Integrand size = 26, antiderivative size = 94 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=-\frac {15 a^3 x}{2 c}+\frac {15 a^3 \cos (e+f x)}{2 c f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac {5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))} \] Output:
-15/2*a^3*x/c+15/2*a^3*cos(f*x+e)/c/f+2*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f* x+e))^3+5/2*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))
Time = 8.96 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.63 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (\cos \left (\frac {1}{2} (e+f x)\right ) (30 (e+f x)-16 \cos (e+f x)-\sin (2 (e+f x)))+\sin \left (\frac {1}{2} (e+f x)\right ) (-64-30 e-30 f x+16 \cos (e+f x)+\sin (2 (e+f x)))\right )}{4 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (-1+\sin (e+f x))} \] Input:
Integrate[(a + a*Sin[e + f*x])^3/(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*(e + f*x) - 16*Cos[e + f*x] - Sin[2*(e + f*x)]) + Sin[(e + f*x )/2]*(-64 - 30*e - 30*f*x + 16*Cos[e + f*x] + Sin[2*(e + f*x)])))/(4*c*f*( Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(-1 + Sin[e + f*x]))
Time = 0.55 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{c-c \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{c-c \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^2}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^2}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)}dx}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \int \frac {\cos (e+f x)^2}{c-c \sin (e+f x)}dx}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \left (\frac {\int 1dx}{c}-\frac {\cos (e+f x)}{c f}\right )}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{c f (c-c \sin (e+f x))^3}-\frac {5 \left (\frac {3 \left (\frac {x}{c}-\frac {\cos (e+f x)}{c f}\right )}{2 c}-\frac {\cos ^3(e+f x)}{2 f \left (c^2-c^2 \sin (e+f x)\right )}\right )}{c^2}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x]),x]
Output:
a^3*c^3*((2*Cos[e + f*x]^5)/(c*f*(c - c*Sin[e + f*x])^3) - (5*((3*(x/c - C os[e + f*x]/(c*f)))/(2*c) - Cos[e + f*x]^3/(2*f*(c^2 - c^2*Sin[e + f*x]))) )/c^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 = 1.56 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72
| method | result | size |
| parallelrisch | \(\frac {a^{3} \left (80+65 \sin \left (f x +e \right )+\sin \left (3 f x +3 e \right )+16 \cos \left (2 f x +2 e \right )+96 \cos \left (f x +e \right )-60 \cos \left (f x +e \right ) f x \right )}{8 c f \cos \left (f x +e \right )}\) | \(68\) |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-4}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f c}\) | \(96\) |
| default | \(\frac {2 a^{3} \left (-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}-4}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f c}\) | \(96\) |
| risch | \(-\frac {15 a^{3} x}{2 c}+\frac {2 a^{3} {\mathrm e}^{i \left (f x +e \right )}}{c f}+\frac {2 a^{3} {\mathrm e}^{-i \left (f x +e \right )}}{c f}+\frac {16 a^{3}}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}+\frac {a^{3} \sin \left (2 f x +2 e \right )}{4 c f}\) | \(96\) |
| norman | \(\frac {\frac {15 a^{3} x}{2 c}-\frac {7 a^{3}}{c f}-\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c}+\frac {45 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 c}-\frac {45 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 c}+\frac {45 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2 c}-\frac {45 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 c}+\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{2 c}-\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{2 c}-\frac {12 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}-\frac {5 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}-\frac {10 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {42 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}-\frac {17 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c f}-\frac {35 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) | \(319\) |
Input:
int((a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e)),x,method=_RETURNVERBOSE)
Output:
1/8*a^3/c/f*(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 = 129, normalized size of antiderivative = 1.37 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=\frac {a^{3} \cos \left (f x + e\right )^{3} - 15 \, a^{3} f x + 8 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} - {\left (15 \, a^{3} f x - 23 \, a^{3}\right )} \cos \left (f x + e\right ) + {\left (15 \, a^{3} f x + a^{3} \cos \left (f x + e\right )^{2} - 7 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3}\right )} \sin \left (f x + e\right )}{2 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorithm="fricas")
Output:
1/2*(a^3*cos(f*x + e)^3 - 15*a^3*f*x + 8*a^3*cos(f*x + e)^2 + 16*a^3 - (15 *a^3*f*x - 23*a^3)*cos(f*x + e) + (15*a^3*f*x + a^3*cos(f*x + e)^2 - 7*a^3 *cos(f*x + e) + 16*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. 1168 vs. \(2 (83) = 166\).
Time = 1.99 (sec) , antiderivative size = 1168, normalized size of antiderivative = 12.43 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e)),x)
Output:
Piecewise((-15*a**3*f*x*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 + f*x/2)**5 - 2 *c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x /2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 15*a**3*f*x*tan(e/2 + f*x/2)**4 /(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 30*a**3*f*x*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c *f*tan(e/2 + f*x/2) - 2*c*f) + 30*a**3*f*x*tan(e/2 + f*x/2)**2/(2*c*f*tan( e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 15*a**3*f*x* tan(e/2 + f*x/2)/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f* x/2) - 2*c*f) + 15*a**3*f*x/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f *x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*t an(e/2 + f*x/2) - 2*c*f) - 34*a**3*tan(e/2 + f*x/2)**4/(2*c*f*tan(e/2 + f* x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*ta n(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 18*a**3*tan(e/2 + f* x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan (e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2* c*f) - 78*a**3*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*t...
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (90) = 180\).
Time = 0.12 (sec) , antiderivative size = 433, normalized size of antiderivative = 4.61 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=-\frac {6 \, a^{3} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + a^{3} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 4}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + 6 \, a^{3} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {2 \, a^{3}}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{f} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorithm="maxima")
Output:
-(6*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) + a^3*((sin(f*x + e)/(cos(f*x + e) + 1) - 5*si n(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*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c) + 6*a^3*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(f*x + e) /(cos(f*x + e) + 1))) - 2*a^3/(c - c*sin(f*x + e)/(cos(f*x + e) + 1)))/f
Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c} + \frac {32 \, a^{3}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} c}}{2 \, f} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorithm="giac")
Output:
-1/2*(15*(f*x + e)*a^3/c + 32*a^3/(c*(tan(1/2*f*x + 1/2*e) - 1)) + 2*(a^3* tan(1/2*f*x + 1/2*e)^3 - 8*a^3*tan(1/2*f*x + 1/2*e)^2 - a^3*tan(1/2*f*x + 1/2*e) - 8*a^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*c))/f
Time = 19.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.33 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=-\frac {15\,a^3\,x}{2\,c}-\frac {\frac {15\,a^3\,\left (e+f\,x\right )}{2}-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {15\,a^3\,\left (e+f\,x\right )}{2}-\frac {a^3\,\left (15\,e+15\,f\,x-14\right )}{2}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-48\right )}{2}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {15\,a^3\,\left (e+f\,x\right )}{2}-\frac {a^3\,\left (15\,e+15\,f\,x-34\right )}{2}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (30\,e+30\,f\,x-18\right )}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (30\,e+30\,f\,x-78\right )}{2}\right )}{c\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^2} \] Input:
int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x)),x)
Output:
- (15*a^3*x)/(2*c) - ((15*a^3*(e + f*x))/2 - tan(e/2 + (f*x)/2)*((15*a^3*( e + f*x))/2 - (a^3*(15*e + 15*f*x - 14))/2) - (a^3*(15*e + 15*f*x - 48))/2 + tan(e/2 + (f*x)/2)^4*((15*a^3*(e + f*x))/2 - (a^3*(15*e + 15*f*x - 34)) /2) - tan(e/2 + (f*x)/2)^3*(15*a^3*(e + f*x) - (a^3*(30*e + 30*f*x - 18))/ 2) + tan(e/2 + (f*x)/2)^2*(15*a^3*(e + f*x) - (a^3*(30*e + 30*f*x - 78))/2 ))/(c*f*(tan(e/2 + (f*x)/2) - 1)*(tan(e/2 + (f*x)/2)^2 + 1)^2)
Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26 \[ \int \frac {(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx=\frac {a^{3} \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+7 \cos \left (f x +e \right ) \sin \left (f x +e \right )-15 \cos \left (f x +e \right ) f x -34 \cos \left (f x +e \right )-\sin \left (f x +e \right )^{3}-8 \sin \left (f x +e \right )^{2}-15 \sin \left (f x +e \right ) f x +7 \sin \left (f x +e \right )+15 f x +34\right )}{2 c f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right )} \] Input:
int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x)
Output:
(a**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*c*f*(cos(e + f*x) + sin(e + f*x) - 1))