Integrand size = 35, antiderivative size = 164 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {B d^2 x}{a^3}-\frac {(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2+14 c d-29 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3} \] Output:
B*d^2*x/a^3-1/15*(c-d)*(B*(3*c-7*d)+2*A*(c+d))*cos(f*x+e)/a/f/(a+a*sin(f*x +e))^2-1/15*(B*(3*c^2+14*c*d-29*d^2)+2*A*(c^2+3*c*d+2*d^2))*cos(f*x+e)/f/( a^3+a^3*sin(f*x+e))-1/5*(A-B)*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x +e))^3
Leaf count is larger than twice the leaf count of optimal. \(514\) vs. \(2(164)=328\).
Time = 1.70 (sec) , antiderivative size = 514, normalized size of antiderivative = 3.13 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (30 \left (2 A d (c+d)+B \left (c^2+4 c d+d^2 (-9+5 e+5 f x)\right )\right ) \cos \left (\frac {1}{2} (e+f x)\right )-5 \left (4 A \left (c^2+3 c d+2 d^2\right )+B \left (6 c^2+16 c d+d^2 (-46+15 e+15 f x)\right )\right ) \cos \left (\frac {3}{2} (e+f x)\right )-15 B d^2 e \cos \left (\frac {5}{2} (e+f x)\right )-15 B d^2 f x \cos \left (\frac {5}{2} (e+f x)\right )+40 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )+30 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )+60 A c d \sin \left (\frac {1}{2} (e+f x)\right )+160 B c d \sin \left (\frac {1}{2} (e+f x)\right )+80 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )-370 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )+150 B d^2 e \sin \left (\frac {1}{2} (e+f x)\right )+150 B d^2 f x \sin \left (\frac {1}{2} (e+f x)\right )+60 B c d \sin \left (\frac {3}{2} (e+f x)\right )+30 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )-90 B d^2 \sin \left (\frac {3}{2} (e+f x)\right )+75 B d^2 e \sin \left (\frac {3}{2} (e+f x)\right )+75 B d^2 f x \sin \left (\frac {3}{2} (e+f x)\right )-4 A c^2 \sin \left (\frac {5}{2} (e+f x)\right )-6 B c^2 \sin \left (\frac {5}{2} (e+f x)\right )-12 A c d \sin \left (\frac {5}{2} (e+f x)\right )-28 B c d \sin \left (\frac {5}{2} (e+f x)\right )-14 A d^2 \sin \left (\frac {5}{2} (e+f x)\right )+64 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )-15 B d^2 e \sin \left (\frac {5}{2} (e+f x)\right )-15 B d^2 f x \sin \left (\frac {5}{2} (e+f x)\right )\right )}{60 a^3 f (1+\sin (e+f x))^3} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x ])^3,x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(30*(2*A*d*(c + d) + B*(c^2 + 4*c*d + d^2*(-9 + 5*e + 5*f*x)))*Cos[(e + f*x)/2] - 5*(4*A*(c^2 + 3*c*d + 2*d^2 ) + B*(6*c^2 + 16*c*d + d^2*(-46 + 15*e + 15*f*x)))*Cos[(3*(e + f*x))/2] - 15*B*d^2*e*Cos[(5*(e + f*x))/2] - 15*B*d^2*f*x*Cos[(5*(e + f*x))/2] + 40* A*c^2*Sin[(e + f*x)/2] + 30*B*c^2*Sin[(e + f*x)/2] + 60*A*c*d*Sin[(e + f*x )/2] + 160*B*c*d*Sin[(e + f*x)/2] + 80*A*d^2*Sin[(e + f*x)/2] - 370*B*d^2* Sin[(e + f*x)/2] + 150*B*d^2*e*Sin[(e + f*x)/2] + 150*B*d^2*f*x*Sin[(e + f *x)/2] + 60*B*c*d*Sin[(3*(e + f*x))/2] + 30*A*d^2*Sin[(3*(e + f*x))/2] - 9 0*B*d^2*Sin[(3*(e + f*x))/2] + 75*B*d^2*e*Sin[(3*(e + f*x))/2] + 75*B*d^2* f*x*Sin[(3*(e + f*x))/2] - 4*A*c^2*Sin[(5*(e + f*x))/2] - 6*B*c^2*Sin[(5*( e + f*x))/2] - 12*A*c*d*Sin[(5*(e + f*x))/2] - 28*B*c*d*Sin[(5*(e + f*x))/ 2] - 14*A*d^2*Sin[(5*(e + f*x))/2] + 64*B*d^2*Sin[(5*(e + f*x))/2] - 15*B* d^2*e*Sin[(5*(e + f*x))/2] - 15*B*d^2*f*x*Sin[(5*(e + f*x))/2]))/(60*a^3*f *(1 + Sin[e + f*x])^3)
Time = 1.04 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3456, 3042, 3447, 3042, 3498, 25, 3042, 3214, 3042, 3127}
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.
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (B (3 c-2 d)+2 A (c+d))+5 a B d \sin (e+f x))}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (B (3 c-2 d)+2 A (c+d))+5 a B d \sin (e+f x))}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {5 a B d^2 \sin ^2(e+f x)+(5 a B c d+a (B (3 c-2 d)+2 A (c+d)) d) \sin (e+f x)+a c (B (3 c-2 d)+2 A (c+d))}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a B d^2 \sin (e+f x)^2+(5 a B c d+a (B (3 c-2 d)+2 A (c+d)) d) \sin (e+f x)+a c (B (3 c-2 d)+2 A (c+d))}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {-\frac {\int -\frac {\left (B \left (3 c^2+14 d c-14 d^2\right )+2 A \left (c^2+3 d c+2 d^2\right )\right ) a^2+15 B d^2 \sin (e+f x) a^2}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\left (B \left (3 c^2+14 d c-14 d^2\right )+2 A \left (c^2+3 d c+2 d^2\right )\right ) a^2+15 B d^2 \sin (e+f x) a^2}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (B \left (3 c^2+14 d c-14 d^2\right )+2 A \left (c^2+3 d c+2 d^2\right )\right ) a^2+15 B d^2 \sin (e+f x) a^2}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {a^2 \left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \int \frac {1}{\sin (e+f x) a+a}dx+15 a B d^2 x}{3 a^2}-\frac {a (c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 \left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \int \frac {1}{\sin (e+f x) a+a}dx+15 a B d^2 x}{3 a^2}-\frac {a (c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {\frac {15 a B d^2 x-\frac {a^2 \left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \cos (e+f x)}{f (a \sin (e+f x)+a)}}{3 a^2}-\frac {a (c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
Input:
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^3,x ]
Output:
-1/5*((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]) ^3) + (-1/3*(a*(c - d)*(B*(3*c - 7*d) + 2*A*(c + d))*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^2) + (15*a*B*d^2*x - (a^2*(B*(3*c^2 + 14*c*d - 29*d^2) + 2*A*(c^2 + 3*c*d + 2*d^2))*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])))/(3*a^2) )/(5*a^2)
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.76 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.46
| method | result | size |
| parallelrisch | \(\frac {15 B x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} d^{2} f +\left (\left (75 f x +30\right ) B \,d^{2}-30 A \,c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (\left (150 f x +150\right ) B \,d^{2}-60 A c d -60 c^{2} \left (A +\frac {B}{2}\right )\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (\left (150 f x B -40 A +290 B \right ) d^{2}-60 c \left (A +\frac {4 B}{3}\right ) d -80 c^{2} \left (A +\frac {3 B}{8}\right )\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (\left (75 f x B -20 A +190 B \right ) d^{2}-60 c \left (A +\frac {2 B}{3}\right ) d -40 c^{2} \left (A +\frac {3 B}{4}\right )\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (15 f x B -4 A +44 B \right ) d^{2}-12 c \left (A +\frac {2 B}{3}\right ) d -14 c^{2} \left (A +\frac {3 B}{7}\right )}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(240\) |
| derivativedivides | \(\frac {2 B \,d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 \left (A \,c^{2}-B \,d^{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 A \,c^{2}+4 A c d +2 B \,c^{2}-2 B \,d^{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-8 A \,c^{2}+16 A c d -8 A \,d^{2}+8 B \,c^{2}-16 B c d +8 B \,d^{2}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A \,c^{2}-8 A c d +4 A \,d^{2}-4 B \,c^{2}+8 B c d -4 B \,d^{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 A \,c^{2}-12 A c d +4 A \,d^{2}-6 B \,c^{2}+8 B c d -2 B \,d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} f}\) | \(241\) |
| default | \(\frac {2 B \,d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 \left (A \,c^{2}-B \,d^{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 A \,c^{2}+4 A c d +2 B \,c^{2}-2 B \,d^{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-8 A \,c^{2}+16 A c d -8 A \,d^{2}+8 B \,c^{2}-16 B c d +8 B \,d^{2}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A \,c^{2}-8 A c d +4 A \,d^{2}-4 B \,c^{2}+8 B c d -4 B \,d^{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 A \,c^{2}-12 A c d +4 A \,d^{2}-6 B \,c^{2}+8 B c d -2 B \,d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} f}\) | \(241\) |
| risch | \(\frac {B \,d^{2} x}{a^{3}}-\frac {2 \left (2 A \,c^{2}+3 B \,c^{2}+7 A \,d^{2}+6 A c d -32 B \,d^{2}-10 i A \,c^{2} {\mathrm e}^{i \left (f x +e \right )}-15 i B \,c^{2} {\mathrm e}^{i \left (f x +e \right )}+14 B c d -20 A \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-15 B \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+185 B \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-135 i B \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-40 i B c d \,{\mathrm e}^{i \left (f x +e \right )}+60 i B c d \,{\mathrm e}^{3 i \left (f x +e \right )}+30 i A c d \,{\mathrm e}^{3 i \left (f x +e \right )}-40 A \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-45 B \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+15 A \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-30 A c d \,{\mathrm e}^{2 i \left (f x +e \right )}-20 i A \,d^{2} {\mathrm e}^{i \left (f x +e \right )}+30 i A \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+30 B c d \,{\mathrm e}^{4 i \left (f x +e \right )}+115 i B \,d^{2} {\mathrm e}^{i \left (f x +e \right )}+15 i B \,c^{2} {\mathrm e}^{3 i \left (f x +e \right )}-80 B c d \,{\mathrm e}^{2 i \left (f x +e \right )}-30 i A c d \,{\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(372\) |
| norman | \(\frac {\frac {B \,d^{2} x}{a}+\frac {B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{a}-\frac {14 A \,c^{2}+12 A c d +4 A \,d^{2}+6 B \,c^{2}+8 B c d -44 B \,d^{2}}{15 a f}-\frac {\left (2 A \,c^{2}-2 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a f}-\frac {\left (4 A \,c^{2}+4 A c d +2 B \,c^{2}-10 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a f}-\frac {\left (8 A \,c^{2}+12 A c d +4 A \,d^{2}+6 B \,c^{2}+8 B c d -38 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {2 \left (10 A \,c^{2}+12 A c d +2 A \,d^{2}+6 B \,c^{2}+4 B c d -34 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {\left (12 A \,c^{2}+16 A c d +4 A \,d^{2}+8 B \,c^{2}+8 B c d -48 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}-\frac {2 \left (22 A \,c^{2}+24 A c d +2 A \,d^{2}+12 B \,c^{2}+4 B c d -64 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 a f}-\frac {\left (34 A \,c^{2}+12 A c d +8 A \,d^{2}+6 B \,c^{2}+16 B c d -76 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 a f}-\frac {2 \left (52 A \,c^{2}+36 A c d +22 A \,d^{2}+18 B \,c^{2}+44 B c d -172 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 a f}-\frac {\left (122 A \,c^{2}+96 A c d +52 A \,d^{2}+48 B \,c^{2}+104 B c d -422 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 a f}-\frac {2 \left (172 A \,c^{2}+96 A c d +62 A \,d^{2}+48 B \,c^{2}+124 B c d -502 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 a f}+\frac {5 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {13 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}+\frac {25 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}+\frac {38 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}+\frac {46 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}+\frac {46 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}+\frac {38 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}+\frac {25 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a}+\frac {13 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a}+\frac {5 B \,d^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(819\) |
Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x,method=_RETUR NVERBOSE)
Output:
1/15*(15*B*x*tan(1/2*f*x+1/2*e)^5*d^2*f+((75*f*x+30)*B*d^2-30*A*c^2)*tan(1 /2*f*x+1/2*e)^4+((150*f*x+150)*B*d^2-60*A*c*d-60*c^2*(A+1/2*B))*tan(1/2*f* x+1/2*e)^3+((150*B*f*x-40*A+290*B)*d^2-60*c*(A+4/3*B)*d-80*c^2*(A+3/8*B))* tan(1/2*f*x+1/2*e)^2+((75*B*f*x-20*A+190*B)*d^2-60*c*(A+2/3*B)*d-40*c^2*(A +3/4*B))*tan(1/2*f*x+1/2*e)+(15*B*f*x-4*A+44*B)*d^2-12*c*(A+2/3*B)*d-14*c^ 2*(A+3/7*B))/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5
Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (158) = 316\).
Time = 0.10 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.63 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {60 \, B d^{2} f x - {\left (15 \, B d^{2} f x - {\left (2 \, A + 3 \, B\right )} c^{2} - 2 \, {\left (3 \, A + 7 \, B\right )} c d - {\left (7 \, A - 32 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (A - B\right )} c^{2} + 6 \, {\left (A - B\right )} c d - 3 \, {\left (A - B\right )} d^{2} - {\left (45 \, B d^{2} f x + 2 \, {\left (2 \, A + 3 \, B\right )} c^{2} + 2 \, {\left (6 \, A - B\right )} c d - {\left (A + 19 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, B d^{2} f x - {\left (3 \, A + 2 \, B\right )} c^{2} - 2 \, {\left (2 \, A + 3 \, B\right )} c d - 3 \, {\left (A - 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) + {\left (60 \, B d^{2} f x + 3 \, {\left (A - B\right )} c^{2} - 6 \, {\left (A - B\right )} c d + 3 \, {\left (A - B\right )} d^{2} - {\left (15 \, B d^{2} f x + {\left (2 \, A + 3 \, B\right )} c^{2} + 2 \, {\left (3 \, A + 7 \, B\right )} c d + {\left (7 \, A - 32 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, B d^{2} f x - {\left (2 \, A + 3 \, B\right )} c^{2} - 2 \, {\left (3 \, A + 2 \, B\right )} c d - {\left (2 \, A - 17 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algori thm="fricas")
Output:
-1/15*(60*B*d^2*f*x - (15*B*d^2*f*x - (2*A + 3*B)*c^2 - 2*(3*A + 7*B)*c*d - (7*A - 32*B)*d^2)*cos(f*x + e)^3 - 3*(A - B)*c^2 + 6*(A - B)*c*d - 3*(A - B)*d^2 - (45*B*d^2*f*x + 2*(2*A + 3*B)*c^2 + 2*(6*A - B)*c*d - (A + 19*B )*d^2)*cos(f*x + e)^2 + 3*(10*B*d^2*f*x - (3*A + 2*B)*c^2 - 2*(2*A + 3*B)* c*d - 3*(A - 6*B)*d^2)*cos(f*x + e) + (60*B*d^2*f*x + 3*(A - B)*c^2 - 6*(A - B)*c*d + 3*(A - B)*d^2 - (15*B*d^2*f*x + (2*A + 3*B)*c^2 + 2*(3*A + 7*B )*c*d + (7*A - 32*B)*d^2)*cos(f*x + e)^2 + 3*(10*B*d^2*f*x - (2*A + 3*B)*c ^2 - 2*(3*A + 2*B)*c*d - (2*A - 17*B)*d^2)*cos(f*x + e))*sin(f*x + e))/(a^ 3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3 *f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 3468 vs. \(2 (151) = 302\).
Time = 8.82 (sec) , antiderivative size = 3468, normalized size of antiderivative = 21.15 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-30*A*c**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3 *f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c* *2*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2) **2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*A*c**2*tan(e/2 + f*x/2) **2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a **3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan (e/2 + f*x/2) + 15*a**3*f) - 40*A*c**2*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2) **3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a** 3*f) - 14*A*c**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/ 2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c*d*tan(e/2 + f*x/2)**3/(15 *a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*t an(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c*d*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f* x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c*d*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*...
Leaf count of result is larger than twice the leaf count of optimal. 1132 vs. \(2 (158) = 316\).
Time = 0.14 (sec) , antiderivative size = 1132, normalized size of antiderivative = 6.90 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algori thm="maxima")
Output:
2/15*(B*d^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos (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)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1 ) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(co s(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f* x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1) )/a^3) - A*c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(co s(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)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1 ) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(co s(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f* x + e)^5/(cos(f*x + e) + 1)^5) - 4*B*c*d*(5*sin(f*x + e)/(cos(f*x + e) + 1 ) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/ (cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*s in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 2*A*d^2*(5*sin(f*x + e)/ (cos(f*x + e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5* a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*...
Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (158) = 316\).
Time = 0.27 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.21 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} B d^{2}}{a^{3}} - \frac {2 \, {\left (15 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 75 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 30 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 40 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 145 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 20 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 95 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A c^{2} + 3 \, B c^{2} + 6 \, A c d + 4 \, B c d + 2 \, A d^{2} - 22 \, B d^{2}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algori thm="giac")
Output:
1/15*(15*(f*x + e)*B*d^2/a^3 - 2*(15*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*d ^2*tan(1/2*f*x + 1/2*e)^4 + 30*A*c^2*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^2*tan (1/2*f*x + 1/2*e)^3 + 30*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 75*B*d^2*tan(1/2*f *x + 1/2*e)^3 + 40*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 1 /2*e)^2 + 30*A*c*d*tan(1/2*f*x + 1/2*e)^2 + 40*B*c*d*tan(1/2*f*x + 1/2*e)^ 2 + 20*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 145*B*d^2*tan(1/2*f*x + 1/2*e)^2 + 2 0*A*c^2*tan(1/2*f*x + 1/2*e) + 15*B*c^2*tan(1/2*f*x + 1/2*e) + 30*A*c*d*ta n(1/2*f*x + 1/2*e) + 20*B*c*d*tan(1/2*f*x + 1/2*e) + 10*A*d^2*tan(1/2*f*x + 1/2*e) - 95*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 + 6*A*c*d + 4 *B*c*d + 2*A*d^2 - 22*B*d^2)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
Time = 40.05 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {B\,d^2\,x}{a^3}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {16\,A\,c^2}{3}+\frac {8\,A\,d^2}{3}+2\,B\,c^2-\frac {58\,B\,d^2}{3}+4\,A\,c\,d+\frac {16\,B\,c\,d}{3}\right )+\frac {14\,A\,c^2}{15}+\frac {4\,A\,d^2}{15}+\frac {2\,B\,c^2}{5}-\frac {44\,B\,d^2}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (4\,A\,c^2+2\,B\,c^2-10\,B\,d^2+4\,A\,c\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,A\,c^2-2\,B\,d^2\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {8\,A\,c^2}{3}+\frac {4\,A\,d^2}{3}+2\,B\,c^2-\frac {38\,B\,d^2}{3}+4\,A\,c\,d+\frac {8\,B\,c\,d}{3}\right )+\frac {4\,A\,c\,d}{5}+\frac {8\,B\,c\,d}{15}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \] Input:
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^3,x )
Output:
(B*d^2*x)/a^3 - (tan(e/2 + (f*x)/2)^2*((16*A*c^2)/3 + (8*A*d^2)/3 + 2*B*c^ 2 - (58*B*d^2)/3 + 4*A*c*d + (16*B*c*d)/3) + (14*A*c^2)/15 + (4*A*d^2)/15 + (2*B*c^2)/5 - (44*B*d^2)/15 + tan(e/2 + (f*x)/2)^3*(4*A*c^2 + 2*B*c^2 - 10*B*d^2 + 4*A*c*d) + tan(e/2 + (f*x)/2)^4*(2*A*c^2 - 2*B*d^2) + tan(e/2 + (f*x)/2)*((8*A*c^2)/3 + (4*A*d^2)/3 + 2*B*c^2 - (38*B*d^2)/3 + 4*A*c*d + (8*B*c*d)/3) + (4*A*c*d)/5 + (8*B*c*d)/15)/(f*(10*a^3*tan(e/2 + (f*x)/2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/2 + (f*x)/2)^4 + a^3*tan(e/2 + (f*x)/2)^5 + a^3 + 5*a^3*tan(e/2 + (f*x)/2)))
Time = 0.18 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.93 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a \,c^{2}-6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b \,d^{2}-30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b \,c^{2}+90 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b \,d^{2}-20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a \,c^{2}-40 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a \,d^{2}-30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b \,c^{2}+230 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b \,d^{2}-20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a \,d^{2}-10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a \,c^{2}-30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b \,c^{2}+160 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b \,d^{2}+15 b \,d^{2} f x -4 a \,d^{2}-6 b \,c^{2}+38 b \,d^{2}-8 a \,c^{2}-60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a c d -60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a c d -80 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b c d -60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a c d -40 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b c d +15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b \,d^{2} f x +75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} b \,d^{2} f x +150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b \,d^{2} f x +150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b \,d^{2} f x +75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b \,d^{2} f x -12 a c d -8 b c d}{15 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x)
Output:
(6*tan((e + f*x)/2)**5*a*c**2 + 15*tan((e + f*x)/2)**5*b*d**2*f*x - 6*tan( (e + f*x)/2)**5*b*d**2 + 75*tan((e + f*x)/2)**4*b*d**2*f*x - 60*tan((e + f *x)/2)**3*a*c*d - 30*tan((e + f*x)/2)**3*b*c**2 + 150*tan((e + f*x)/2)**3* b*d**2*f*x + 90*tan((e + f*x)/2)**3*b*d**2 - 20*tan((e + f*x)/2)**2*a*c**2 - 60*tan((e + f*x)/2)**2*a*c*d - 40*tan((e + f*x)/2)**2*a*d**2 - 30*tan(( e + f*x)/2)**2*b*c**2 - 80*tan((e + f*x)/2)**2*b*c*d + 150*tan((e + f*x)/2 )**2*b*d**2*f*x + 230*tan((e + f*x)/2)**2*b*d**2 - 10*tan((e + f*x)/2)*a*c **2 - 60*tan((e + f*x)/2)*a*c*d - 20*tan((e + f*x)/2)*a*d**2 - 30*tan((e + f*x)/2)*b*c**2 - 40*tan((e + f*x)/2)*b*c*d + 75*tan((e + f*x)/2)*b*d**2*f *x + 160*tan((e + f*x)/2)*b*d**2 - 8*a*c**2 - 12*a*c*d - 4*a*d**2 - 6*b*c* *2 - 8*b*c*d + 15*b*d**2*f*x + 38*b*d**2)/(15*a**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))