Integrand size = 36, antiderivative size = 163 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {5 a^3 (2 A+5 B) x}{2 c^2}-\frac {5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )} \] Output:
5/2*a^3*(2*A+5*B)*x/c^2-5/2*a^3*(2*A+5*B)*cos(f*x+e)/c^2/f+1/3*a^3*(A+B)*c ^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^5-2/3*a^3*(2*A+5*B)*c*cos(f*x+e)^5/f/(c -c*sin(f*x+e))^3-5/6*a^3*(2*A+5*B)*cos(f*x+e)^3/f/(c^2-c^2*sin(f*x+e))
Time = 11.55 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.72 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (32 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+30 (2 A+5 B) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-12 (A+5 B) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+64 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-32 (7 A+13 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sin (2 (e+f x))\right )}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^2} \] Input:
Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^2,x]
Output:
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(32*(A + B )*(Cos[(e + f*x)/2] - Sin[(e + f*x)/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 + 64*(A + B)*Sin[(e + f*x)/2] - 32*(7*A + 13*B) *(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] - 3*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^2)
Time = 0.85 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.91, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^4}dx}{3 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^4}dx}{3 c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^2}dx}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^2}dx}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)}dx}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \int \frac {\cos (e+f x)^2}{c-c \sin (e+f x)}dx}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \left (\frac {\int 1dx}{c}-\frac {\cos (e+f x)}{c f}\right )}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )}{3 c}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {(2 A+5 B) \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \left (\frac {x}{c}-\frac {\cos (e+f x)}{c f}\right )}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )}{3 c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x ]
Output:
a^3*c^3*(((A + B)*Cos[e + f*x]^7)/(3*f*(c - c*Sin[e + f*x])^5) - ((2*A + 5 *B)*((2*Cos[e + f*x]^5)/(c*f*(c - c*Sin[e + f*x])^3) - (5*((3*(x/c - Cos[e + f*x]/(c*f)))/(2*c) - Cos[e + f*x]^3/(2*f*(c^2 - c^2*Sin[e + f*x]))))/c^ 2))/(3*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 _)])^(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 = 1.89 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {-4 A -12 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {16 A +16 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (-A -5 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-A -5 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {5 \left (2 A +5 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f \,c^{2}}\) | \(166\) |
| default | \(\frac {2 a^{3} \left (-\frac {-4 A -12 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {16 A +16 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (-A -5 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-A -5 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {5 \left (2 A +5 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f \,c^{2}}\) | \(166\) |
| parallelrisch | \(-\frac {\left (\left (30 f x A +75 f x B -56 A -\frac {607}{4} B \right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (-10 f x A -25 f x B -\frac {19}{3} A -\frac {115}{12} B \right ) \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (-30 f x A -75 f x B +36 A +\frac {337}{4} B \right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (-10 f x A -25 f x B +37 A +\frac {353}{4} B \right ) \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (A +\frac {17 B}{4}\right ) \cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\left (A +\frac {17 B}{4}\right ) \sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )-\frac {B \left (\cos \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )-\sin \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )\right )}{4}\right ) a^{3}}{2 f \,c^{2} \left (-3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+3 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}\) | \(218\) |
| risch | \(\frac {5 a^{3} x A}{c^{2}}+\frac {25 a^{3} x B}{2 c^{2}}+\frac {i a^{3} B \,{\mathrm e}^{2 i \left (f x +e \right )}}{8 c^{2} f}-\frac {a^{3} {\mathrm e}^{i \left (f x +e \right )} A}{2 c^{2} f}-\frac {5 a^{3} {\mathrm e}^{i \left (f x +e \right )} B}{2 c^{2} f}-\frac {a^{3} {\mathrm e}^{-i \left (f x +e \right )} A}{2 c^{2} f}-\frac {5 a^{3} {\mathrm e}^{-i \left (f x +e \right )} B}{2 c^{2} f}-\frac {i a^{3} B \,{\mathrm e}^{-2 i \left (f x +e \right )}}{8 c^{2} f}-\frac {8 \left (-12 i A \,a^{3} {\mathrm e}^{i \left (f x +e \right )}+9 A \,a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-24 i B \,a^{3} {\mathrm e}^{i \left (f x +e \right )}+15 B \,a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-7 a^{3} A -13 a^{3} B \right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\) | \(247\) |
| norman | \(\frac {\frac {\left (8 a^{3} A +25 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{c f}+\frac {46 a^{3} A +118 a^{3} B}{3 c f}-\frac {5 a^{3} \left (2 A +5 B \right ) x}{2 c}-\frac {\left (34 a^{3} A +77 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c f}-\frac {\left (38 a^{3} A +93 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}+\frac {2 \left (68 a^{3} A +194 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}-\frac {2 \left (70 a^{3} A +160 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c f}-\frac {2 \left (108 a^{3} A +251 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}-\frac {\left (148 a^{3} A +352 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}+\frac {\left (154 a^{3} A +478 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 c f}+\frac {2 \left (182 a^{3} A +545 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f}+\frac {\left (220 a^{3} A +595 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 c f}+\frac {15 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c}-\frac {35 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 c}+\frac {65 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 c}-\frac {45 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}+\frac {55 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}-\frac {55 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}+\frac {45 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}-\frac {65 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 c}+\frac {35 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 c}-\frac {15 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{2 c}+\frac {5 a^{3} \left (2 A +5 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{2 c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) | \(683\) |
Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x,method=_RETUR NVERBOSE)
Output:
2/f*a^3/c^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*(16*A+16*B)/(tan(1/2*f*x+1/2*e)-1)^2+(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 = 0.13 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.75 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {3 \, B a^{3} \cos \left (f x + e\right )^{4} - 6 \, {\left (A + 4 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 30 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x - 16 \, {\left (A + B\right )} a^{3} + {\left (15 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x + {\left (62 \, A + 131 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x - 2 \, {\left (26 \, A + 71 \, B\right )} a^{3}\right )} \cos \left (f x + e\right ) - {\left (3 \, B a^{3} \cos \left (f x + e\right )^{3} - 30 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x + 3 \, {\left (2 \, A + 9 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 16 \, {\left (A + B\right )} a^{3} - {\left (15 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x - 2 \, {\left (34 \, A + 79 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algori thm="fricas")
Output:
1/6*(3*B*a^3*cos(f*x + e)^4 - 6*(A + 4*B)*a^3*cos(f*x + e)^3 - 30*(2*A + 5 *B)*a^3*f*x - 16*(A + B)*a^3 + (15*(2*A + 5*B)*a^3*f*x + (62*A + 131*B)*a^ 3)*cos(f*x + e)^2 - (15*(2*A + 5*B)*a^3*f*x - 2*(26*A + 71*B)*a^3)*cos(f*x + e) - (3*B*a^3*cos(f*x + e)^3 - 30*(2*A + 5*B)*a^3*f*x + 3*(2*A + 9*B)*a ^3*cos(f*x + e)^2 + 16*(A + B)*a^3 - (15*(2*A + 5*B)*a^3*f*x - 2*(34*A + 7 9*B)*a^3)*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. 4665 vs. \(2 (151) = 302\).
Time = 8.21 (sec) , antiderivative size = 4665, normalized size of antiderivative = 28.62 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)
Output:
Piecewise((30*A*a**3*f*x*tan(e/2 + f*x/2)**7/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2 *f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 90*A*a**3*f*x*tan( e/2 + f*x/2)**6/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2) **6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c **2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan( e/2 + f*x/2) - 6*c**2*f) + 150*A*a**3*f*x*tan(e/2 + f*x/2)**5/(6*c**2*f*ta n(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f* x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 2 10*A*a**3*f*x*tan(e/2 + f*x/2)**4/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2* f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)** 2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 210*A*a**3*f*x*tan(e/2 + f*x/ 2)**3/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c **2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan( e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/ 2) - 6*c**2*f) - 150*A*a**3*f*x*tan(e/2 + f*x/2)**2/(6*c**2*f*tan(e/2 + f* x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5...
Leaf count of result is larger than twice the leaf count of optimal. 1386 vs. \(2 (156) = 312\).
Time = 0.16 (sec) , antiderivative size = 1386, normalized size of antiderivative = 8.50 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algori thm="maxima")
Output:
1/3*(B*a^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)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 7*c^2*sin(f*x + e)^ 3/(cos(f*x + e) + 1)^3 + 7*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*c^2 *sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*c^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e) /(cos(f*x + e) + 1))/c^2) + 4*A*a^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)/(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) + 12*B*a^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)/(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*s in(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)...
Time = 0.25 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.36 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {\frac {15 \, {\left (2 \, A a^{3} + 5 \, B a^{3}\right )} {\left (f x + e\right )}}{c^{2}} + \frac {6 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A a^{3} - 10 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} c^{2}} + \frac {16 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 24 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, A a^{3} + 11 \, B a^{3}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{6 \, f} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algori thm="giac")
Output:
1/6*(15*(2*A*a^3 + 5*B*a^3)*(f*x + e)/c^2 + 6*(B*a^3*tan(1/2*f*x + 1/2*e)^ 3 - 2*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 10*B*a^3*tan(1/2*f*x + 1/2*e)^2 - B*a ^3*tan(1/2*f*x + 1/2*e) - 2*A*a^3 - 10*B*a^3)/((tan(1/2*f*x + 1/2*e)^2 + 1 )^2*c^2) + 16*(3*A*a^3*tan(1/2*f*x + 1/2*e)^2 + 9*B*a^3*tan(1/2*f*x + 1/2* e)^2 - 12*A*a^3*tan(1/2*f*x + 1/2*e) - 24*B*a^3*tan(1/2*f*x + 1/2*e) + 5*A *a^3 + 11*B*a^3)/(c^2*(tan(1/2*f*x + 1/2*e) - 1)^3))/f
Time = 36.73 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.09 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {5\,a^3\,\mathrm {atan}\left (\frac {5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A+5\,B\right )}{10\,A\,a^3+25\,B\,a^3}\right )\,\left (2\,A+5\,B\right )}{c^2\,f}-\frac {\frac {46\,A\,a^3}{3}-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (38\,A\,a^3+93\,B\,a^3\right )+\frac {118\,B\,a^3}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (8\,A\,a^3+25\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (34\,A\,a^3+77\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (72\,A\,a^3+166\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {106\,A\,a^3}{3}+\frac {328\,B\,a^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {128\,A\,a^3}{3}+\frac {359\,B\,a^3}{3}\right )}{f\,\left (-c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+3\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-5\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+7\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-7\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+5\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c^2\right )} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^2,x )
Output:
(5*a^3*atan((5*a^3*tan(e/2 + (f*x)/2)*(2*A + 5*B))/(10*A*a^3 + 25*B*a^3))* (2*A + 5*B))/(c^2*f) - ((46*A*a^3)/3 - tan(e/2 + (f*x)/2)*(38*A*a^3 + 93*B *a^3) + (118*B*a^3)/3 + tan(e/2 + (f*x)/2)^6*(8*A*a^3 + 25*B*a^3) - tan(e/ 2 + (f*x)/2)^5*(34*A*a^3 + 77*B*a^3) - tan(e/2 + (f*x)/2)^3*(72*A*a^3 + 16 6*B*a^3) + tan(e/2 + (f*x)/2)^4*((106*A*a^3)/3 + (328*B*a^3)/3) + tan(e/2 + (f*x)/2)^2*((128*A*a^3)/3 + (359*B*a^3)/3))/(f*(5*c^2*tan(e/2 + (f*x)/2) ^2 - 7*c^2*tan(e/2 + (f*x)/2)^3 + 7*c^2*tan(e/2 + (f*x)/2)^4 - 5*c^2*tan(e /2 + (f*x)/2)^5 + 3*c^2*tan(e/2 + (f*x)/2)^6 - c^2*tan(e/2 + (f*x)/2)^7 + c^2 - 3*c^2*tan(e/2 + (f*x)/2)))
Time = 0.18 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.12 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx=\frac {a^{3} \left (16 a +50 b +30 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a f x +75 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b f x -50 \cos \left (f x +e \right ) b +93 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +30 a f x +75 b f x -30 \cos \left (f x +e \right ) a f x -75 \cos \left (f x +e \right ) b f x -60 \sin \left (f x +e \right ) a f x -150 \sin \left (f x +e \right ) b f x -24 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +38 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +38 a \sin \left (f x +e \right )-3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +3 \sin \left (f x +e \right )^{4} b +6 \sin \left (f x +e \right )^{3} a -16 \cos \left (f x +e \right ) a +27 \sin \left (f x +e \right )^{3} b -92 \sin \left (f x +e \right )^{2} a -205 \sin \left (f x +e \right )^{2} b +30 \sin \left (f x +e \right )^{2} a f x +75 \sin \left (f x +e \right )^{2} b f x +93 \sin \left (f x +e \right ) b \right )}{6 c^{2} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )-\cos \left (f x +e \right )+\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1\right )} \] Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x)
Output:
(a**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*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))