**2*f*tan(e/2 + f*x/2) + 6*a**2*f) + 210*A*c**3*f*x*tan(e/2 + f*x/ 2)**3/(6*a**2*f*tan(e/2 + f*x/2)**7 + 18*a**2*f*tan(e/2 + f*x/2)**6 + 30*a **2*f*tan(e/2 + f*x/2)**5 + 42*a**2*f*tan(e/2 + f*x/2)**4 + 42*a**2*f*tan( e/2 + f*x/2)**3 + 30*a**2*f*tan(e/2 + f*x/2)**2 + 18*a**2*f*tan(e/2 + f*x/ 2) + 6*a**2*f) + 150*A*c**3*f*x*tan(e/2 + f*x/2)**2/(6*a**2*f*tan(e/2 + f* x/2)**7 + 18*a**2*f*tan(e/2 + f*x/2)**6 + 30*a**2*f*tan(e/2 + f*x/2)**5...
Leaf count of result is larger than twice the leaf count of optimal. 1378 vs. \(2 (152) = 304\).
Time = 0.15 (sec) , antiderivative size = 1378, normalized size of antiderivative = 8.51 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algori thm="maxima")
Output:
-1/3*(B*c^3*((75*sin(f*x + e)/(cos(f*x + e) + 1) + 97*sin(f*x + e)^2/(cos( f*x + e) + 1)^2 + 126*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 98*sin(f*x + e )^4/(cos(f*x + e) + 1)^4 + 63*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 21*sin (f*x + e)^6/(cos(f*x + e) + 1)^6 + 32)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*a^2*sin(f*x + e) ^3/(cos(f*x + e) + 1)^3 + 7*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5*a^ 2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e )/(cos(f*x + e) + 1))/a^2) - 4*A*c^3*((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)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2 *sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/ (cos(f*x + e) + 1))/a^2) + 12*B*c^3*((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)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2* sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e...
Time = 0.24 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {\frac {15 \, {\left (2 \, A c^{3} - 5 \, B c^{3}\right )} {\left (f x + e\right )}}{a^{2}} - \frac {6 \, {\left (B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 10 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A c^{3} + 10 \, B c^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} a^{2}} + \frac {16 \, {\left (3 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 24 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, A c^{3} - 11 \, B c^{3}\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{6 \, f} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algori thm="giac")
Output:
1/6*(15*(2*A*c^3 - 5*B*c^3)*(f*x + e)/a^2 - 6*(B*c^3*tan(1/2*f*x + 1/2*e)^ 3 - 2*A*c^3*tan(1/2*f*x + 1/2*e)^2 + 10*B*c^3*tan(1/2*f*x + 1/2*e)^2 - B*c ^3*tan(1/2*f*x + 1/2*e) - 2*A*c^3 + 10*B*c^3)/((tan(1/2*f*x + 1/2*e)^2 + 1 )^2*a^2) + 16*(3*A*c^3*tan(1/2*f*x + 1/2*e)^2 - 9*B*c^3*tan(1/2*f*x + 1/2* e)^2 + 12*A*c^3*tan(1/2*f*x + 1/2*e) - 24*B*c^3*tan(1/2*f*x + 1/2*e) + 5*A *c^3 - 11*B*c^3)/(a^2*(tan(1/2*f*x + 1/2*e) + 1)^3))/f
Time = 38.71 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.07 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (38\,A\,c^3-93\,B\,c^3\right )+\frac {46\,A\,c^3}{3}-\frac {118\,B\,c^3}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (8\,A\,c^3-25\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (34\,A\,c^3-77\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (72\,A\,c^3-166\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {106\,A\,c^3}{3}-\frac {328\,B\,c^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {128\,A\,c^3}{3}-\frac {359\,B\,c^3}{3}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+5\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+7\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+7\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+5\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^2\right )}+\frac {5\,c^3\,\mathrm {atan}\left (\frac {5\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A-5\,B\right )}{10\,A\,c^3-25\,B\,c^3}\right )\,\left (2\,A-5\,B\right )}{a^2\,f} \] Input:
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^3)/(a + a*sin(e + f*x))^2,x )
Output:
(tan(e/2 + (f*x)/2)*(38*A*c^3 - 93*B*c^3) + (46*A*c^3)/3 - (118*B*c^3)/3 + tan(e/2 + (f*x)/2)^6*(8*A*c^3 - 25*B*c^3) + tan(e/2 + (f*x)/2)^5*(34*A*c^ 3 - 77*B*c^3) + tan(e/2 + (f*x)/2)^3*(72*A*c^3 - 166*B*c^3) + tan(e/2 + (f *x)/2)^4*((106*A*c^3)/3 - (328*B*c^3)/3) + tan(e/2 + (f*x)/2)^2*((128*A*c^ 3)/3 - (359*B*c^3)/3))/(f*(5*a^2*tan(e/2 + (f*x)/2)^2 + 7*a^2*tan(e/2 + (f *x)/2)^3 + 7*a^2*tan(e/2 + (f*x)/2)^4 + 5*a^2*tan(e/2 + (f*x)/2)^5 + 3*a^2 *tan(e/2 + (f*x)/2)^6 + a^2*tan(e/2 + (f*x)/2)^7 + a^2 + 3*a^2*tan(e/2 + ( f*x)/2))) + (5*c^3*atan((5*c^3*tan(e/2 + (f*x)/2)*(2*A - 5*B))/(10*A*c^3 - 25*B*c^3))*(2*A - 5*B))/(a^2*f)
Time = 0.18 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.13 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {c^{3} \left (16 a -50 b +30 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a f x -75 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b f x +50 \cos \left (f x +e \right ) b +93 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -30 a f x +75 b f x +30 \cos \left (f x +e \right ) a f x -75 \cos \left (f x +e \right ) b f x -60 \sin \left (f x +e \right ) a f x +150 \sin \left (f x +e \right ) b f x +24 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b -38 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -38 a \sin \left (f x +e \right )-3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a -3 \sin \left (f x +e \right )^{4} b -6 \sin \left (f x +e \right )^{3} a -16 \cos \left (f x +e \right ) a +27 \sin \left (f x +e \right )^{3} b -92 \sin \left (f x +e \right )^{2} a +205 \sin \left (f x +e \right )^{2} b -30 \sin \left (f x +e \right )^{2} a f x +75 \sin \left (f x +e \right )^{2} b f x +93 \sin \left (f x +e \right ) b \right )}{6 a^{2} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )-\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )-1\right )} \] Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x)
Output:
(c**3*( - 3*cos(e + f*x)*sin(e + f*x)**3*b - 6*cos(e + f*x)*sin(e + f*x)** 2*a + 24*cos(e + f*x)*sin(e + f*x)**2*b + 30*cos(e + f*x)*sin(e + f*x)*a*f *x - 38*cos(e + f*x)*sin(e + f*x)*a - 75*cos(e + f*x)*sin(e + f*x)*b*f*x + 93*cos(e + f*x)*sin(e + f*x)*b + 30*cos(e + f*x)*a*f*x - 16*cos(e + f*x)* a - 75*cos(e + f*x)*b*f*x + 50*cos(e + f*x)*b - 3*sin(e + f*x)**4*b - 6*si n(e + f*x)**3*a + 27*sin(e + f*x)**3*b - 30*sin(e + f*x)**2*a*f*x - 92*sin (e + f*x)**2*a + 75*sin(e + f*x)**2*b*f*x + 205*sin(e + f*x)**2*b - 60*sin (e + f*x)*a*f*x - 38*sin(e + f*x)*a + 150*sin(e + f*x)*b*f*x + 93*sin(e + f*x)*b - 30*a*f*x + 16*a + 75*b*f*x - 50*b))/(6*a**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))