Integrand size = 26, antiderivative size = 106 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=-\frac {a^3 x}{c^3}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \] Output:
-a^3*x/c^3+2/5*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^5-2/3*a^3*cos(f*x+e )^3/f/(c-c*sin(f*x+e))^3+2*a^3*cos(f*x+e)/f/(c^3-c^3*sin(f*x+e))
Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(106)=212\).
Time = 8.67 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.35 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-44 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-15 (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+48 \sin \left (\frac {1}{2} (e+f x)\right )-88 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+92 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (a+a \sin (e+f x))^3}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^3} \] Input:
Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^3,x]
Output:
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(24*(Cos[(e + f*x)/2] - Sin[(e + f* x)/2]) - 44*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 15*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 48*Sin[(e + f*x)/2] - 88*(Cos[(e + f*x) /2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 92*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(a + a*Sin[e + f*x])^3)/(15*f*(Cos[(e + f* x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^3)
Time = 0.57 (sec) , antiderivative size = 116, 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, 3159, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{(c-c \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{(c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^6}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^6}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^5}-\frac {\int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^4}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^5}-\frac {\int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^4}dx}{c^2}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^5}-\frac {\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^2}dx}{c^2}}{c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^5}-\frac {\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^2}dx}{c^2}}{c^2}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^5}-\frac {\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\frac {2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}-\frac {\int 1dx}{c^2}}{c^2}}{c^2}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^3 c^3 \left (\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^5}-\frac {\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac {\frac {2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}-\frac {x}{c^2}}{c^2}}{c^2}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^3,x]
Output:
a^3*c^3*((2*Cos[e + f*x]^5)/(5*c*f*(c - c*Sin[e + f*x])^5) - ((2*Cos[e + f *x]^3)/(3*c*f*(c - c*Sin[e + f*x])^3) - (-(x/c^2) + (2*Cos[e + f*x])/(f*(c ^2 - c^2*Sin[e + f*x])))/c^2)/c^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[((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]))
Result contains complex when optimal does not.
Time = 1.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(-\frac {a^{3} x}{c^{3}}+\frac {-24 i a^{3} {\mathrm e}^{3 i \left (f x +e \right )}+12 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+\frac {56 i a^{3} {\mathrm e}^{i \left (f x +e \right )}}{3}-\frac {112 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{3}+\frac {92 a^{3}}{15}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\) | \(97\) |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {32}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {40}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\) | \(100\) |
| default | \(\frac {2 a^{3} \left (-\frac {32}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {40}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\) | \(100\) |
| parallelrisch | \(-\frac {a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} x f +\left (-5 f x +4\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (10 f x -8\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (-10 f x +\frac {80}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (5 f x -\frac {40}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-f x +\frac {52}{15}\right )}{f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(115\) |
| norman | \(\frac {\frac {a^{3} x}{c}+\frac {8 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f c}+\frac {48 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c f}-\frac {52 a^{3}}{15 c f}-\frac {5 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}+\frac {13 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c}-\frac {25 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c}+\frac {38 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c}-\frac {46 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}+\frac {46 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c}-\frac {38 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}+\frac {25 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c}-\frac {13 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c}+\frac {5 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{c}-\frac {a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{c}-\frac {4 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{f c}+\frac {40 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}+\frac {64 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c f}+\frac {112 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 c f}-\frac {116 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 c f}-\frac {472 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 c f}-\frac {556 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 c f}-\frac {1432 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 c f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(489\) |
Input:
int((a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
-a^3*x/c^3+4/15*(-90*I*a^3*exp(3*I*(f*x+e))+45*a^3*exp(4*I*(f*x+e))+70*I*a ^3*exp(I*(f*x+e))-140*a^3*exp(2*I*(f*x+e))+23*a^3)/(exp(I*(f*x+e))-I)^5/f/ c^3
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (105) = 210\).
Time = 0.09 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.22 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\frac {60 \, a^{3} f x - {\left (15 \, a^{3} f x - 46 \, a^{3}\right )} \cos \left (f x + e\right )^{3} - 24 \, a^{3} - {\left (45 \, a^{3} f x + 2 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, a^{3} f x - 12 \, a^{3}\right )} \cos \left (f x + e\right ) - {\left (60 \, a^{3} f x + 24 \, a^{3} - {\left (15 \, a^{3} f x + 46 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, a^{3} f x - 8 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^3,x, algorithm="fricas")
Output:
1/15*(60*a^3*f*x - (15*a^3*f*x - 46*a^3)*cos(f*x + e)^3 - 24*a^3 - (45*a^3 *f*x + 2*a^3)*cos(f*x + e)^2 + 6*(5*a^3*f*x - 12*a^3)*cos(f*x + e) - (60*a ^3*f*x + 24*a^3 - (15*a^3*f*x + 46*a^3)*cos(f*x + e)^2 + 6*(5*a^3*f*x - 8* a^3)*cos(f*x + e))*sin(f*x + e))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*c os(f*x + e) - 4*c^3*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 1282 vs. \(2 (95) = 190\).
Time = 7.93 (sec) , antiderivative size = 1282, normalized size of antiderivative = 12.09 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**3,x)
Output:
Piecewise((-15*a**3*f*x*tan(e/2 + f*x/2)**5/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c* *3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 75*a* *3*f*x*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan( e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f* x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 150*a**3*f*x*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c** 3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 150*a**3*f*x*tan(e/2 + f*x/2)**2/(15*c **3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan (e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f* x/2) - 15*c**3*f) - 75*a**3*f*x*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x/ 2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 1 50*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 15*a**3*f*x/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c* *3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 60*a**3*tan(e/2 + f*x/2)**4/(15*c**3* f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 120*a**3*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2...
Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (105) = 210\).
Time = 0.14 (sec) , antiderivative size = 785, normalized size of antiderivative = 7.41 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^3,x, algorithm="maxima")
Output:
-2/15*(a^3*((95*sin(f*x + e)/(cos(f*x + e) + 1) - 145*sin(f*x + e)^2/(cos( f*x + e) + 1)^2 + 75*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e) ^4/(cos(f*x + e) + 1)^4 - 22)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos (f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1)) /c^3) + a^3*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f *x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^ 4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f *x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 9*a^3*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 5 *sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1) ^3 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e) ^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c ^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 6*a^3*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c ^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e...
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{3}} + \frac {4 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 30 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 100 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 50 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, a^{3}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{15 \, f} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^3,x, algorithm="giac")
Output:
-1/15*(15*(f*x + e)*a^3/c^3 + 4*(15*a^3*tan(1/2*f*x + 1/2*e)^4 - 30*a^3*ta n(1/2*f*x + 1/2*e)^3 + 100*a^3*tan(1/2*f*x + 1/2*e)^2 - 50*a^3*tan(1/2*f*x + 1/2*e) + 13*a^3)/(c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f
Time = 18.80 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.92 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=-\frac {a^3\,x}{c^3}-\frac {a^3\,\left (e+f\,x\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (75\,e+75\,f\,x-200\right )}{15}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-52\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (5\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (75\,e+75\,f\,x-60\right )}{15}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (10\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (150\,e+150\,f\,x-120\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (10\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (150\,e+150\,f\,x-400\right )}{15}\right )}{c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5} \] Input:
int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^3,x)
Output:
- (a^3*x)/c^3 - (a^3*(e + f*x) - tan(e/2 + (f*x)/2)*(5*a^3*(e + f*x) - (a^ 3*(75*e + 75*f*x - 200))/15) - (a^3*(15*e + 15*f*x - 52))/15 + tan(e/2 + ( f*x)/2)^4*(5*a^3*(e + f*x) - (a^3*(75*e + 75*f*x - 60))/15) - tan(e/2 + (f *x)/2)^3*(10*a^3*(e + f*x) - (a^3*(150*e + 150*f*x - 120))/15) + tan(e/2 + (f*x)/2)^2*(10*a^3*(e + f*x) - (a^3*(150*e + 150*f*x - 400))/15))/(c^3*f* (tan(e/2 + (f*x)/2) - 1)^5)
Time = 0.19 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.81 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\frac {a^{3} \left (-15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} f x -12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} f x -150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} f x +150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} f x -280 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x +140 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+15 f x -40\right )}{15 c^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:
int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^3,x)
Output:
(a**3*( - 15*tan((e + f*x)/2)**5*f*x - 12*tan((e + f*x)/2)**5 + 75*tan((e + f*x)/2)**4*f*x - 150*tan((e + f*x)/2)**3*f*x + 150*tan((e + f*x)/2)**2*f *x - 280*tan((e + f*x)/2)**2 - 75*tan((e + f*x)/2)*f*x + 140*tan((e + f*x) /2) + 15*f*x - 40))/(15*c**3*f*(tan((e + f*x)/2)**5 - 5*tan((e + f*x)/2)** 4 + 10*tan((e + f*x)/2)**3 - 10*tan((e + f*x)/2)**2 + 5*tan((e + f*x)/2) - 1))