Integrand size = 33, antiderivative size = 127 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(2 A c+3 B c+3 A d+7 B d) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \] Output:
-1/5*(A-B)*(c-d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^3-1/15*(2*A*c+3*A*d+3*B*c-8 *B*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(2*A*c+3*A*d+3*B*c+7*B*d)*cos (f*x+e)/f/(a^3+a^3*sin(f*x+e))
Time = 3.39 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (15 (A d+B (c+2 d)) \cos \left (\frac {1}{2} (e+f x)\right )-5 (2 A c+3 B c+3 A d+4 B d) \cos \left (\frac {3}{2} (e+f x)\right )-2 (-3 (3 A c+2 B c+2 A d+8 B d)+(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)+(2 A c+3 B c+3 A d+7 B d) \cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{30 a^3 f (1+\sin (e+f x))^3} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x]) ^3,x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(15*(A*d + B*(c + 2*d))*Cos[(e + f* x)/2] - 5*(2*A*c + 3*B*c + 3*A*d + 4*B*d)*Cos[(3*(e + f*x))/2] - 2*(-3*(3* A*c + 2*B*c + 2*A*d + 8*B*d) + (2*A*c + 3*B*c + 3*A*d - 8*B*d)*Cos[e + f*x ] + (2*A*c + 3*B*c + 3*A*d + 7*B*d)*Cos[2*(e + f*x)])*Sin[(e + f*x)/2]))/( 30*a^3*f*(1 + Sin[e + f*x])^3)
Time = 0.62 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3447, 3042, 3498, 25, 3042, 3229, 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.
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a \sin (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \frac {(A d+B c) \sin (e+f x)+A c+B d \sin ^2(e+f x)}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(A d+B c) \sin (e+f x)+A c+B d \sin (e+f x)^2}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle -\frac {\int -\frac {a (2 A c+3 B c+3 A d-3 B d)+5 a B d \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a (2 A c+3 B c+3 A d-3 B d)+5 a B d \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (2 A c+3 B c+3 A d-3 B d)+5 a B d \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {\frac {1}{3} (2 A c+3 A d+3 B c+7 B d) \int \frac {1}{\sin (e+f x) a+a}dx-\frac {a (2 A c+3 A d+3 B c-8 B d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} (2 A c+3 A d+3 B c+7 B d) \int \frac {1}{\sin (e+f x) a+a}dx-\frac {a (2 A c+3 A d+3 B c-8 B d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {-\frac {(2 A c+3 A d+3 B c+7 B d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)}-\frac {a (2 A c+3 A d+3 B c-8 B d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
Input:
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x])^3,x]
Output:
-1/5*((A - B)*(c - d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^3) + (-1/3*(a* (2*A*c + 3*B*c + 3*A*d - 8*B*d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^2) - ((2*A*c + 3*B*c + 3*A*d + 7*B*d)*Cos[e + f*x])/(3*f*(a + a*Sin[e + f*x])) )/(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_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
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_.)*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 = 0.60 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.09
| method | result | size |
| parallelrisch | \(\frac {-30 A c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (\left (-60 c -30 d \right ) A -30 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (\left (-80 c -30 d \right ) A -30 B \left (c +\frac {4 d}{3}\right )\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (\left (-40 c -30 d \right ) A -30 \left (c +\frac {2 d}{3}\right ) B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (-14 c -6 d \right ) A -6 \left (c +\frac {2 d}{3}\right ) B}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(139\) |
| derivativedivides | \(\frac {-\frac {-8 A c +8 A d +8 B c -8 B d}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A c -4 A d -4 B c +4 B d \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 A c -6 A d -6 B c +4 B d \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 A c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 A c +2 A d +2 B c}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\) | \(151\) |
| default | \(\frac {-\frac {-8 A c +8 A d +8 B c -8 B d}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A c -4 A d -4 B c +4 B d \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 A c -6 A d -6 B c +4 B d \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 A c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 A c +2 A d +2 B c}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\) | \(151\) |
| risch | \(-\frac {2 \left (2 A c +3 A d +7 B d +3 B c +30 i B d \,{\mathrm e}^{3 i \left (f x +e \right )}+15 i B c \,{\mathrm e}^{3 i \left (f x +e \right )}-20 i B d \,{\mathrm e}^{i \left (f x +e \right )}-10 i A c \,{\mathrm e}^{i \left (f x +e \right )}+15 i A d \,{\mathrm e}^{3 i \left (f x +e \right )}-15 i A d \,{\mathrm e}^{i \left (f x +e \right )}-15 i B c \,{\mathrm e}^{i \left (f x +e \right )}+15 B d \,{\mathrm e}^{4 i \left (f x +e \right )}-20 A c \,{\mathrm e}^{2 i \left (f x +e \right )}-15 A d \,{\mathrm e}^{2 i \left (f x +e \right )}-15 B c \,{\mathrm e}^{2 i \left (f x +e \right )}-40 B d \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(203\) |
| norman | \(\frac {-\frac {14 A c +6 A d +6 B c +4 B d}{15 a f}-\frac {2 A c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f}-\frac {2 \left (2 A c +A d +B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}-\frac {2 \left (34 A c +11 A d +11 B c +14 B d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 a f}-\frac {2 \left (16 A c +9 A d +9 B c +2 B d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 a f}-\frac {2 \left (14 A c +3 A d +3 B c +4 B d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 a f}-\frac {2 \left (14 A c +9 A d +9 B c +4 B d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {2 \left (18 A c +7 A d +7 B c +8 B d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 a f}-\frac {\left (8 A c +6 A d +6 B c +4 B d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(324\) |
Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x,method=_RETURNV ERBOSE)
Output:
1/15*(-30*A*c*tan(1/2*f*x+1/2*e)^4+((-60*c-30*d)*A-30*B*c)*tan(1/2*f*x+1/2 *e)^3+((-80*c-30*d)*A-30*B*(c+4/3*d))*tan(1/2*f*x+1/2*e)^2+((-40*c-30*d)*A -30*(c+2/3*d)*B)*tan(1/2*f*x+1/2*e)+(-14*c-6*d)*A-6*(c+2/3*d)*B)/f/a^3/(ta n(1/2*f*x+1/2*e)+1)^5
Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (121) = 242\).
Time = 0.07 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.13 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {{\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{3} - {\left (2 \, {\left (2 \, A + 3 \, B\right )} c + {\left (6 \, A - B\right )} d\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 3 \, {\left (A - B\right )} d - 3 \, {\left ({\left (3 \, A + 2 \, B\right )} c + {\left (2 \, A + 3 \, B\right )} d\right )} \cos \left (f x + e\right ) - {\left ({\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 3 \, {\left (A - B\right )} d + 3 \, {\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 2 \, B\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorith m="fricas")
Output:
-1/15*(((2*A + 3*B)*c + (3*A + 7*B)*d)*cos(f*x + e)^3 - (2*(2*A + 3*B)*c + (6*A - B)*d)*cos(f*x + e)^2 - 3*(A - B)*c + 3*(A - B)*d - 3*((3*A + 2*B)* c + (2*A + 3*B)*d)*cos(f*x + e) - (((2*A + 3*B)*c + (3*A + 7*B)*d)*cos(f*x + e)^2 - 3*(A - B)*c + 3*(A - B)*d + 3*((2*A + 3*B)*c + (3*A + 2*B)*d)*co s(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. 1819 vs. \(2 (121) = 242\).
Time = 4.94 (sec) , antiderivative size = 1819, normalized size of antiderivative = 14.32 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-30*A*c*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*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*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*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/(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) - 30*A*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*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) - 30*A*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) - 30*A*d*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 +...
Leaf count of result is larger than twice the leaf count of optimal. 733 vs. \(2 (121) = 242\).
Time = 0.06 (sec) , antiderivative size = 733, normalized size of antiderivative = 5.77 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorith m="maxima")
Output:
-2/15*(A*c*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(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/(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*B*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*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*c*(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)/(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*A*d*(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 )/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(co s(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 ))/f
Time = 0.20 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A c + 3 \, B c + 3 \, A d + 2 \, B d\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorith m="giac")
Output:
-2/15*(15*A*c*tan(1/2*f*x + 1/2*e)^4 + 30*A*c*tan(1/2*f*x + 1/2*e)^3 + 15* B*c*tan(1/2*f*x + 1/2*e)^3 + 15*A*d*tan(1/2*f*x + 1/2*e)^3 + 40*A*c*tan(1/ 2*f*x + 1/2*e)^2 + 15*B*c*tan(1/2*f*x + 1/2*e)^2 + 15*A*d*tan(1/2*f*x + 1/ 2*e)^2 + 20*B*d*tan(1/2*f*x + 1/2*e)^2 + 20*A*c*tan(1/2*f*x + 1/2*e) + 15* B*c*tan(1/2*f*x + 1/2*e) + 15*A*d*tan(1/2*f*x + 1/2*e) + 10*B*d*tan(1/2*f* x + 1/2*e) + 7*A*c + 3*B*c + 3*A*d + 2*B*d)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)
Time = 37.42 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.93 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {53\,A\,c}{4}+3\,A\,d+3\,B\,c+\frac {13\,B\,d}{4}-4\,A\,c\,\cos \left (e+f\,x\right )+\frac {3\,A\,d\,\cos \left (e+f\,x\right )}{2}+\frac {3\,B\,c\,\cos \left (e+f\,x\right )}{2}+B\,d\,\cos \left (e+f\,x\right )+\frac {25\,A\,c\,\sin \left (e+f\,x\right )}{2}+\frac {15\,A\,d\,\sin \left (e+f\,x\right )}{2}+\frac {15\,B\,c\,\sin \left (e+f\,x\right )}{2}+\frac {5\,B\,d\,\sin \left (e+f\,x\right )}{2}-\frac {9\,A\,c\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {3\,A\,d\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {3\,B\,c\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {9\,B\,d\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,A\,c\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,d\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \] Input:
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x)))/(a + a*sin(e + f*x))^3,x)
Output:
(2*cos(e/2 + (f*x)/2)*((53*A*c)/4 + 3*A*d + 3*B*c + (13*B*d)/4 - 4*A*c*cos (e + f*x) + (3*A*d*cos(e + f*x))/2 + (3*B*c*cos(e + f*x))/2 + B*d*cos(e + f*x) + (25*A*c*sin(e + f*x))/2 + (15*A*d*sin(e + f*x))/2 + (15*B*c*sin(e + f*x))/2 + (5*B*d*sin(e + f*x))/2 - (9*A*c*cos(2*e + 2*f*x))/4 - (3*A*d*co s(2*e + 2*f*x))/2 - (3*B*c*cos(2*e + 2*f*x))/2 - (9*B*d*cos(2*e + 2*f*x))/ 4 - (5*A*c*sin(2*e + 2*f*x))/4 + (5*B*d*sin(2*e + 2*f*x))/4))/(15*a^3*f*(( 5*2^(1/2)*cos((3*e)/2 + pi/4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 + (2^(1/2)*cos((5*e)/2 - pi/4 + (5*f*x)/2))/4))
Time = 0.18 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.94 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a c}{5}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a d -2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b c -\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a c}{3}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a d -2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b c -\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b d}{3}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a c}{3}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a d -2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b c -\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b d}{3}-\frac {8 a c}{15}-\frac {2 a d}{5}-\frac {2 b c}{5}-\frac {4 b d}{15}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x)
Output:
(2*(3*tan((e + f*x)/2)**5*a*c - 15*tan((e + f*x)/2)**3*a*d - 15*tan((e + f *x)/2)**3*b*c - 10*tan((e + f*x)/2)**2*a*c - 15*tan((e + f*x)/2)**2*a*d - 15*tan((e + f*x)/2)**2*b*c - 20*tan((e + f*x)/2)**2*b*d - 5*tan((e + f*x)/ 2)*a*c - 15*tan((e + f*x)/2)*a*d - 15*tan((e + f*x)/2)*b*c - 10*tan((e + f *x)/2)*b*d - 4*a*c - 3*a*d - 3*b*c - 2*b*d))/(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))