Integrand size = 36, antiderivative size = 156 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {5 a^3 (3 A+4 B) x}{2 c}+\frac {5 a^3 (3 A+4 B) \cos ^3(e+f x)}{3 c f}-\frac {5 a^3 (3 A+4 B) \cos (e+f x) \sin (e+f x)}{2 c f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}+\frac {2 a^3 (3 A+4 B) c^3 \cos ^5(e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )^2} \] Output:
-5/2*a^3*(3*A+4*B)*x/c+5/3*a^3*(3*A+4*B)*cos(f*x+e)^3/c/f-5/2*a^3*(3*A+4*B )*cos(f*x+e)*sin(f*x+e)/c/f+a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^ 4+2*a^3*(3*A+4*B)*c^3*cos(f*x+e)^5/f/(c^2-c^2*sin(f*x+e))^2
Time = 12.14 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.43 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (\cos \left (\frac {1}{2} (e+f x)\right ) (30 (3 A+4 B) (e+f x)-3 (16 A+31 B) \cos (e+f x)+B \cos (3 (e+f x))-3 (A+4 B) \sin (2 (e+f x)))-\sin \left (\frac {1}{2} (e+f x)\right ) (24 B (8+5 e+5 f x)+6 A (32+15 e+15 f x)-3 (16 A+31 B) \cos (e+f x)+B \cos (3 (e+f x))-3 (A+4 B) \sin (2 (e+f x)))\right )}{12 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (-1+\sin (e+f x))} \] Input:
Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ]),x]
Output:
(a^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) - 3*(16*A + 31*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) - 3*(16*A + 31*B)*Cos[e + f*x] + B *Cos[3*(e + f*x)] - 3*(A + 4*B)*Sin[2*(e + f*x)])))/(12*c*f*(Cos[(e + f*x) /2] + Sin[(e + f*x)/2])^6*(-1 + Sin[e + f*x]))
Time = 0.75 (sec) , antiderivative size = 131, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^3}dx}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^3}dx}{c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \left (\frac {5 \int \frac {\cos ^4(e+f x)}{c-c \sin (e+f x)}dx}{c^2}-\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \left (\frac {5 \int \frac {\cos (e+f x)^4}{c-c \sin (e+f x)}dx}{c^2}-\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \left (\frac {5 \left (\frac {\int \cos ^2(e+f x)dx}{c}-\frac {\cos ^3(e+f x)}{3 c f}\right )}{c^2}-\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \left (\frac {5 \left (\frac {\int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx}{c}-\frac {\cos ^3(e+f x)}{3 c f}\right )}{c^2}-\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \left (\frac {5 \left (\frac {\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}}{c}-\frac {\cos ^3(e+f x)}{3 c f}\right )}{c^2}-\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^2}\right )}{c}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{f (c-c \sin (e+f x))^4}-\frac {(3 A+4 B) \left (\frac {5 \left (\frac {\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}}{c}-\frac {\cos ^3(e+f x)}{3 c f}\right )}{c^2}-\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^2}\right )}{c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x]),x]
Output:
a^3*c^3*(((A + B)*Cos[e + f*x]^7)/(f*(c - c*Sin[e + f*x])^4) - ((3*A + 4*B )*((-2*Cos[e + f*x]^5)/(c*f*(c - c*Sin[e + f*x])^2) + (5*(-1/3*Cos[e + f*x ]^3/(c*f) + (x/2 + (Cos[e + f*x]*Sin[e + f*x])/(2*f))/c))/c^2))/c)
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 = 1.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71
| method | result | size |
| parallelrisch | \(\frac {65 \left (\frac {4 \left (4 A +\frac {23 B}{3}\right ) \cos \left (2 f x +2 e \right )}{65}+\frac {\left (A +4 B \right ) \sin \left (3 f x +3 e \right )}{65}-\frac {B \cos \left (4 f x +4 e \right )}{195}+\frac {4 \left (-3 f x A -4 f x B +\frac {24}{5} A +\frac {94}{15} B \right ) \cos \left (f x +e \right )}{13}+\left (A +\frac {68 B}{65}\right ) \sin \left (f x +e \right )+\frac {16 A}{13}+\frac {19 B}{13}\right ) a^{3}}{8 c f \cos \left (f x +e \right )}\) | \(110\) |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {\left (\frac {A}{2}+2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (-4 A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-8 A -16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {A}{2}-2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-4 A -\frac {23 B}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}-\frac {5 \left (3 A +4 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {8 A +8 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f c}\) | \(152\) |
| default | \(\frac {2 a^{3} \left (-\frac {\left (\frac {A}{2}+2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (-4 A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-8 A -16 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {A}{2}-2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-4 A -\frac {23 B}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}-\frac {5 \left (3 A +4 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {8 A +8 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f c}\) | \(152\) |
| risch | \(-\frac {15 a^{3} x A}{2 c}-\frac {10 a^{3} x B}{c}+\frac {2 a^{3} {\mathrm e}^{i \left (f x +e \right )} A}{c f}+\frac {31 a^{3} {\mathrm e}^{i \left (f x +e \right )} B}{8 c f}+\frac {2 a^{3} {\mathrm e}^{-i \left (f x +e \right )} A}{c f}+\frac {31 a^{3} {\mathrm e}^{-i \left (f x +e \right )} B}{8 c f}+\frac {16 a^{3} A}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}+\frac {16 a^{3} B}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}-\frac {B \,a^{3} \cos \left (3 f x +3 e \right )}{12 c f}+\frac {a^{3} \sin \left (2 f x +2 e \right ) A}{4 c f}+\frac {a^{3} \sin \left (2 f x +2 e \right ) B}{c f}\) | \(220\) |
| norman | \(\frac {-\frac {17 a^{3} A +20 a^{3} B}{c f}+\frac {5 a^{3} \left (3 A +4 B \right ) x}{2 c}-\frac {\left (5 a^{3} A +2 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}-\frac {\left (17 a^{3} A +18 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}-\frac {\left (21 a^{3} A +34 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 c f}-\frac {\left (30 a^{3} A +26 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 c f}-\frac {\left (57 a^{3} A +82 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 c f}-\frac {\left (59 a^{3} A +62 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}-\frac {\left (77 a^{3} A +70 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}-\frac {\left (135 a^{3} A +110 a^{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f}-\frac {5 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c}+\frac {10 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c}-\frac {10 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c}+\frac {15 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}-\frac {15 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}+\frac {10 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}-\frac {10 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}+\frac {5 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 c}-\frac {5 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) | \(565\) |
Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x,method=_RETURNV ERBOSE)
Output:
65/8*(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^3/c/f/cos(f*x+e)
Time = 0.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.40 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {2 \, B a^{3} \cos \left (f x + e\right )^{4} - {\left (3 \, A + 10 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (3 \, A + 4 \, B\right )} a^{3} f x - 24 \, {\left (A + 2 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 48 \, {\left (A + B\right )} a^{3} + 3 \, {\left (5 \, {\left (3 \, A + 4 \, B\right )} a^{3} f x - {\left (23 \, A + 28 \, B\right )} a^{3}\right )} \cos \left (f x + e\right ) - {\left (2 \, B a^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (3 \, A + 4 \, B\right )} a^{3} f x + 3 \, {\left (A + 4 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 3 \, {\left (7 \, A + 12 \, B\right )} a^{3} \cos \left (f x + e\right ) + 48 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )}{6 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorith m="fricas")
Output:
-1/6*(2*B*a^3*cos(f*x + e)^4 - (3*A + 10*B)*a^3*cos(f*x + e)^3 + 15*(3*A + 4*B)*a^3*f*x - 24*(A + 2*B)*a^3*cos(f*x + e)^2 - 48*(A + B)*a^3 + 3*(5*(3 *A + 4*B)*a^3*f*x - (23*A + 28*B)*a^3)*cos(f*x + e) - (2*B*a^3*cos(f*x + e )^3 + 15*(3*A + 4*B)*a^3*f*x + 3*(A + 4*B)*a^3*cos(f*x + e)^2 - 3*(7*A + 1 2*B)*a^3*cos(f*x + e) + 48*(A + B)*a^3)*sin(f*x + e))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)
Leaf count of result is larger than twice the leaf count of optimal. 4255 vs. \(2 (144) = 288\).
Time = 4.15 (sec) , antiderivative size = 4255, normalized size of antiderivative = 27.28 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x)
Output:
Piecewise((-45*A*a**3*f*x*tan(e/2 + f*x/2)**7/(6*c*f*tan(e/2 + f*x/2)**7 - 6*c*f*tan(e/2 + f*x/2)**6 + 18*c*f*tan(e/2 + f*x/2)**5 - 18*c*f*tan(e/2 + f*x/2)**4 + 18*c*f*tan(e/2 + f*x/2)**3 - 18*c*f*tan(e/2 + f*x/2)**2 + 6*c *f*tan(e/2 + f*x/2) - 6*c*f) + 45*A*a**3*f*x*tan(e/2 + f*x/2)**6/(6*c*f*ta n(e/2 + f*x/2)**7 - 6*c*f*tan(e/2 + f*x/2)**6 + 18*c*f*tan(e/2 + f*x/2)**5 - 18*c*f*tan(e/2 + f*x/2)**4 + 18*c*f*tan(e/2 + f*x/2)**3 - 18*c*f*tan(e/ 2 + f*x/2)**2 + 6*c*f*tan(e/2 + f*x/2) - 6*c*f) - 135*A*a**3*f*x*tan(e/2 + f*x/2)**5/(6*c*f*tan(e/2 + f*x/2)**7 - 6*c*f*tan(e/2 + f*x/2)**6 + 18*c*f *tan(e/2 + f*x/2)**5 - 18*c*f*tan(e/2 + f*x/2)**4 + 18*c*f*tan(e/2 + f*x/2 )**3 - 18*c*f*tan(e/2 + f*x/2)**2 + 6*c*f*tan(e/2 + f*x/2) - 6*c*f) + 135* A*a**3*f*x*tan(e/2 + f*x/2)**4/(6*c*f*tan(e/2 + f*x/2)**7 - 6*c*f*tan(e/2 + f*x/2)**6 + 18*c*f*tan(e/2 + f*x/2)**5 - 18*c*f*tan(e/2 + f*x/2)**4 + 18 *c*f*tan(e/2 + f*x/2)**3 - 18*c*f*tan(e/2 + f*x/2)**2 + 6*c*f*tan(e/2 + f* x/2) - 6*c*f) - 135*A*a**3*f*x*tan(e/2 + f*x/2)**3/(6*c*f*tan(e/2 + f*x/2) **7 - 6*c*f*tan(e/2 + f*x/2)**6 + 18*c*f*tan(e/2 + f*x/2)**5 - 18*c*f*tan( e/2 + f*x/2)**4 + 18*c*f*tan(e/2 + f*x/2)**3 - 18*c*f*tan(e/2 + f*x/2)**2 + 6*c*f*tan(e/2 + f*x/2) - 6*c*f) + 135*A*a**3*f*x*tan(e/2 + f*x/2)**2/(6* c*f*tan(e/2 + f*x/2)**7 - 6*c*f*tan(e/2 + f*x/2)**6 + 18*c*f*tan(e/2 + f*x /2)**5 - 18*c*f*tan(e/2 + f*x/2)**4 + 18*c*f*tan(e/2 + f*x/2)**3 - 18*c*f* tan(e/2 + f*x/2)**2 + 6*c*f*tan(e/2 + f*x/2) - 6*c*f) - 45*A*a**3*f*x*t...
Leaf count of result is larger than twice the leaf count of optimal. 1139 vs. \(2 (152) = 304\).
Time = 0.14 (sec) , antiderivative size = 1139, normalized size of antiderivative = 7.30 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorith m="maxima")
Output:
-1/3*(B*a^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)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 3*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3*c*sin(f*x + e)^5/( cos(f*x + e) + 1)^5 + c*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 9*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c) + 18*A*a^3*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e ) + 1)^2 - 2)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(c os(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f *x + e)/(cos(f*x + e) + 1))/c) + 18*B*a^3*((sin(f*x + e)/(cos(f*x + e) + 1 ) - sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos( f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c) + 3*A*a^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)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + 2*c*sin(f*x + e)^2/(cos( f*x + e) + 1)^2 - 2*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c*sin(f*x + e) ^4/(cos(f*x + e) + 1)^4 - c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*ar...
Time = 0.24 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.43 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=-\frac {\frac {15 \, {\left (3 \, A a^{3} + 4 \, B a^{3}\right )} {\left (f x + e\right )}}{c} + \frac {96 \, {\left (A a^{3} + B a^{3}\right )}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {2 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 12 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 24 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 42 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 48 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 96 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 24 \, A a^{3} - 46 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{3} c}}{6 \, f} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorith m="giac")
Output:
-1/6*(15*(3*A*a^3 + 4*B*a^3)*(f*x + e)/c + 96*(A*a^3 + B*a^3)/(c*(tan(1/2* f*x + 1/2*e) - 1)) + 2*(3*A*a^3*tan(1/2*f*x + 1/2*e)^5 + 12*B*a^3*tan(1/2* f*x + 1/2*e)^5 - 24*A*a^3*tan(1/2*f*x + 1/2*e)^4 - 42*B*a^3*tan(1/2*f*x + 1/2*e)^4 - 48*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 96*B*a^3*tan(1/2*f*x + 1/2*e) ^2 - 3*A*a^3*tan(1/2*f*x + 1/2*e) - 12*B*a^3*tan(1/2*f*x + 1/2*e) - 24*A*a ^3 - 46*B*a^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^3*c))/f
Time = 37.02 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.07 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {24\,A\,a^3-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (7\,A\,a^3+\frac {34\,B\,a^3}{3}\right )+\frac {94\,B\,a^3}{3}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (9\,A\,a^3+18\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (17\,A\,a^3+20\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (16\,A\,a^3+32\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (56\,A\,a^3+62\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (63\,A\,a^3+76\,B\,a^3\right )}{f\,\left (-c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )}-\frac {5\,a^3\,\mathrm {atan}\left (\frac {5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,A+4\,B\right )}{15\,A\,a^3+20\,B\,a^3}\right )\,\left (3\,A+4\,B\right )}{c\,f} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x)),x)
Output:
(24*A*a^3 - tan(e/2 + (f*x)/2)*(7*A*a^3 + (34*B*a^3)/3) + (94*B*a^3)/3 - t an(e/2 + (f*x)/2)^5*(9*A*a^3 + 18*B*a^3) + tan(e/2 + (f*x)/2)^6*(17*A*a^3 + 20*B*a^3) - tan(e/2 + (f*x)/2)^3*(16*A*a^3 + 32*B*a^3) + tan(e/2 + (f*x) /2)^4*(56*A*a^3 + 62*B*a^3) + tan(e/2 + (f*x)/2)^2*(63*A*a^3 + 76*B*a^3))/ (f*(c - c*tan(e/2 + (f*x)/2) + 3*c*tan(e/2 + (f*x)/2)^2 - 3*c*tan(e/2 + (f *x)/2)^3 + 3*c*tan(e/2 + (f*x)/2)^4 - 3*c*tan(e/2 + (f*x)/2)^5 + c*tan(e/2 + (f*x)/2)^6 - c*tan(e/2 + (f*x)/2)^7)) - (5*a^3*atan((5*a^3*tan(e/2 + (f *x)/2)*(3*A + 4*B))/(15*A*a^3 + 20*B*a^3))*(3*A + 4*B))/(c*f)
Time = 0.18 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.67 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {a^{3} \left (2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +10 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +21 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +34 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -45 \cos \left (f x +e \right ) a f x -102 \cos \left (f x +e \right ) a -60 \cos \left (f x +e \right ) b f x -120 \cos \left (f x +e \right ) b -2 \sin \left (f x +e \right )^{4} b -3 \sin \left (f x +e \right )^{3} a -12 \sin \left (f x +e \right )^{3} b -24 \sin \left (f x +e \right )^{2} a -44 \sin \left (f x +e \right )^{2} b -45 \sin \left (f x +e \right ) a f x +21 a \sin \left (f x +e \right )-60 \sin \left (f x +e \right ) b f x +34 \sin \left (f x +e \right ) b +45 a f x +102 a +60 b f x +120 b \right )}{6 c f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right )} \] Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e)),x)
Output:
(a**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*c*f*(co s(e + f*x) + sin(e + f*x) - 1))