Integrand size = 34, antiderivative size = 103 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 (A-B) c \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac {a (A-11 B) c \cos (e+f x)}{15 f \left (a^2+a^2 \sin (e+f x)\right )^2}+\frac {(A+4 B) c \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \] Output:
-2/5*(A-B)*c*cos(f*x+e)/f/(a+a*sin(f*x+e))^3+1/15*a*(A-11*B)*c*cos(f*x+e)/ f/(a^2+a^2*sin(f*x+e))^2+1/15*(A+4*B)*c*cos(f*x+e)/f/(a^3+a^3*sin(f*x+e))
Time = 6.92 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {c \left (-15 (A+B) \cos \left (e+\frac {f x}{2}\right )+5 (A+B) \cos \left (e+\frac {3 f x}{2}\right )+5 A \sin \left (\frac {f x}{2}\right )-25 B \sin \left (\frac {f x}{2}\right )-15 B \sin \left (2 e+\frac {3 f x}{2}\right )+A \sin \left (2 e+\frac {5 f x}{2}\right )+4 B \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 a^3 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]))/(a + a*Sin[e + f*x]) ^3,x]
Output:
(c*(-15*(A + B)*Cos[e + (f*x)/2] + 5*(A + B)*Cos[e + (3*f*x)/2] + 5*A*Sin[ (f*x)/2] - 25*B*Sin[(f*x)/2] - 15*B*Sin[2*e + (3*f*x)/2] + A*Sin[2*e + (5* f*x)/2] + 4*B*Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(Cos[e/2] + Sin[e/2])*(Cos[ (e + f*x)/2] + Sin[(e + f*x)/2])^5)
Time = 0.61 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3446, 3042, 3336, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x)) (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x)) (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a c \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \frac {\cos (e+f x)^2 (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^4}dx\) |
| \(\Big \downarrow \) 3336 |
| \(\displaystyle a c \left (-\frac {\int \frac {a (A-6 B)+5 a B \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^3}-\frac {2 (A-B) \cos (e+f x)}{5 a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (-\frac {\int \frac {a (A-6 B)+5 a B \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^3}-\frac {2 (A-B) \cos (e+f x)}{5 a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle a c \left (-\frac {\frac {1}{3} (A+4 B) \int \frac {1}{\sin (e+f x) a+a}dx-\frac {a (A-11 B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^3}-\frac {2 (A-B) \cos (e+f x)}{5 a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (-\frac {\frac {1}{3} (A+4 B) \int \frac {1}{\sin (e+f x) a+a}dx-\frac {a (A-11 B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^3}-\frac {2 (A-B) \cos (e+f x)}{5 a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle a c \left (-\frac {-\frac {(A+4 B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)}-\frac {a (A-11 B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^3}-\frac {2 (A-B) \cos (e+f x)}{5 a f (a \sin (e+f x)+a)^3}\right )\) |
Input:
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]))/(a + a*Sin[e + f*x])^3,x]
Output:
a*c*((-2*(A - B)*Cos[e + f*x])/(5*a*f*(a + a*Sin[e + f*x])^3) - (-1/3*(a*( A - 11*B)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^2) - ((A + 4*B)*Cos[e + f* x])/(3*f*(a + a*Sin[e + f*x])))/(5*a^3))
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[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*( (c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[2*(b*c - a*d)*Cos [e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(2*m + 3))), x] + Simp[1/(b^ 3*(2*m + 3)) Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d* (2*m + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 0.53 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(-\frac {2 \left (A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\frac {\left (5 A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}+\frac {\left (A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}+\frac {4 A}{15}+\frac {B}{15}\right ) c}{f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(92\) |
| derivativedivides | \(\frac {2 c \left (-\frac {8 A -8 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {14 A -10 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-16 A +16 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-6 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f \,a^{3}}\) | \(115\) |
| default | \(\frac {2 c \left (-\frac {8 A -8 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {14 A -10 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-16 A +16 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-6 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f \,a^{3}}\) | \(115\) |
| risch | \(\frac {\frac {2 A c \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}-\frac {2 i A c \,{\mathrm e}^{i \left (f x +e \right )}}{3}-\frac {10 B c \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}+2 i B c \,{\mathrm e}^{3 i \left (f x +e \right )}-\frac {2 i B c \,{\mathrm e}^{i \left (f x +e \right )}}{3}+2 B c \,{\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 A c}{15}+\frac {8 B c}{15}+2 i A c \,{\mathrm e}^{3 i \left (f x +e \right )}}{f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(126\) |
| norman | \(\frac {-\frac {8 A c +2 B c}{15 a f}-\frac {2 A c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f}-\frac {2 \left (23 A c -3 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 a f}-\frac {2 \left (7 A c +7 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 a f}-\frac {2 \left (11 A c -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 a f}-\frac {2 \left (11 A c -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 a f}-\frac {\left (2 A c +2 B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {10 \left (A c +B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {2 \left (A c +B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(262\) |
Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x,method=_RETURNV ERBOSE)
Output:
-2*(A*tan(1/2*f*x+1/2*e)^4+(A+B)*tan(1/2*f*x+1/2*e)^3+1/3*(5*A-B)*tan(1/2* f*x+1/2*e)^2+1/3*(A+B)*tan(1/2*f*x+1/2*e)+4/15*A+1/15*B)*c/f/a^3/(tan(1/2* f*x+1/2*e)+1)^5
Time = 0.08 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.85 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {{\left (A + 4 \, B\right )} c \cos \left (f x + e\right )^{3} - {\left (2 \, A - 7 \, B\right )} c \cos \left (f x + e\right )^{2} + 3 \, {\left (A - B\right )} c \cos \left (f x + e\right ) + 6 \, {\left (A - B\right )} c - {\left ({\left (A + 4 \, B\right )} c \cos \left (f x + e\right )^{2} + 3 \, {\left (A - B\right )} c \cos \left (f x + e\right ) + 6 \, {\left (A - B\right )} c\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorith m="fricas")
Output:
1/15*((A + 4*B)*c*cos(f*x + e)^3 - (2*A - 7*B)*c*cos(f*x + e)^2 + 3*(A - B )*c*cos(f*x + e) + 6*(A - B)*c - ((A + 4*B)*c*cos(f*x + e)^2 + 3*(A - B)*c *cos(f*x + e) + 6*(A - B)*c)*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. 1035 vs. \(2 (97) = 194\).
Time = 4.85 (sec) , antiderivative size = 1035, normalized size of antiderivative = 10.05 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*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) - 30*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) - 50*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) - 10*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) - 8*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*B*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 ) + 10*B*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) - 10*B*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 + ...
Leaf count of result is larger than twice the leaf count of optimal. 733 vs. \(2 (97) = 194\).
Time = 0.05 (sec) , antiderivative size = 733, normalized size of antiderivative = 7.12 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*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*c*(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*A*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*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/(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 = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.26 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, A c + B c\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*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 + 15*A*c*tan(1/2*f*x + 1/2*e)^3 + 15* B*c*tan(1/2*f*x + 1/2*e)^3 + 25*A*c*tan(1/2*f*x + 1/2*e)^2 - 5*B*c*tan(1/2 *f*x + 1/2*e)^2 + 5*A*c*tan(1/2*f*x + 1/2*e) + 5*B*c*tan(1/2*f*x + 1/2*e) + 4*A*c + B*c)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)
Time = 37.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {41\,A\,c}{4}-\frac {B\,c}{4}-\frac {11\,A\,c\,\cos \left (e+f\,x\right )}{2}+\frac {B\,c\,\cos \left (e+f\,x\right )}{2}+5\,A\,c\,\sin \left (e+f\,x\right )+5\,B\,c\,\sin \left (e+f\,x\right )-\frac {3\,A\,c\,\cos \left (2\,e+2\,f\,x\right )}{4}+\frac {3\,B\,c\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,A\,c\,\sin \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,B\,c\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \] Input:
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x)))/(a + a*sin(e + f*x))^3,x)
Output:
(2*cos(e/2 + (f*x)/2)*((41*A*c)/4 - (B*c)/4 - (11*A*c*cos(e + f*x))/2 + (B *c*cos(e + f*x))/2 + 5*A*c*sin(e + f*x) + 5*B*c*sin(e + f*x) - (3*A*c*cos( 2*e + 2*f*x))/4 + (3*B*c*cos(2*e + 2*f*x))/4 - (5*A*c*sin(2*e + 2*f*x))/4 - (5*B*c*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 = 175, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx=\frac {2 c \left (3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a +15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a -15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b +5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b +10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b -a -b \right )}{15 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x)
Output:
(2*c*(3*tan((e + f*x)/2)**5*a + 15*tan((e + f*x)/2)**3*a - 15*tan((e + f*x )/2)**3*b + 5*tan((e + f*x)/2)**2*a + 5*tan((e + f*x)/2)**2*b + 10*tan((e + f*x)/2)*a - 5*tan((e + f*x)/2)*b - a - b))/(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))