Integrand size = 26, antiderivative size = 124 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=-\frac {7 c^4 x}{a^3}-\frac {7 c^4 \cos (e+f x)}{a^3 f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^6}+\frac {14 a c^4 \cos ^5(e+f x)}{15 f (a+a \sin (e+f x))^4}-\frac {14 c^4 \cos ^3(e+f x)}{3 a f (a+a \sin (e+f x))^2} \] Output:
-7*c^4*x/a^3-7*c^4*cos(f*x+e)/a^3/f-2/5*a^3*c^4*cos(f*x+e)^7/f/(a+a*sin(f* x+e))^6+14/15*a*c^4*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^4-14/3*c^4*cos(f*x+e)^ 3/a/f/(a+a*sin(f*x+e))^2
Leaf count is larger than twice the leaf count of optimal. \(270\) vs. \(2(124)=248\).
Time = 13.30 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.18 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (96 \sin \left (\frac {1}{2} (e+f x)\right )-48 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-256 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+128 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+464 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-105 (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-15 \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right ) (c-c \sin (e+f x))^4}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8 (a+a \sin (e+f x))^3} \] Input:
Integrate[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x])^3,x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(96*Sin[(e + f*x)/2] - 48*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 256*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Si n[(e + f*x)/2])^2 + 128*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 464*Sin[ (e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 105*(e + f*x)*(Cos[ (e + f*x)/2] + Sin[(e + f*x)/2])^5 - 15*Cos[e + f*x]*(Cos[(e + f*x)/2] + S in[(e + f*x)/2])^5)*(c - c*Sin[e + f*x])^4)/(15*f*(Cos[(e + f*x)/2] - Sin[ (e + f*x)/2])^8*(a + a*Sin[e + f*x])^3)
Time = 0.66 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10, 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, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^4}{(a \sin (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^4}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^4 c^4 \int \frac {\cos ^8(e+f x)}{(\sin (e+f x) a+a)^7}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \int \frac {\cos (e+f x)^8}{(\sin (e+f x) a+a)^7}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^5}dx}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^5}dx}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (-\frac {5 \int \frac {\cos ^4(e+f x)}{(\sin (e+f x) a+a)^3}dx}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (-\frac {5 \int \frac {\cos (e+f x)^4}{(\sin (e+f x) a+a)^3}dx}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (-\frac {5 \left (-\frac {3 \int \frac {\cos ^2(e+f x)}{\sin (e+f x) a+a}dx}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (-\frac {5 \left (-\frac {3 \int \frac {\cos (e+f x)^2}{\sin (e+f x) a+a}dx}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (e+f x)}{a f}\right )}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a^4 c^4 \left (-\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\cos (e+f x)}{a f}+\frac {x}{a}\right )}{a^2}-\frac {2 \cos ^3(e+f x)}{a f (a \sin (e+f x)+a)^2}\right )}{3 a^2}-\frac {2 \cos ^5(e+f x)}{3 a f (a \sin (e+f x)+a)^4}\right )}{5 a^2}-\frac {2 \cos ^7(e+f x)}{5 a f (a \sin (e+f x)+a)^6}\right )\) |
Input:
Int[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x])^3,x]
Output:
a^4*c^4*((-2*Cos[e + f*x]^7)/(5*a*f*(a + a*Sin[e + f*x])^6) - (7*((-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)))/( 5*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 = 10.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {2 c^{4} \left (-\frac {1}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}-7 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {64}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {32}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {64}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f \,a^{3}}\) | \(102\) |
| default | \(\frac {2 c^{4} \left (-\frac {1}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}-7 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {64}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {32}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {64}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f \,a^{3}}\) | \(102\) |
| risch | \(-\frac {7 c^{4} x}{a^{3}}-\frac {c^{4} {\mathrm e}^{i \left (f x +e \right )}}{2 a^{3} f}-\frac {c^{4} {\mathrm e}^{-i \left (f x +e \right )}}{2 a^{3} f}-\frac {16 \left (120 i c^{4} {\mathrm e}^{3 i \left (f x +e \right )}+45 c^{4} {\mathrm e}^{4 i \left (f x +e \right )}-100 i c^{4} {\mathrm e}^{i \left (f x +e \right )}-170 c^{4} {\mathrm e}^{2 i \left (f x +e \right )}+29 c^{4}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(137\) |
| parallelrisch | \(-\frac {c^{4} \left (105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7} x f +525 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6} x f +1155 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} x f +240 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+1575 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} x f +990 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+1575 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} x f +2470 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+1155 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} x f +2540 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+525 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x +2684 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+105 f x +1430 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+334\right )}{15 f \,a^{3} \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(225\) |
| norman | \(\frac {-\frac {35 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}-\frac {98 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}-\frac {210 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}-\frac {357 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}-\frac {497 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}-\frac {588 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}-\frac {588 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}-\frac {497 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a}-\frac {357 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a}-\frac {11524 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{15 a f}-\frac {16036 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 a f}-\frac {10814 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{15 a f}-\frac {2404 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 a f}-\frac {3686 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 a f}-\frac {210 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a}-\frac {98 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{a}-\frac {1102 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 a f}-\frac {638 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{3 a f}-\frac {1366 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {286 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {860 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {35 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}}{a}-\frac {7 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{a}-\frac {66 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{a f}-\frac {16 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}}{a f}-\frac {334 c^{4}}{15 a f}-\frac {7 c^{4} x}{a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(574\) |
Input:
int((c-c*sin(f*x+e))^4/(a+sin(f*x+e)*a)^3,x,method=_RETURNVERBOSE)
Output:
2/f*c^4/a^3*(-1/(1+tan(1/2*f*x+1/2*e)^2)-7*arctan(tan(1/2*f*x+1/2*e))-64/5 /(tan(1/2*f*x+1/2*e)+1)^5+32/(tan(1/2*f*x+1/2*e)+1)^4-64/3/(tan(1/2*f*x+1/ 2*e)+1)^3-8/(tan(1/2*f*x+1/2*e)+1))
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (118) = 236\).
Time = 0.09 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.06 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=-\frac {15 \, c^{4} \cos \left (f x + e\right )^{4} - 420 \, c^{4} f x - 48 \, c^{4} + {\left (105 \, c^{4} f x + 277 \, c^{4}\right )} \cos \left (f x + e\right )^{3} + {\left (315 \, c^{4} f x - 134 \, c^{4}\right )} \cos \left (f x + e\right )^{2} - 6 \, {\left (35 \, c^{4} f x + 74 \, c^{4}\right )} \cos \left (f x + e\right ) + {\left (15 \, c^{4} \cos \left (f x + e\right )^{3} - 420 \, c^{4} f x + 48 \, c^{4} + {\left (105 \, c^{4} f x - 262 \, c^{4}\right )} \cos \left (f x + e\right )^{2} - 6 \, {\left (35 \, c^{4} f x + 66 \, c^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
Output:
-1/15*(15*c^4*cos(f*x + e)^4 - 420*c^4*f*x - 48*c^4 + (105*c^4*f*x + 277*c ^4)*cos(f*x + e)^3 + (315*c^4*f*x - 134*c^4)*cos(f*x + e)^2 - 6*(35*c^4*f* x + 74*c^4)*cos(f*x + e) + (15*c^4*cos(f*x + e)^3 - 420*c^4*f*x + 48*c^4 + (105*c^4*f*x - 262*c^4)*cos(f*x + e)^2 - 6*(35*c^4*f*x + 66*c^4)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3 *f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 2314 vs. \(2 (119) = 238\).
Time = 14.22 (sec) , antiderivative size = 2314, normalized size of antiderivative = 18.66 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))**4/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-105*c**4*f*x*tan(e/2 + f*x/2)**7/(15*a**3*f*tan(e/2 + f*x/2)** 7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a **3*f*tan(e/2 + f*x/2)**4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*ta n(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 525*c**4*f*x *tan(e/2 + f*x/2)**6/(15*a**3*f*tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a**3*f*tan(e/2 + f*x/2)** 4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e/2 + f*x/2)**2 + 75*a **3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 1155*c**4*f*x*tan(e/2 + f*x/2)**5/(1 5*a**3*f*tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f* tan(e/2 + f*x/2)**5 + 225*a**3*f*tan(e/2 + f*x/2)**4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 1575*c**4*f*x*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/ 2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 2 25*a**3*f*tan(e/2 + f*x/2)**4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3* f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 1575*c** 4*f*x*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e /2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a**3*f*tan(e/2 + f*x /2)**4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 1155*c**4*f*x*tan(e/2 + f*x/2)* *2/(15*a**3*f*tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165...
Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (118) = 236\).
Time = 0.14 (sec) , antiderivative size = 1096, normalized size of antiderivative = 8.84 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
Output:
-2/15*(3*c^4*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/(c os(f*x + e) + 1)^2 + 200*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 160*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 75*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15 *sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(cos( f*x + e) + 1) + 11*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin(f* x + e)^3/(cos(f*x + e) + 1)^3 + 15*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 11*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(cos( f*x + e) + 1)^6 + a^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(sin (f*x + e)/(cos(f*x + e) + 1))/a^3) + 4*c^4*((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)/(a^3 + 5*a^3 *sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1 )^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(c os(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan( sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + c^4*(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)/(a^3 + 5*a^3*s in(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^ 2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos (f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 12*c^4*(5...
Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=-\frac {\frac {105 \, {\left (f x + e\right )} c^{4}}{a^{3}} + \frac {30 \, c^{4}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} a^{3}} + \frac {16 \, {\left (15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 60 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 130 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 80 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 19 \, c^{4}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \] Input:
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="giac")
Output:
-1/15*(105*(f*x + e)*c^4/a^3 + 30*c^4/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^3) + 16*(15*c^4*tan(1/2*f*x + 1/2*e)^4 + 60*c^4*tan(1/2*f*x + 1/2*e)^3 + 130*c ^4*tan(1/2*f*x + 1/2*e)^2 + 80*c^4*tan(1/2*f*x + 1/2*e) + 19*c^4)/(a^3*(ta n(1/2*f*x + 1/2*e) + 1)^5))/f
Time = 20.17 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.34 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\frac {7\,c^4\,\left (e+f\,x\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (35\,c^4\,\left (e+f\,x\right )-\frac {c^4\,\left (525\,e+525\,f\,x+1430\right )}{15}\right )-\frac {c^4\,\left (105\,e+105\,f\,x+334\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (35\,c^4\,\left (e+f\,x\right )-\frac {c^4\,\left (525\,e+525\,f\,x+240\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (77\,c^4\,\left (e+f\,x\right )-\frac {c^4\,\left (1155\,e+1155\,f\,x+990\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (77\,c^4\,\left (e+f\,x\right )-\frac {c^4\,\left (1155\,e+1155\,f\,x+2684\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (105\,c^4\,\left (e+f\,x\right )-\frac {c^4\,\left (1575\,e+1575\,f\,x+2470\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (105\,c^4\,\left (e+f\,x\right )-\frac {c^4\,\left (1575\,e+1575\,f\,x+2540\right )}{15}\right )}{a^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {7\,c^4\,x}{a^3} \] Input:
int((c - c*sin(e + f*x))^4/(a + a*sin(e + f*x))^3,x)
Output:
(7*c^4*(e + f*x) + tan(e/2 + (f*x)/2)*(35*c^4*(e + f*x) - (c^4*(525*e + 52 5*f*x + 1430))/15) - (c^4*(105*e + 105*f*x + 334))/15 + tan(e/2 + (f*x)/2) ^6*(35*c^4*(e + f*x) - (c^4*(525*e + 525*f*x + 240))/15) + tan(e/2 + (f*x) /2)^5*(77*c^4*(e + f*x) - (c^4*(1155*e + 1155*f*x + 990))/15) + tan(e/2 + (f*x)/2)^2*(77*c^4*(e + f*x) - (c^4*(1155*e + 1155*f*x + 2684))/15) + tan( e/2 + (f*x)/2)^4*(105*c^4*(e + f*x) - (c^4*(1575*e + 1575*f*x + 2470))/15) + tan(e/2 + (f*x)/2)^3*(105*c^4*(e + f*x) - (c^4*(1575*e + 1575*f*x + 254 0))/15))/(a^3*f*(tan(e/2 + (f*x)/2) + 1)^5*(tan(e/2 + (f*x)/2)^2 + 1)) - ( 7*c^4*x)/a^3
Time = 0.18 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.05 \[ \int \frac {(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\frac {c^{4} \left (15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3}-105 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} f x +158 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}-210 \cos \left (f x +e \right ) \sin \left (f x +e \right ) f x +143 \cos \left (f x +e \right ) \sin \left (f x +e \right )-105 \cos \left (f x +e \right ) f x +48 \cos \left (f x +e \right )+15 \sin \left (f x +e \right )^{4}+105 \sin \left (f x +e \right )^{3} f x +381 \sin \left (f x +e \right )^{3}+315 \sin \left (f x +e \right )^{2} f x +461 \sin \left (f x +e \right )^{2}+315 \sin \left (f x +e \right ) f x +143 \sin \left (f x +e \right )+105 f x -48\right )}{15 a^{3} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )-\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )-1\right )} \] Input:
int((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x)
Output:
(c**4*(15*cos(e + f*x)*sin(e + f*x)**3 - 105*cos(e + f*x)*sin(e + f*x)**2* f*x + 158*cos(e + f*x)*sin(e + f*x)**2 - 210*cos(e + f*x)*sin(e + f*x)*f*x + 143*cos(e + f*x)*sin(e + f*x) - 105*cos(e + f*x)*f*x + 48*cos(e + f*x) + 15*sin(e + f*x)**4 + 105*sin(e + f*x)**3*f*x + 381*sin(e + f*x)**3 + 315 *sin(e + f*x)**2*f*x + 461*sin(e + f*x)**2 + 315*sin(e + f*x)*f*x + 143*si n(e + f*x) + 105*f*x - 48))/(15*a**3*f*(cos(e + f*x)*sin(e + f*x)**2 + 2*c os(e + f*x)*sin(e + f*x) + cos(e + f*x) - sin(e + f*x)**3 - 3*sin(e + f*x) **2 - 3*sin(e + f*x) - 1))