Integrand size = 26, antiderivative size = 118 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {35 c^4 x}{2 a}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {35 c^4 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2} \] Output:
-35/2*c^4*x/a-35/3*c^4*cos(f*x+e)^3/a/f-35/2*c^4*cos(f*x+e)*sin(f*x+e)/a/f -2*a^3*c^4*cos(f*x+e)^7/f/(a+a*sin(f*x+e))^4-14*a*c^4*cos(f*x+e)^5/f/(a+a* sin(f*x+e))^2
Time = 14.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {c^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \left (\sin \left (\frac {1}{2} (e+f x)\right ) (-384+210 e+210 f x+141 \cos (e+f x)-\cos (3 (e+f x))-15 \sin (2 (e+f x)))+\cos \left (\frac {1}{2} (e+f x)\right ) (210 e+210 f x+141 \cos (e+f x)-\cos (3 (e+f x))-15 \sin (2 (e+f x)))\right )}{12 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8 (1+\sin (e+f x))} \] Input:
Integrate[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x]),x]
Output:
-1/12*(c^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^4*(Si n[(e + f*x)/2]*(-384 + 210*e + 210*f*x + 141*Cos[e + f*x] - Cos[3*(e + f*x )] - 15*Sin[2*(e + f*x)]) + Cos[(e + f*x)/2]*(210*e + 210*f*x + 141*Cos[e + f*x] - Cos[3*(e + f*x)] - 15*Sin[2*(e + f*x)])))/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*(1 + Sin[e + f*x]))
Time = 0.66 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 3042, 3161, 3042, 3115, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^4}{a \sin (e+f x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^4}{a \sin (e+f x)+a}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^4 c^4 \int \frac {\cos ^8(e+f x)}{(\sin (e+f x) a+a)^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \int \frac {\cos (e+f x)^8}{(\sin (e+f x) a+a)^5}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^3}dx}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^3}dx}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (\frac {5 \int \frac {\cos ^4(e+f x)}{\sin (e+f x) a+a}dx}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (\frac {5 \int \frac {\cos (e+f x)^4}{\sin (e+f x) a+a}dx}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (\frac {5 \left (\frac {\int \cos ^2(e+f x)dx}{a}+\frac {\cos ^3(e+f x)}{3 a f}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (\frac {5 \left (\frac {\int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx}{a}+\frac {\cos ^3(e+f x)}{3 a f}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (\frac {5 \left (\frac {\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}}{a}+\frac {\cos ^3(e+f x)}{3 a f}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (\frac {5 \left (\frac {\cos ^3(e+f x)}{3 a f}+\frac {\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}}{a}\right )}{a^2}+\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{a^2}-\frac {2 \cos ^7(e+f x)}{a f (a \sin (e+f x)+a)^4}\right )\) |
Input:
Int[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x]),x]
Output:
a^4*c^4*((-2*Cos[e + f*x]^7)/(a*f*(a + a*Sin[e + f*x])^4) - (7*((2*Cos[e + f*x]^5)/(a*f*(a + a*Sin[e + f*x])^2) + (5*(Cos[e + f*x]^3/(3*a*f) + (x/2 + (Cos[e + f*x]*Sin[e + f*x])/(2*f))/a))/a^2))/a^2)
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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 = 9.96 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.67
| method | result | size |
| parallelrisch | \(\frac {c^{4} \left (-525+\cos \left (4 f x +4 e \right )-140 \cos \left (2 f x +2 e \right )-664 \cos \left (f x +e \right )+399 \sin \left (f x +e \right )+15 \sin \left (3 f x +3 e \right )-420 \cos \left (f x +e \right ) f x \right )}{24 f a \cos \left (f x +e \right )}\) | \(79\) |
| derivativedivides | \(\frac {2 c^{4} \left (-\frac {16}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2}+11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+24 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+\frac {35}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}-\frac {35 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f a}\) | \(109\) |
| default | \(\frac {2 c^{4} \left (-\frac {16}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2}+11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+24 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+\frac {35}{3}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}-\frac {35 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f a}\) | \(109\) |
| risch | \(-\frac {35 c^{4} x}{2 a}-\frac {47 c^{4} {\mathrm e}^{i \left (f x +e \right )}}{8 a f}-\frac {47 c^{4} {\mathrm e}^{-i \left (f x +e \right )}}{8 a f}-\frac {32 c^{4}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {c^{4} \cos \left (3 f x +3 e \right )}{12 a f}+\frac {5 c^{4} \sin \left (2 f x +2 e \right )}{4 a f}\) | \(116\) |
| norman | \(\frac {-\frac {257 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f}-\frac {155 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f}-\frac {37 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f}-\frac {27 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}-\frac {35 c^{4} x}{2 a}-\frac {166 c^{4}}{3 a f}-\frac {35 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}-\frac {70 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}-\frac {70 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}-\frac {105 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}-\frac {105 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}-\frac {70 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}-\frac {70 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}-\frac {35 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{2 a}-\frac {35 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 a}-\frac {583 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 a f}-\frac {199 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {55 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {75 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\) | \(403\) |
Input:
int((c-c*sin(f*x+e))^4/(a+sin(f*x+e)*a),x,method=_RETURNVERBOSE)
Output:
1/24*c^4/f/a*(-525+cos(4*f*x+4*e)-140*cos(2*f*x+2*e)-664*cos(f*x+e)+399*si n(f*x+e)+15*sin(3*f*x+3*e)-420*cos(f*x+e)*f*x)/cos(f*x+e)
Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.32 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\frac {2 \, c^{4} \cos \left (f x + e\right )^{4} - 13 \, c^{4} \cos \left (f x + e\right )^{3} - 105 \, c^{4} f x - 72 \, c^{4} \cos \left (f x + e\right )^{2} - 96 \, c^{4} - 3 \, {\left (35 \, c^{4} f x + 51 \, c^{4}\right )} \cos \left (f x + e\right ) + {\left (2 \, c^{4} \cos \left (f x + e\right )^{3} - 105 \, c^{4} f x + 15 \, c^{4} \cos \left (f x + e\right )^{2} - 57 \, c^{4} \cos \left (f x + e\right ) + 96 \, c^{4}\right )} \sin \left (f x + e\right )}{6 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \] Input:
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x, algorithm="fricas")
Output:
1/6*(2*c^4*cos(f*x + e)^4 - 13*c^4*cos(f*x + e)^3 - 105*c^4*f*x - 72*c^4*c os(f*x + e)^2 - 96*c^4 - 3*(35*c^4*f*x + 51*c^4)*cos(f*x + e) + (2*c^4*cos (f*x + e)^3 - 105*c^4*f*x + 15*c^4*cos(f*x + e)^2 - 57*c^4*cos(f*x + e) + 96*c^4)*sin(f*x + e))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)
Leaf count of result is larger than twice the leaf count of optimal. 2108 vs. \(2 (112) = 224\).
Time = 3.80 (sec) , antiderivative size = 2108, normalized size of antiderivative = 17.86 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))**4/(a+a*sin(f*x+e)),x)
Output:
Piecewise((-105*c**4*f*x*tan(e/2 + f*x/2)**7/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a* f*tan(e/2 + f*x/2) + 6*a*f) - 105*c**4*f*x*tan(e/2 + f*x/2)**6/(6*a*f*tan( e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x /2)**5/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan (e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4 *f*x*tan(e/2 + f*x/2)**4/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/ 2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*t an(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x/2)**3/(6*a*f*tan(e/2 + f*x/2)**7 + 6* a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f* x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f* tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x/2)**2/(6*a*f*tan(e/ 2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 1 8*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 105*c**4*f*x*tan(e/2 + f*...
Leaf count of result is larger than twice the leaf count of optimal. 720 vs. \(2 (112) = 224\).
Time = 0.13 (sec) , antiderivative size = 720, normalized size of antiderivative = 6.10 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x, algorithm="maxima")
Output:
-1/3*(c^4*((7*sin(f*x + e)/(cos(f*x + e) + 1) + 39*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 24*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 24*sin(f*x + e)^4/ (cos(f*x + e) + 1)^4 + 9*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 16)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3*a*sin(f*x + e)^5/(co s(f*x + e) + 1)^5 + a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a*sin(f*x + e) ^7/(cos(f*x + e) + 1)^7) + 9*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 12*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(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)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e )^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*sin (f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 36*c^4*((sin(f*x + e)/(co s(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f* x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a ) + 24*c^4*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) + 6*c^4/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/ f
Time = 0.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=-\frac {\frac {105 \, {\left (f x + e\right )} c^{4}}{a} + \frac {192 \, c^{4}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 66 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 144 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 70 \, c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{3} a}}{6 \, f} \] Input:
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x, algorithm="giac")
Output:
-1/6*(105*(f*x + e)*c^4/a + 192*c^4/(a*(tan(1/2*f*x + 1/2*e) + 1)) + 2*(15 *c^4*tan(1/2*f*x + 1/2*e)^5 + 66*c^4*tan(1/2*f*x + 1/2*e)^4 + 144*c^4*tan( 1/2*f*x + 1/2*e)^2 - 15*c^4*tan(1/2*f*x + 1/2*e) + 70*c^4)/((tan(1/2*f*x + 1/2*e)^2 + 1)^3*a))/f
Time = 20.51 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.46 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\frac {\frac {35\,c^4\,\left (e+f\,x\right )}{2}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {35\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (105\,e+105\,f\,x+110\right )}{6}\right )-\frac {c^4\,\left (105\,e+105\,f\,x+332\right )}{6}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {35\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (105\,e+105\,f\,x+222\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+162\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+288\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+708\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+834\right )}{6}\right )}{a\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^3}-\frac {35\,c^4\,x}{2\,a} \] Input:
int((c - c*sin(e + f*x))^4/(a + a*sin(e + f*x)),x)
Output:
((35*c^4*(e + f*x))/2 + tan(e/2 + (f*x)/2)*((35*c^4*(e + f*x))/2 - (c^4*(1 05*e + 105*f*x + 110))/6) - (c^4*(105*e + 105*f*x + 332))/6 + tan(e/2 + (f *x)/2)^6*((35*c^4*(e + f*x))/2 - (c^4*(105*e + 105*f*x + 222))/6) + tan(e/ 2 + (f*x)/2)^5*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 162))/6) + tan(e/2 + (f*x)/2)^3*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 288 ))/6) + tan(e/2 + (f*x)/2)^4*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f* x + 708))/6) + tan(e/2 + (f*x)/2)^2*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 834))/6))/(a*f*(tan(e/2 + (f*x)/2) + 1)*(tan(e/2 + (f*x)/2)^2 + 1)^3) - (35*c^4*x)/(2*a)
Time = 0.19 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.25 \[ \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx=\frac {c^{4} \left (2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3}-13 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+55 \cos \left (f x +e \right ) \sin \left (f x +e \right )-105 \cos \left (f x +e \right ) f x +222 \cos \left (f x +e \right )+2 \sin \left (f x +e \right )^{4}-15 \sin \left (f x +e \right )^{3}+68 \sin \left (f x +e \right )^{2}+105 \sin \left (f x +e \right ) f x +55 \sin \left (f x +e \right )+105 f x -222\right )}{6 a f \left (\cos \left (f x +e \right )-\sin \left (f x +e \right )-1\right )} \] Input:
int((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x)
Output:
(c**4*(2*cos(e + f*x)*sin(e + f*x)**3 - 13*cos(e + f*x)*sin(e + f*x)**2 + 55*cos(e + f*x)*sin(e + f*x) - 105*cos(e + f*x)*f*x + 222*cos(e + f*x) + 2 *sin(e + f*x)**4 - 15*sin(e + f*x)**3 + 68*sin(e + f*x)**2 + 105*sin(e + f *x)*f*x + 55*sin(e + f*x) + 105*f*x - 222))/(6*a*f*(cos(e + f*x) - sin(e + f*x) - 1))