Integrand size = 24, antiderivative size = 92 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=\frac {a c \cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}+\frac {2 a \cos ^3(e+f x)}{35 f (c-c \sin (e+f x))^4}+\frac {2 a \cos ^3(e+f x)}{105 c f (c-c \sin (e+f x))^3} \] Output:
1/7*a*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^5+2/35*a*cos(f*x+e)^3/f/(c-c*sin(f *x+e))^4+2/105*a*cos(f*x+e)^3/c/f/(c-c*sin(f*x+e))^3
Time = 5.70 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.18 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=\frac {a \left (70 \cos \left (e+\frac {f x}{2}\right )-21 \cos \left (e+\frac {3 f x}{2}\right )+\cos \left (3 e+\frac {7 f x}{2}\right )+35 \sin \left (\frac {f x}{2}\right )+7 \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{210 c^4 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \] Input:
Integrate[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^4,x]
Output:
(a*(70*Cos[e + (f*x)/2] - 21*Cos[e + (3*f*x)/2] + Cos[3*e + (7*f*x)/2] + 3 5*Sin[(f*x)/2] + 7*Sin[2*e + (5*f*x)/2]))/(210*c^4*f*(Cos[e/2] - Sin[e/2]) *(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7)
Time = 0.51 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3215, 3042, 3151, 3042, 3151, 3042, 3150}
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.
| \(\displaystyle \int \frac {a \sin (e+f x)+a}{(c-c \sin (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \sin (e+f x)+a}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a c \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle a c \left (\frac {2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^4}dx}{7 c}+\frac {\cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^4}dx}{7 c}+\frac {\cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}\right )\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle a c \left (\frac {2 \left (\frac {\int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^3}dx}{5 c}+\frac {\cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}\right )}{7 c}+\frac {\cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 \left (\frac {\int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^3}dx}{5 c}+\frac {\cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}\right )}{7 c}+\frac {\cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}\right )\) |
| \(\Big \downarrow \) 3150 |
| \(\displaystyle a c \left (\frac {\cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}+\frac {2 \left (\frac {\cos ^3(e+f x)}{15 c f (c-c \sin (e+f x))^3}+\frac {\cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}\right )}{7 c}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^4,x]
Output:
a*c*(Cos[e + f*x]^3/(7*f*(c - c*Sin[e + f*x])^5) + (2*(Cos[e + f*x]^3/(5*f *(c - c*Sin[e + f*x])^4) + Cos[e + f*x]^3/(15*c*f*(c - c*Sin[e + f*x])^3)) )/(7*c))
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*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[Simplify[m + p + 1], 0] && !ILtQ[p, 0]
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]))
Result contains complex when optimal does not.
Time = 0.75 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82
| method | result | size |
| risch | \(\frac {\frac {4 a}{105}+\frac {4 i a \,{\mathrm e}^{3 i \left (f x +e \right )}}{3}+\frac {8 a \,{\mathrm e}^{4 i \left (f x +e \right )}}{3}+\frac {4 i a \,{\mathrm e}^{i \left (f x +e \right )}}{15}-\frac {4 a \,{\mathrm e}^{2 i \left (f x +e \right )}}{5}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} f \,c^{4}}\) | \(75\) |
| parallelrisch | \(-\frac {2 a \left (105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-210 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+455 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-350 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+273 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-56 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+23\right )}{105 f \,c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(101\) |
| derivativedivides | \(\frac {2 a \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {68}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {16}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {28}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {14}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{4}}\) | \(116\) |
| default | \(\frac {2 a \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {68}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {16}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {28}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {14}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{4}}\) | \(116\) |
| norman | \(\frac {-\frac {46 a}{105 c f}-\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{f c}+\frac {4 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f c}+\frac {32 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 f c}-\frac {32 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 f c}+\frac {16 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 f c}-\frac {208 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{15 c f}+\frac {116 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{15 f c}-\frac {592 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{105 c f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(201\) |
Input:
int((a+sin(f*x+e)*a)/(c-c*sin(f*x+e))^4,x,method=_RETURNVERBOSE)
Output:
4/105*(a+35*I*a*exp(3*I*(f*x+e))+70*a*exp(4*I*(f*x+e))+7*I*a*exp(I*(f*x+e) )-21*a*exp(2*I*(f*x+e)))/(exp(I*(f*x+e))-I)^7/f/c^4
Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (89) = 178\).
Time = 0.07 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.24 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=\frac {2 \, a \cos \left (f x + e\right )^{4} + 8 \, a \cos \left (f x + e\right )^{3} - 9 \, a \cos \left (f x + e\right )^{2} + 15 \, a \cos \left (f x + e\right ) - {\left (2 \, a \cos \left (f x + e\right )^{3} - 6 \, a \cos \left (f x + e\right )^{2} - 15 \, a \cos \left (f x + e\right ) - 30 \, a\right )} \sin \left (f x + e\right ) + 30 \, a}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")
Output:
1/105*(2*a*cos(f*x + e)^4 + 8*a*cos(f*x + e)^3 - 9*a*cos(f*x + e)^2 + 15*a *cos(f*x + e) - (2*a*cos(f*x + e)^3 - 6*a*cos(f*x + e)^2 - 15*a*cos(f*x + e) - 30*a)*sin(f*x + e) + 30*a)/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^4*f + (c^4*f*co s(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e) - 8*c^4*f)*si n(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 1061 vs. \(2 (82) = 164\).
Time = 5.28 (sec) , antiderivative size = 1061, normalized size of antiderivative = 11.53 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)
Output:
Piecewise((-210*a*tan(e/2 + f*x/2)**6/(105*c**4*f*tan(e/2 + f*x/2)**7 - 73 5*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4 *f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan (e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) + 420*a*tan(e /2 + f*x/2)**5/(105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/ 2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735* c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) - 910*a*tan(e/2 + f*x/2)**4/(105*c** 4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan (e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) + 700*a*tan(e/2 + f*x/2)**3/(105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675 *c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4* f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) - 546*a* tan(e/2 + f*x/2)**2/(105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2 )**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) + 112*a*tan(e/2 + f*x/2)/(105*c **4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*...
Leaf count of result is larger than twice the leaf count of optimal. 561 vs. \(2 (89) = 178\).
Time = 0.05 (sec) , antiderivative size = 561, normalized size of antiderivative = 6.10 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")
Output:
2/105*(a*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 175*sin(f*x + e)^ 4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 13)/(c^ 4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 3*a*(49*sin(f*x + e)/(cos(f*x + e) + 1) - 147*sin(f*x + e)^ 2/(cos(f*x + e) + 1)^2 + 210*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 210*sin (f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 12)/(c^4 - 7*c^4*sin(f*x + e)/ (cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*s in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + e)^6/ (cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7))/f
Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=-\frac {2 \, {\left (105 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 210 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 455 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 350 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 273 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 56 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 23 \, a\right )}}{105 \, c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}} \] Input:
integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="giac")
Output:
-2/105*(105*a*tan(1/2*f*x + 1/2*e)^6 - 210*a*tan(1/2*f*x + 1/2*e)^5 + 455* a*tan(1/2*f*x + 1/2*e)^4 - 350*a*tan(1/2*f*x + 1/2*e)^3 + 273*a*tan(1/2*f* x + 1/2*e)^2 - 56*a*tan(1/2*f*x + 1/2*e) + 23*a)/(c^4*f*(tan(1/2*f*x + 1/2 *e) - 1)^7)
Time = 18.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=\frac {\sqrt {2}\,a\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {25\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {595\,\sin \left (e+f\,x\right )}{8}-\frac {43\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {353\,\cos \left (e+f\,x\right )}{8}+\frac {77\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {21\,\sin \left (3\,e+3\,f\,x\right )}{8}+\frac {171}{2}\right )}{840\,c^4\,f\,{\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}^7} \] Input:
int((a + a*sin(e + f*x))/(c - c*sin(e + f*x))^4,x)
Output:
(2^(1/2)*a*cos(e/2 + (f*x)/2)*((25*cos(3*e + 3*f*x))/8 - (595*sin(e + f*x) )/8 - (43*cos(2*e + 2*f*x))/2 - (353*cos(e + f*x))/8 + (77*sin(2*e + 2*f*x ))/4 + (21*sin(3*e + 3*f*x))/8 + 171/2))/(840*c^4*f*cos(e/2 + pi/4 + (f*x) /2)^7)
Time = 0.18 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.93 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx=\frac {2 a \left (-15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+70 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-175 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+42 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-49 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-8\right )}{105 c^{4} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:
int((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)
Output:
(2*a*( - 15*tan((e + f*x)/2)**7 - 105*tan((e + f*x)/2)**5 + 70*tan((e + f* x)/2)**4 - 175*tan((e + f*x)/2)**3 + 42*tan((e + f*x)/2)**2 - 49*tan((e + f*x)/2) - 8))/(105*c**4*f*(tan((e + f*x)/2)**7 - 7*tan((e + f*x)/2)**6 + 2 1*tan((e + f*x)/2)**5 - 35*tan((e + f*x)/2)**4 + 35*tan((e + f*x)/2)**3 - 21*tan((e + f*x)/2)**2 + 7*tan((e + f*x)/2) - 1))