Integrand size = 26, antiderivative size = 75 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {4 \tan (e+f x)}{5 a^3 c^2 f}+\frac {4 \tan ^3(e+f x)}{15 a^3 c^2 f} \] Output:
-1/5*sec(f*x+e)^3/c^2/f/(a^3+a^3*sin(f*x+e))+4/5*tan(f*x+e)/a^3/c^2/f+4/15 *tan(f*x+e)^3/a^3/c^2/f
Leaf count is larger than twice the leaf count of optimal. \(154\) vs. \(2(75)=150\).
Time = 2.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1050 \cos (e+f x)+256 \cos (2 (e+f x))+350 \cos (3 (e+f x))+128 \cos (4 (e+f x))-768 \sin (e+f x)+350 \sin (2 (e+f x))-256 \sin (3 (e+f x))+175 \sin (4 (e+f x)))}{1920 a^3 c^2 f (-1+\sin (e+f x))^2 (1+\sin (e+f x))^3} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^2),x]
Output:
-1/1920*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(1050*Cos[e + f*x] + 256*Cos[2*(e + f*x)] + 350*Cos[3*(e + f*x) ] + 128*Cos[4*(e + f*x)] - 768*Sin[e + f*x] + 350*Sin[2*(e + f*x)] - 256*S in[3*(e + f*x)] + 175*Sin[4*(e + f*x)]))/(a^3*c^2*f*(-1 + Sin[e + f*x])^2* (1 + Sin[e + f*x])^3)
Time = 0.41 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3215, 3042, 3151, 3042, 4254, 2009}
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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle \frac {\int \frac {\sec ^4(e+f x)}{\sin (e+f x) a+a}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\cos (e+f x)^4 (\sin (e+f x) a+a)}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {\frac {4 \int \sec ^4(e+f x)dx}{5 a}-\frac {\sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4 \int \csc \left (e+f x+\frac {\pi }{2}\right )^4dx}{5 a}-\frac {\sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {4 \int \left (\tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{5 a f}-\frac {\sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {4 \left (-\frac {1}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{5 a f}-\frac {\sec ^3(e+f x)}{5 f (a \sin (e+f x)+a)}}{a^2 c^2}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^2),x]
Output:
(-1/5*Sec[e + f*x]^3/(f*(a + a*Sin[e + f*x])) - (4*(-Tan[e + f*x] - Tan[e + f*x]^3/3))/(5*a*f))/(a^2*c^2)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
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]))
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Result contains complex when optimal does not.
Time = 1.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03
| method | result | size |
| risch | \(-\frac {16 \left (6 \,{\mathrm e}^{3 i \left (f x +e \right )}+2 i {\mathrm e}^{2 i \left (f x +e \right )}+2 \,{\mathrm e}^{i \left (f x +e \right )}+i\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2} a^{3}}\) | \(77\) |
| parallelrisch | \(\frac {\frac {2}{5}-\frac {26 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{15}+\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3}-\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5}-\frac {14 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5}}{f \,c^{2} a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(129\) |
| derivativedivides | \(\frac {-\frac {2}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {5}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {11}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {1}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {5}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{a^{3} c^{2} f}\) | \(133\) |
| default | \(\frac {-\frac {2}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {5}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {11}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {1}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {5}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{a^{3} c^{2} f}\) | \(133\) |
| norman | \(\frac {\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 a f c}+\frac {2}{5 a c f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f c}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f c}-\frac {14 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 a c f}-\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 a c f}+\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 a f c}-\frac {26 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{15 a c f}}{a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\) | \(198\) |
Input:
int(1/(a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
-16/15*(6*exp(3*I*(f*x+e))+2*I*exp(2*I*(f*x+e))+2*exp(I*(f*x+e))+I)/(exp(I *(f*x+e))+I)^5/(exp(I*(f*x+e))-I)^3/f/c^2/a^3
Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {8 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right ) - 1}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="fricas")
Output:
-1/15*(8*cos(f*x + e)^4 - 4*cos(f*x + e)^2 - 4*(2*cos(f*x + e)^2 + 1)*sin( f*x + e) - 1)/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*f*cos(f*x + e)^3)
Leaf count of result is larger than twice the leaf count of optimal. 1418 vs. \(2 (66) = 132\).
Time = 5.10 (sec) , antiderivative size = 1418, normalized size of antiderivative = 18.91 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**2,x)
Output:
Piecewise((-30*tan(e/2 + f*x/2)**7/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 3 0*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 9 0*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 3 0*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a **3*c**2*f) - 30*tan(e/2 + f*x/2)**6/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15 *a**3*c**2*f) + 10*tan(e/2 + f*x/2)**5/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) + 50*tan(e/2 + f*x/2)**4/(15*a**3*c**2*f*tan(e/2 + f*x/2)* *8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)* *6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)* *3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 26*tan(e/2 + f*x/2)**3/(15*a**3*c**2*f*tan(e/2 + f*x/2 )**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2 )**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2 )**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*...
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (69) = 138\).
Time = 0.04 (sec) , antiderivative size = 335, normalized size of antiderivative = 4.47 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {13 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {25 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {15 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - 3\right )}}{15 \, {\left (a^{3} c^{2} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{3} c^{2} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} f} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="maxima")
Output:
2/15*(9*sin(f*x + e)/(cos(f*x + e) + 1) + 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 25*sin(f*x + e)^4/(cos(f *x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f*x + e)^6 /(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3)/((a^3* c^2 + 2*a^3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e)^2 /(cos(f*x + e) + 1)^2 - 6*a^3*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6* a^3*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)^6/(co s(f*x + e) + 1)^6 - 2*a^3*c^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3*c^ 2*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*f)
Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\frac {5 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13\right )}}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}} + \frac {165 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 480 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 400 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{120 \, f} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="giac")
Output:
-1/120*(5*(15*tan(1/2*f*x + 1/2*e)^2 - 24*tan(1/2*f*x + 1/2*e) + 13)/(a^3* c^2*(tan(1/2*f*x + 1/2*e) - 1)^3) + (165*tan(1/2*f*x + 1/2*e)^4 + 480*tan( 1/2*f*x + 1/2*e)^3 + 650*tan(1/2*f*x + 1/2*e)^2 + 400*tan(1/2*f*x + 1/2*e) + 113)/(a^3*c^2*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
Time = 17.79 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {2\,\left (15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-25\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+13\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+9\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-3\right )}{15\,a^3\,c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5} \] Input:
int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^2),x)
Output:
-(2*(9*tan(e/2 + (f*x)/2) + 21*tan(e/2 + (f*x)/2)^2 + 13*tan(e/2 + (f*x)/2 )^3 - 25*tan(e/2 + (f*x)/2)^4 - 5*tan(e/2 + (f*x)/2)^5 + 15*tan(e/2 + (f*x )/2)^6 + 15*tan(e/2 + (f*x)/2)^7 - 3))/(15*a^3*c^2*f*(tan(e/2 + (f*x)/2) - 1)^3*(tan(e/2 + (f*x)/2) + 1)^5)
Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3}+12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}-12 \cos \left (f x +e \right ) \sin \left (f x +e \right )-12 \cos \left (f x +e \right )+8 \sin \left (f x +e \right )^{4}+8 \sin \left (f x +e \right )^{3}-12 \sin \left (f x +e \right )^{2}-12 \sin \left (f x +e \right )+3}{15 \cos \left (f x +e \right ) a^{3} c^{2} f \left (\sin \left (f x +e \right )^{3}+\sin \left (f x +e \right )^{2}-\sin \left (f x +e \right )-1\right )} \] Input:
int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x)
Output:
(12*cos(e + f*x)*sin(e + f*x)**3 + 12*cos(e + f*x)*sin(e + f*x)**2 - 12*co s(e + f*x)*sin(e + f*x) - 12*cos(e + f*x) + 8*sin(e + f*x)**4 + 8*sin(e + f*x)**3 - 12*sin(e + f*x)**2 - 12*sin(e + f*x) + 3)/(15*cos(e + f*x)*a**3* c**2*f*(sin(e + f*x)**3 + sin(e + f*x)**2 - sin(e + f*x) - 1))