Integrand size = 26, antiderivative size = 90 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {5 c^3 x}{a^2}+\frac {5 c^3 \cos (e+f x)}{a^2 f}-\frac {2 a^2 c^3 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^4}+\frac {10 c^3 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^2} \] Output:
5*c^3*x/a^2+5*c^3*cos(f*x+e)/a^2/f-2/3*a^2*c^3*cos(f*x+e)^5/f/(a+a*sin(f*x +e))^4+10/3*c^3*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^2
Leaf count is larger than twice the leaf count of optimal. \(210\) vs. \(2(90)=180\).
Time = 12.03 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.33 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (16 \sin \left (\frac {1}{2} (e+f x)\right )-8 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-56 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+15 (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+3 \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right ) (c-c \sin (e+f x))^3}{3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (a+a \sin (e+f x))^2} \] Input:
Integrate[(c - c*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^2,x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(16*Sin[(e + f*x)/2] - 8*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 56*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[ (e + f*x)/2])^2 + 15*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 3 *Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)*(c - c*Sin[e + f*x] )^3)/(3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(a + a*Sin[e + f*x])^2)
Time = 0.54 (sec) , antiderivative size = 98, 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, 3159, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3}{(a \sin (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^3}{(a \sin (e+f x)+a)^2}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^5}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \int \frac {\cos ^4(e+f x)}{(\sin (e+f x) a+a)^3}dx}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \int \frac {\cos (e+f x)^4}{(\sin (e+f x) a+a)^3}dx}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (-\frac {3 \int \frac {\cos ^2(e+f x)}{\sin (e+f x) a+a}dx}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (-\frac {3 \int \frac {\cos (e+f x)^2}{\sin (e+f x) a+a}dx}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (-\frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (e+f x)}{a f}\right )}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (-\frac {5 \left (-\frac {3 \left (\frac {\cos (e+f x)}{a f}+\frac {x}{a}\right )}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )\) |
Input:
Int[(c - c*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^2,x]
Output:
a^3*c^3*((-2*Cos[e + f*x]^5)/(3*a*f*(a + a*Sin[e + f*x])^4) - (5*((-3*(x/a + Cos[e + f*x]/(a*f)))/a^2 - (2*Cos[e + f*x]^3)/(a*f*(a + a*Sin[e + f*x]) ^2)))/(3*a^2))
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 = 1.86 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {2 c^{3} \left (-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {1}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+5 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,a^{2}}\) | \(85\) |
| default | \(\frac {2 c^{3} \left (-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {1}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+5 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,a^{2}}\) | \(85\) |
| risch | \(\frac {5 c^{3} x}{a^{2}}+\frac {c^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 a^{2} f}+\frac {c^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 a^{2} f}+\frac {32 i c^{3} {\mathrm e}^{i \left (f x +e \right )}+24 c^{3} {\mathrm e}^{2 i \left (f x +e \right )}-\frac {56 c^{3}}{3}}{f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\) | \(108\) |
| parallelrisch | \(\frac {c^{3} \left (15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} x f +45 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} x f +60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} x f +24 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} x f +102 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+45 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x +82 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+15 f x +114 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+46\right )}{3 f \,a^{2} \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\) | \(169\) |
| norman | \(\frac {\frac {8 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f}+\frac {38 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {5 c^{3} x}{a}+\frac {46 c^{3}}{3 a f}+\frac {15 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {30 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}+\frac {50 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}+\frac {60 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}+\frac {60 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}+\frac {50 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}+\frac {30 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}+\frac {15 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a}+\frac {5 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a}+\frac {110 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}+\frac {78 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f}+\frac {130 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 a f}+\frac {34 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}+\frac {58 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a f}+\frac {106 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\) | \(406\) |
Input:
int((c-c*sin(f*x+e))^3/(a+sin(f*x+e)*a)^2,x,method=_RETURNVERBOSE)
Output:
2/f*c^3/a^2*(-16/3/(tan(1/2*f*x+1/2*e)+1)^3+8/(tan(1/2*f*x+1/2*e)+1)^2+4/( tan(1/2*f*x+1/2*e)+1)+1/(1+tan(1/2*f*x+1/2*e)^2)+5*arctan(tan(1/2*f*x+1/2* e)))
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (86) = 172\).
Time = 0.08 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.06 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {3 \, c^{3} \cos \left (f x + e\right )^{3} - 30 \, c^{3} f x + 8 \, c^{3} + {\left (15 \, c^{3} f x - 31 \, c^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, c^{3} f x + 26 \, c^{3}\right )} \cos \left (f x + e\right ) - {\left (30 \, c^{3} f x + 3 \, c^{3} \cos \left (f x + e\right )^{2} + 8 \, c^{3} + {\left (15 \, c^{3} f x + 34 \, c^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
Output:
1/3*(3*c^3*cos(f*x + e)^3 - 30*c^3*f*x + 8*c^3 + (15*c^3*f*x - 31*c^3)*cos (f*x + e)^2 - (15*c^3*f*x + 26*c^3)*cos(f*x + e) - (30*c^3*f*x + 3*c^3*cos (f*x + e)^2 + 8*c^3 + (15*c^3*f*x + 34*c^3)*cos(f*x + e))*sin(f*x + e))/(a ^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 1282 vs. \(2 (87) = 174\).
Time = 4.04 (sec) , antiderivative size = 1282, normalized size of antiderivative = 14.24 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))**3/(a+a*sin(f*x+e))**2,x)
Output:
Piecewise((15*c**3*f*x*tan(e/2 + f*x/2)**5/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f* tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 45*c**3*f*x* tan(e/2 + f*x/2)**4/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x /2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9 *a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 60*c**3*f*x*tan(e/2 + f*x/2)**3/(3* a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan( e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2 ) + 3*a**2*f) + 60*c**3*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2) **5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a* *2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 45*c**3 *f*x*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f *x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 15*c**3*f*x/(3*a**2*f*tan(e/2 + f *x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 24 *c**3*tan(e/2 + f*x/2)**4/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)* *2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 102*c**3*tan(e/2 + f*x/2)**3/ (3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*...
Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (86) = 172\).
Time = 0.13 (sec) , antiderivative size = 590, normalized size of antiderivative = 6.56 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
Output:
2/3*(2*c^3*((12*sin(f*x + e)/(cos(f*x + e) + 1) + 11*sin(f*x + e)^2/(cos(f *x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/ (cos(f*x + e) + 1)^4 + 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4 *a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^ 5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 3*c^3*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e ) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - c^3*(3*sin(f*x + e)/(cos( f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*si n(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*c^3*(3*sin(f*x + e)/(cos(f* x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin( f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) )/f
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} c^{3}}{a^{2}} + \frac {6 \, c^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} a^{2}} + \frac {8 \, {\left (3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, c^{3}\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \] Input:
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="giac")
Output:
1/3*(15*(f*x + e)*c^3/a^2 + 6*c^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^2) + 8*( 3*c^3*tan(1/2*f*x + 1/2*e)^2 + 12*c^3*tan(1/2*f*x + 1/2*e) + 5*c^3)/(a^2*( tan(1/2*f*x + 1/2*e) + 1)^3))/f
Time = 19.58 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.41 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {5\,c^3\,x}{a^2}-\frac {5\,c^3\,\left (e+f\,x\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (45\,e+45\,f\,x+114\right )}{3}\right )-\frac {c^3\,\left (15\,e+15\,f\,x+46\right )}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (15\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (45\,e+45\,f\,x+24\right )}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (20\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (60\,e+60\,f\,x+82\right )}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (20\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (60\,e+60\,f\,x+102\right )}{3}\right )}{a^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^3\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \] Input:
int((c - c*sin(e + f*x))^3/(a + a*sin(e + f*x))^2,x)
Output:
(5*c^3*x)/a^2 - (5*c^3*(e + f*x) + tan(e/2 + (f*x)/2)*(15*c^3*(e + f*x) - (c^3*(45*e + 45*f*x + 114))/3) - (c^3*(15*e + 15*f*x + 46))/3 + tan(e/2 + (f*x)/2)^4*(15*c^3*(e + f*x) - (c^3*(45*e + 45*f*x + 24))/3) + tan(e/2 + ( f*x)/2)^2*(20*c^3*(e + f*x) - (c^3*(60*e + 60*f*x + 82))/3) + tan(e/2 + (f *x)/2)^3*(20*c^3*(e + f*x) - (c^3*(60*e + 60*f*x + 102))/3))/(a^2*f*(tan(e /2 + (f*x)/2) + 1)^3*(tan(e/2 + (f*x)/2)^2 + 1))
Time = 0.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.91 \[ \int \frac {(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx=\frac {c^{3} \left (-3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+15 \cos \left (f x +e \right ) \sin \left (f x +e \right ) f x -19 \cos \left (f x +e \right ) \sin \left (f x +e \right )+15 \cos \left (f x +e \right ) f x -8 \cos \left (f x +e \right )-3 \sin \left (f x +e \right )^{3}-15 \sin \left (f x +e \right )^{2} f x -46 \sin \left (f x +e \right )^{2}-30 \sin \left (f x +e \right ) f x -19 \sin \left (f x +e \right )-15 f x +8\right )}{3 a^{2} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )-\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )-1\right )} \] Input:
int((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x)
Output:
(c**3*( - 3*cos(e + f*x)*sin(e + f*x)**2 + 15*cos(e + f*x)*sin(e + f*x)*f* x - 19*cos(e + f*x)*sin(e + f*x) + 15*cos(e + f*x)*f*x - 8*cos(e + f*x) - 3*sin(e + f*x)**3 - 15*sin(e + f*x)**2*f*x - 46*sin(e + f*x)**2 - 30*sin(e + f*x)*f*x - 19*sin(e + f*x) - 15*f*x + 8))/(3*a**2*f*(cos(e + f*x)*sin(e + f*x) + cos(e + f*x) - sin(e + f*x)**2 - 2*sin(e + f*x) - 1))