Integrand size = 26, antiderivative size = 92 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {15 c^3 x}{2 a}-\frac {15 c^3 \cos (e+f x)}{2 a f}-\frac {2 a^2 c^3 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^3}-\frac {5 c^3 \cos ^3(e+f x)}{2 f (a+a \sin (e+f x))} \] Output:
-15/2*c^3*x/a-15/2*c^3*cos(f*x+e)/a/f-2*a^2*c^3*cos(f*x+e)^5/f/(a+a*sin(f* x+e))^3-5/2*c^3*cos(f*x+e)^3/f/(a+a*sin(f*x+e))
Time = 12.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.68 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^3 \left (\sin \left (\frac {1}{2} (e+f x)\right ) (-64+30 e+30 f x+16 \cos (e+f x)-\sin (2 (e+f x)))+\cos \left (\frac {1}{2} (e+f x)\right ) (30 (e+f x)+16 \cos (e+f x)-\sin (2 (e+f x)))\right )}{4 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (1+\sin (e+f x))} \] Input:
Integrate[(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*(Sin[(e + f*x)/2]*(-64 + 30*e + 30*f*x + 16*Cos[e + f*x] - Sin[2*(e + f*x)]) + Cos[ (e + f*x)/2]*(30*(e + f*x) + 16*Cos[e + f*x] - Sin[2*(e + f*x)])))/(4*a*f* (Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(1 + Sin[e + f*x]))
Time = 0.54 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3215, 3042, 3159, 3042, 3158, 3042, 3161, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3}{a \sin (e+f x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3}{a \sin (e+f x)+a}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^4}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \int \frac {\cos ^4(e+f x)}{(\sin (e+f x) a+a)^2}dx}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \int \frac {\cos (e+f x)^4}{(\sin (e+f x) a+a)^2}dx}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{\sin (e+f x) a+a}dx}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (\frac {3 \int \frac {\cos (e+f x)^2}{\sin (e+f x) a+a}dx}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (\frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (e+f x)}{a f}\right )}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac {3 \left (\frac {\cos (e+f x)}{a f}+\frac {x}{a}\right )}{2 a}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )\) |
Input:
Int[(c - c*Sin[e + f*x])^3/(a + a*Sin[e + f*x]),x]
Output:
a^3*c^3*((-2*Cos[e + f*x]^5)/(a*f*(a + a*Sin[e + f*x])^3) - (5*((3*(x/a + Cos[e + f*x]/(a*f)))/(2*a) + Cos[e + f*x]^3/(2*f*(a^2 + a^2*Sin[e + f*x])) ))/a^2)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 1.52 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\frac {c^{3} \left (-80+65 \sin \left (f x +e \right )+\sin \left (3 f x +3 e \right )-16 \cos \left (2 f x +2 e \right )-96 \cos \left (f x +e \right )-60 \cos \left (f x +e \right ) f x \right )}{8 f a \cos \left (f x +e \right )}\) | \(68\) |
| derivativedivides | \(\frac {2 c^{3} \left (-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+4}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f a}\) | \(96\) |
| default | \(\frac {2 c^{3} \left (-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+4}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f a}\) | \(96\) |
| risch | \(-\frac {15 c^{3} x}{2 a}-\frac {2 c^{3} {\mathrm e}^{i \left (f x +e \right )}}{a f}-\frac {2 c^{3} {\mathrm e}^{-i \left (f x +e \right )}}{a f}-\frac {16 c^{3}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {c^{3} \sin \left (2 f x +2 e \right )}{4 a f}\) | \(96\) |
| norman | \(\frac {-\frac {7 c^{3}}{a f}+\frac {10 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {17 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}-\frac {5 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f}+\frac {35 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}-\frac {15 c^{3} x}{2 a}-\frac {15 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}-\frac {45 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 a}-\frac {45 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 a}-\frac {45 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2 a}-\frac {45 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 a}-\frac {15 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{2 a}-\frac {15 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{2 a}-\frac {12 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a f}+\frac {42 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\) | \(319\) |
Input:
int((c-c*sin(f*x+e))^3/(a+sin(f*x+e)*a),x,method=_RETURNVERBOSE)
Output:
1/8*c^3/f/a*(-80+65*sin(f*x+e)+sin(3*f*x+3*e)-16*cos(2*f*x+2*e)-96*cos(f*x +e)-60*cos(f*x+e)*f*x)/cos(f*x+e)
Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.39 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {c^{3} \cos \left (f x + e\right )^{3} + 15 \, c^{3} f x + 8 \, c^{3} \cos \left (f x + e\right )^{2} + 16 \, c^{3} + {\left (15 \, c^{3} f x + 23 \, c^{3}\right )} \cos \left (f x + e\right ) + {\left (15 \, c^{3} f x - c^{3} \cos \left (f x + e\right )^{2} + 7 \, c^{3} \cos \left (f x + e\right ) - 16 \, c^{3}\right )} \sin \left (f x + e\right )}{2 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \] Input:
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x, algorithm="fricas")
Output:
-1/2*(c^3*cos(f*x + e)^3 + 15*c^3*f*x + 8*c^3*cos(f*x + e)^2 + 16*c^3 + (1 5*c^3*f*x + 23*c^3)*cos(f*x + e) + (15*c^3*f*x - c^3*cos(f*x + e)^2 + 7*c^ 3*cos(f*x + e) - 16*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. 1170 vs. \(2 (85) = 170\).
Time = 2.00 (sec) , antiderivative size = 1170, normalized size of antiderivative = 12.72 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))**3/(a+a*sin(f*x+e)),x)
Output:
Piecewise((-15*c**3*f*x*tan(e/2 + f*x/2)**5/(2*a*f*tan(e/2 + f*x/2)**5 + 2 *a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x /2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 15*c**3*f*x*tan(e/2 + f*x/2)**4 /(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 30*c**3*f*x*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a *f*tan(e/2 + f*x/2) + 2*a*f) - 30*c**3*f*x*tan(e/2 + f*x/2)**2/(2*a*f*tan( e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 15*c**3*f*x* tan(e/2 + f*x/2)/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f* x/2) + 2*a*f) - 15*c**3*f*x/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f *x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*t an(e/2 + f*x/2) + 2*a*f) - 34*c**3*tan(e/2 + f*x/2)**4/(2*a*f*tan(e/2 + f* x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*ta n(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 18*c**3*tan(e/2 + f* x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan (e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2* a*f) - 78*c**3*tan(e/2 + f*x/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*t...
Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (86) = 172\).
Time = 0.13 (sec) , antiderivative size = 424, normalized size of antiderivative = 4.61 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {c^{3} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 4}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} + 6 \, c^{3} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} + 6 \, c^{3} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac {2 \, c^{3}}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{f} \] Input:
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x, algorithm="maxima")
Output:
-(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) + 6*c^3*((sin(f*x + e)/(cos( f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 6*c^3*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e) /(cos(f*x + e) + 1))) + 2*c^3/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f
Time = 0.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.21 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} c^{3}}{a} + \frac {32 \, c^{3}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 8 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 \, c^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} a}}{2 \, f} \] Input:
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x, algorithm="giac")
Output:
-1/2*(15*(f*x + e)*c^3/a + 32*c^3/(a*(tan(1/2*f*x + 1/2*e) + 1)) + 2*(c^3* tan(1/2*f*x + 1/2*e)^3 + 8*c^3*tan(1/2*f*x + 1/2*e)^2 - c^3*tan(1/2*f*x + 1/2*e) + 8*c^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*a))/f
Time = 19.14 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.35 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {\frac {15\,c^3\,\left (e+f\,x\right )}{2}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {15\,c^3\,\left (e+f\,x\right )}{2}-\frac {c^3\,\left (15\,e+15\,f\,x+14\right )}{2}\right )-\frac {c^3\,\left (15\,e+15\,f\,x+48\right )}{2}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {15\,c^3\,\left (e+f\,x\right )}{2}-\frac {c^3\,\left (15\,e+15\,f\,x+34\right )}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (15\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (30\,e+30\,f\,x+18\right )}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (15\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (30\,e+30\,f\,x+78\right )}{2}\right )}{a\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^2}-\frac {15\,c^3\,x}{2\,a} \] Input:
int((c - c*sin(e + f*x))^3/(a + a*sin(e + f*x)),x)
Output:
((15*c^3*(e + f*x))/2 + tan(e/2 + (f*x)/2)*((15*c^3*(e + f*x))/2 - (c^3*(1 5*e + 15*f*x + 14))/2) - (c^3*(15*e + 15*f*x + 48))/2 + tan(e/2 + (f*x)/2) ^4*((15*c^3*(e + f*x))/2 - (c^3*(15*e + 15*f*x + 34))/2) + tan(e/2 + (f*x) /2)^3*(15*c^3*(e + f*x) - (c^3*(30*e + 30*f*x + 18))/2) + tan(e/2 + (f*x)/ 2)^2*(15*c^3*(e + f*x) - (c^3*(30*e + 30*f*x + 78))/2))/(a*f*(tan(e/2 + (f *x)/2) + 1)*(tan(e/2 + (f*x)/2)^2 + 1)^2) - (15*c^3*x)/(2*a)
Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.32 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {c^{3} \left (-\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+7 \cos \left (f x +e \right ) \sin \left (f x +e \right )-15 \cos \left (f x +e \right ) f x +34 \cos \left (f x +e \right )-\sin \left (f x +e \right )^{3}+8 \sin \left (f x +e \right )^{2}+15 \sin \left (f x +e \right ) f x +7 \sin \left (f x +e \right )+15 f x -34\right )}{2 a f \left (\cos \left (f x +e \right )-\sin \left (f x +e \right )-1\right )} \] Input:
int((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e)),x)
Output:
(c**3*( - cos(e + f*x)*sin(e + f*x)**2 + 7*cos(e + f*x)*sin(e + f*x) - 15* cos(e + f*x)*f*x + 34*cos(e + f*x) - sin(e + f*x)**3 + 8*sin(e + f*x)**2 + 15*sin(e + f*x)*f*x + 7*sin(e + f*x) + 15*f*x - 34))/(2*a*f*(cos(e + f*x) - sin(e + f*x) - 1))