Integrand size = 36, antiderivative size = 157 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {5 (3 A-4 B) c^3 x}{2 a}-\frac {5 (3 A-4 B) c^3 \cos ^3(e+f x)}{3 a f}-\frac {5 (3 A-4 B) c^3 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {a^3 (A-B) c^3 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {2 a^3 (3 A-4 B) c^3 \cos ^5(e+f x)}{f \left (a^2+a^2 \sin (e+f x)\right )^2} \] Output:
-5/2*(3*A-4*B)*c^3*x/a-5/3*(3*A-4*B)*c^3*cos(f*x+e)^3/a/f-5/2*(3*A-4*B)*c^ 3*cos(f*x+e)*sin(f*x+e)/a/f-a^3*(A-B)*c^3*cos(f*x+e)^7/f/(a+a*sin(f*x+e))^ 4-2*a^3*(3*A-4*B)*c^3*cos(f*x+e)^5/f/(a^2+a^2*sin(f*x+e))^2
Time = 12.19 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^3 \left (\cos \left (\frac {1}{2} (e+f x)\right ) (30 (3 A-4 B) (e+f x)+(48 A-93 B) \cos (e+f x)+B \cos (3 (e+f x))-3 (A-4 B) \sin (2 (e+f x)))+\sin \left (\frac {1}{2} (e+f x)\right ) (-24 B (-8+5 e+5 f x)+6 A (-32+15 e+15 f x)+(48 A-93 B) \cos (e+f x)+B \cos (3 (e+f x))-3 (A-4 B) \sin (2 (e+f x)))\right )}{12 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (1+\sin (e+f x))} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^3)/(a + a*Sin[e + f*x ]),x]
Output:
(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^3*(Cos[(e + f*x)/2]*(30*(3*A - 4*B)*(e + f*x) + (48*A - 93*B)*Cos[e + f*x] + B*Cos[3* (e + f*x)] - 3*(A - 4*B)*Sin[2*(e + f*x)]) + Sin[(e + f*x)/2]*(-24*B*(-8 + 5*e + 5*f*x) + 6*A*(-32 + 15*e + 15*f*x) + (48*A - 93*B)*Cos[e + f*x] + B *Cos[3*(e + f*x)] - 3*(A - 4*B)*Sin[2*(e + f*x)])))/(12*a*f*(Cos[(e + f*x) /2] - Sin[(e + f*x)/2])^6*(1 + Sin[e + f*x]))
Time = 0.77 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84, 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, 3161, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3 (A+B \sin (e+f x))}{a \sin (e+f x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3 (A+B \sin (e+f x))}{a \sin (e+f x)+a}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^4}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^3}dx}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^3}dx}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \left (\frac {5 \int \frac {\cos ^4(e+f x)}{\sin (e+f x) a+a}dx}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \left (\frac {5 \int \frac {\cos (e+f x)^4}{\sin (e+f x) a+a}dx}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \left (\frac {5 \left (\frac {\int \cos ^2(e+f x)dx}{a}+\frac {\cos ^3(e+f x)}{3 a f}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \left (\frac {5 \left (\frac {\int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx}{a}+\frac {\cos ^3(e+f x)}{3 a f}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \left (\frac {5 \left (\frac {\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}}{a}+\frac {\cos ^3(e+f x)}{3 a f}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (-\frac {(3 A-4 B) \left (\frac {5 \left (\frac {\cos ^3(e+f x)}{3 a f}+\frac {\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}}{a}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a}-\frac {(A-B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}\right )\) |
Input:
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^3)/(a + a*Sin[e + f*x]),x]
Output:
a^3*c^3*(-(((A - B)*Cos[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^4)) - ((3*A - 4*B)*((2*Cos[e + f*x]^5)/(a*f*(a + a*Sin[e + f*x])^2) + (5*(Cos[e + f*x]^3 /(3*a*f) + (x/2 + (Cos[e + f*x]*Sin[e + f*x])/(2*f))/a))/a^2))/a)
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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 = 1.68 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70
| method | result | size |
| parallelrisch | \(\frac {65 c^{3} \left (\frac {4 \left (-4 A +\frac {23 B}{3}\right ) \cos \left (2 f x +2 e \right )}{65}+\frac {\left (A -4 B \right ) \sin \left (3 f x +3 e \right )}{65}-\frac {B \cos \left (4 f x +4 e \right )}{195}+\frac {4 \left (-3 f x A +4 f x B -\frac {24}{5} A +\frac {94}{15} B \right ) \cos \left (f x +e \right )}{13}+\left (A -\frac {68 B}{65}\right ) \sin \left (f x +e \right )-\frac {16 A}{13}+\frac {19 B}{13}\right )}{8 a f \cos \left (f x +e \right )}\) | \(110\) |
| derivativedivides | \(\frac {2 c^{3} \left (-\frac {\left (\frac {A}{2}-2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (4 A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (8 A -16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {A}{2}+2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4 A -\frac {23 B}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}-\frac {5 \left (3 A -4 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {8 A -8 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f a}\) | \(152\) |
| default | \(\frac {2 c^{3} \left (-\frac {\left (\frac {A}{2}-2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (4 A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (8 A -16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {A}{2}+2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4 A -\frac {23 B}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}-\frac {5 \left (3 A -4 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {8 A -8 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f a}\) | \(152\) |
| risch | \(-\frac {15 c^{3} x A}{2 a}+\frac {10 c^{3} x B}{a}-\frac {2 c^{3} {\mathrm e}^{i \left (f x +e \right )} A}{a f}+\frac {31 c^{3} {\mathrm e}^{i \left (f x +e \right )} B}{8 a f}-\frac {2 c^{3} {\mathrm e}^{-i \left (f x +e \right )} A}{a f}+\frac {31 c^{3} {\mathrm e}^{-i \left (f x +e \right )} B}{8 a f}-\frac {16 c^{3} A}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {16 c^{3} B}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}-\frac {B \,c^{3} \cos \left (3 f x +3 e \right )}{12 a f}+\frac {c^{3} \sin \left (2 f x +2 e \right ) A}{4 a f}-\frac {c^{3} \sin \left (2 f x +2 e \right ) B}{a f}\) | \(221\) |
| norman | \(\frac {-\frac {72 A \,c^{3}-94 B \,c^{3}}{3 a f}-\frac {5 \left (3 A -4 B \right ) c^{3} x}{2 a}-\frac {\left (9 A \,c^{3}-18 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}-\frac {\left (17 A \,c^{3}-20 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f}-\frac {\left (21 A \,c^{3}-34 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {\left (25 A \,c^{3}-50 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {\left (69 A \,c^{3}-130 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {\left (73 A \,c^{3}-82 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f}-\frac {\left (119 A \,c^{3}-138 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f}-\frac {\left (261 A \,c^{3}-322 B \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 a f}-\frac {5 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}-\frac {10 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}-\frac {10 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}-\frac {15 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}-\frac {15 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}-\frac {10 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}-\frac {10 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}-\frac {5 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 a}-\frac {5 \left (3 A -4 B \right ) c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\) | \(563\) |
Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x,method=_RETURNV ERBOSE)
Output:
65/8*c^3*(4/65*(-4*A+23/3*B)*cos(2*f*x+2*e)+1/65*(A-4*B)*sin(3*f*x+3*e)-1/ 195*B*cos(4*f*x+4*e)+4/13*(-3*f*x*A+4*f*x*B-24/5*A+94/15*B)*cos(f*x+e)+(A- 68/65*B)*sin(f*x+e)-16/13*A+19/13*B)/a/f/cos(f*x+e)
Time = 0.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {2 \, B c^{3} \cos \left (f x + e\right )^{4} + {\left (3 \, A - 10 \, B\right )} c^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (3 \, A - 4 \, B\right )} c^{3} f x + 24 \, {\left (A - 2 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 48 \, {\left (A - B\right )} c^{3} + 3 \, {\left (5 \, {\left (3 \, A - 4 \, B\right )} c^{3} f x + {\left (23 \, A - 28 \, B\right )} c^{3}\right )} \cos \left (f x + e\right ) + {\left (2 \, B c^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (3 \, A - 4 \, B\right )} c^{3} f x - 3 \, {\left (A - 4 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (7 \, A - 12 \, B\right )} c^{3} \cos \left (f x + e\right ) - 48 \, {\left (A - B\right )} c^{3}\right )} \sin \left (f x + e\right )}{6 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x, algorith m="fricas")
Output:
-1/6*(2*B*c^3*cos(f*x + e)^4 + (3*A - 10*B)*c^3*cos(f*x + e)^3 + 15*(3*A - 4*B)*c^3*f*x + 24*(A - 2*B)*c^3*cos(f*x + e)^2 + 48*(A - B)*c^3 + 3*(5*(3 *A - 4*B)*c^3*f*x + (23*A - 28*B)*c^3)*cos(f*x + e) + (2*B*c^3*cos(f*x + e )^3 + 15*(3*A - 4*B)*c^3*f*x - 3*(A - 4*B)*c^3*cos(f*x + e)^2 + 3*(7*A - 1 2*B)*c^3*cos(f*x + e) - 48*(A - B)*c^3)*sin(f*x + e))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)
Leaf count of result is larger than twice the leaf count of optimal. 4255 vs. \(2 (139) = 278\).
Time = 3.98 (sec) , antiderivative size = 4255, normalized size of antiderivative = 27.10 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**3/(a+a*sin(f*x+e)),x)
Output:
Piecewise((-45*A*c**3*f*x*tan(e/2 + f*x/2)**7/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a *f*tan(e/2 + f*x/2) + 6*a*f) - 45*A*c**3*f*x*tan(e/2 + f*x/2)**6/(6*a*f*ta n(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/ 2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 135*A*c**3*f*x*tan(e/2 + f*x/2)**5/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f *tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2 )**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 135* A*c**3*f*x*tan(e/2 + f*x/2)**4/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18 *a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f* x/2) + 6*a*f) - 135*A*c**3*f*x*tan(e/2 + f*x/2)**3/(6*a*f*tan(e/2 + f*x/2) **7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan( e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 135*A*c**3*f*x*tan(e/2 + f*x/2)**2/(6* a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x /2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f* tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 45*A*c**3*f*x*t...
Leaf count of result is larger than twice the leaf count of optimal. 1120 vs. \(2 (151) = 302\).
Time = 0.14 (sec) , antiderivative size = 1120, normalized size of antiderivative = 7.13 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x, algorith m="maxima")
Output:
1/3*(B*c^3*((7*sin(f*x + e)/(cos(f*x + e) + 1) + 39*sin(f*x + e)^2/(cos(f* x + e) + 1)^2 + 24*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 24*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 + 9*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 16)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a*sin(f*x + e)^3/(cos(f*x + e ) + 1)^3 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3*a*sin(f*x + e)^5/(c os(f*x + e) + 1)^5 + a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a*sin(f*x + e )^7/(cos(f*x + e) + 1)^7) + 9*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 3*A*c^3*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e ) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f *x + e) + 1)^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*s in(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 ) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 9*B*c^3*((sin(f*x + e)/ (cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sin(f*x + e )^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1) ^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e )/(cos(f*x + e) + 1))/a) - 18*A*c^3*((sin(f*x + e)/(cos(f*x + e) + 1) +...
Time = 0.24 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {\frac {15 \, {\left (3 \, A c^{3} - 4 \, B c^{3}\right )} {\left (f x + e\right )}}{a} + \frac {96 \, {\left (A c^{3} - B c^{3}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (3 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 12 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 24 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 42 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 48 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 96 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 12 \, B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 24 \, A c^{3} - 46 \, B c^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{3} a}}{6 \, f} \] Input:
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x, algorith m="giac")
Output:
-1/6*(15*(3*A*c^3 - 4*B*c^3)*(f*x + e)/a + 96*(A*c^3 - B*c^3)/(a*(tan(1/2* f*x + 1/2*e) + 1)) + 2*(3*A*c^3*tan(1/2*f*x + 1/2*e)^5 - 12*B*c^3*tan(1/2* f*x + 1/2*e)^5 + 24*A*c^3*tan(1/2*f*x + 1/2*e)^4 - 42*B*c^3*tan(1/2*f*x + 1/2*e)^4 + 48*A*c^3*tan(1/2*f*x + 1/2*e)^2 - 96*B*c^3*tan(1/2*f*x + 1/2*e) ^2 - 3*A*c^3*tan(1/2*f*x + 1/2*e) + 12*B*c^3*tan(1/2*f*x + 1/2*e) + 24*A*c ^3 - 46*B*c^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^3*a))/f
Time = 37.35 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.03 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (7\,A\,c^3-\frac {34\,B\,c^3}{3}\right )+24\,A\,c^3-\frac {94\,B\,c^3}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (9\,A\,c^3-18\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (17\,A\,c^3-20\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (16\,A\,c^3-32\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (56\,A\,c^3-62\,B\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (63\,A\,c^3-76\,B\,c^3\right )}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )}-\frac {5\,c^3\,\mathrm {atan}\left (\frac {5\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,A-4\,B\right )}{15\,A\,c^3-20\,B\,c^3}\right )\,\left (3\,A-4\,B\right )}{a\,f} \] Input:
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^3)/(a + a*sin(e + f*x)),x)
Output:
- (tan(e/2 + (f*x)/2)*(7*A*c^3 - (34*B*c^3)/3) + 24*A*c^3 - (94*B*c^3)/3 + tan(e/2 + (f*x)/2)^5*(9*A*c^3 - 18*B*c^3) + tan(e/2 + (f*x)/2)^6*(17*A*c^ 3 - 20*B*c^3) + tan(e/2 + (f*x)/2)^3*(16*A*c^3 - 32*B*c^3) + tan(e/2 + (f* x)/2)^4*(56*A*c^3 - 62*B*c^3) + tan(e/2 + (f*x)/2)^2*(63*A*c^3 - 76*B*c^3) )/(f*(a + a*tan(e/2 + (f*x)/2) + 3*a*tan(e/2 + (f*x)/2)^2 + 3*a*tan(e/2 + (f*x)/2)^3 + 3*a*tan(e/2 + (f*x)/2)^4 + 3*a*tan(e/2 + (f*x)/2)^5 + a*tan(e /2 + (f*x)/2)^6 + a*tan(e/2 + (f*x)/2)^7)) - (5*c^3*atan((5*c^3*tan(e/2 + (f*x)/2)*(3*A - 4*B))/(15*A*c^3 - 20*B*c^3))*(3*A - 4*B))/(a*f)
Time = 0.17 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {c^{3} \left (-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +10 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +21 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -34 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -45 \cos \left (f x +e \right ) a f x +102 \cos \left (f x +e \right ) a +60 \cos \left (f x +e \right ) b f x -120 \cos \left (f x +e \right ) b -2 \sin \left (f x +e \right )^{4} b -3 \sin \left (f x +e \right )^{3} a +12 \sin \left (f x +e \right )^{3} b +24 \sin \left (f x +e \right )^{2} a -44 \sin \left (f x +e \right )^{2} b +45 \sin \left (f x +e \right ) a f x +21 a \sin \left (f x +e \right )-60 \sin \left (f x +e \right ) b f x -34 \sin \left (f x +e \right ) b +45 a f x -102 a -60 b f x +120 b \right )}{6 a f \left (\cos \left (f x +e \right )-\sin \left (f x +e \right )-1\right )} \] Input:
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x)
Output:
(c**3*( - 2*cos(e + f*x)*sin(e + f*x)**3*b - 3*cos(e + f*x)*sin(e + f*x)** 2*a + 10*cos(e + f*x)*sin(e + f*x)**2*b + 21*cos(e + f*x)*sin(e + f*x)*a - 34*cos(e + f*x)*sin(e + f*x)*b - 45*cos(e + f*x)*a*f*x + 102*cos(e + f*x) *a + 60*cos(e + f*x)*b*f*x - 120*cos(e + f*x)*b - 2*sin(e + f*x)**4*b - 3* sin(e + f*x)**3*a + 12*sin(e + f*x)**3*b + 24*sin(e + f*x)**2*a - 44*sin(e + f*x)**2*b + 45*sin(e + f*x)*a*f*x + 21*sin(e + f*x)*a - 60*sin(e + f*x) *b*f*x - 34*sin(e + f*x)*b + 45*a*f*x - 102*a - 60*b*f*x + 120*b))/(6*a*f* (cos(e + f*x) - sin(e + f*x) - 1))