Integrand size = 36, antiderivative size = 75 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac {a^2 (A-6 B) c \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5} \] Output:
1/7*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^6+1/35*a^2*(A-6*B)*c*cos (f*x+e)^5/f/(c-c*sin(f*x+e))^5
Leaf count is larger than twice the leaf count of optimal. \(191\) vs. \(2(75)=150\).
Time = 11.76 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.55 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=-\frac {a^2 \left (-35 (A+4 B) \cos \left (\frac {1}{2} (e+f x)\right )+7 (2 A+13 B) \cos \left (\frac {3}{2} (e+f x)\right )+35 B \cos \left (\frac {5}{2} (e+f x)\right )+A \cos \left (\frac {7}{2} (e+f x)\right )-6 B \cos \left (\frac {7}{2} (e+f x)\right )-70 A \sin \left (\frac {1}{2} (e+f x)\right )+70 B \sin \left (\frac {1}{2} (e+f x)\right )-35 A \sin \left (\frac {3}{2} (e+f x)\right )+35 B \sin \left (\frac {3}{2} (e+f x)\right )+7 A \sin \left (\frac {5}{2} (e+f x)\right )-7 B \sin \left (\frac {5}{2} (e+f x)\right )\right )}{140 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \] Input:
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^4,x]
Output:
-1/140*(a^2*(-35*(A + 4*B)*Cos[(e + f*x)/2] + 7*(2*A + 13*B)*Cos[(3*(e + f *x))/2] + 35*B*Cos[(5*(e + f*x))/2] + A*Cos[(7*(e + f*x))/2] - 6*B*Cos[(7* (e + f*x))/2] - 70*A*Sin[(e + f*x)/2] + 70*B*Sin[(e + f*x)/2] - 35*A*Sin[( 3*(e + f*x))/2] + 35*B*Sin[(3*(e + f*x))/2] + 7*A*Sin[(5*(e + f*x))/2] - 7 *B*Sin[(5*(e + f*x))/2]))/(c^4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7)
Time = 0.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3446, 3042, 3338, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^6}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^6}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^2 c^2 \left (\frac {(A-6 B) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^5}dx}{7 c}+\frac {(A+B) \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A-6 B) \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^5}dx}{7 c}+\frac {(A+B) \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}\right )\) |
| \(\Big \downarrow \) 3150 |
| \(\displaystyle a^2 c^2 \left (\frac {(A-6 B) \cos ^5(e+f x)}{35 c f (c-c \sin (e+f x))^5}+\frac {(A+B) \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^4,x ]
Output:
a^2*c^2*(((A + B)*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^6) + ((A - 6*B )*Cos[e + f*x]^5)/(35*c*f*(c - c*Sin[e + f*x])^5))
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_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) )), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 ]) && NeQ[2*m + p + 1, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 0.86 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.77
| method | result | size |
| parallelrisch | \(-\frac {2 a^{2} \left (A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (-A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (4 A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \left (-A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\frac {\left (13 A +2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5}+\frac {\left (-A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5}+\frac {6 A}{35}-\frac {B}{35}\right )}{f \,c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(133\) |
| derivativedivides | \(\frac {2 a^{2} \left (-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {128 A +112 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {32 A +32 B}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {96 A +64 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {42 A +18 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {10 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {96 A +96 B}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}\right )}{f \,c^{4}}\) | \(161\) |
| default | \(\frac {2 a^{2} \left (-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {128 A +112 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {32 A +32 B}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {96 A +64 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {42 A +18 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {10 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {96 A +96 B}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}\right )}{f \,c^{4}}\) | \(161\) |
| risch | \(-\frac {2 \left (-6 a^{2} B +a^{2} A -35 A \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-140 B \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+91 B \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+14 A \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-35 i B \,a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+70 i B \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}+35 B \,a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+7 i A \,a^{2} {\mathrm e}^{i \left (f x +e \right )}-7 i B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}-70 i A \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}+35 i A \,a^{2} {\mathrm e}^{5 i \left (f x +e \right )}\right )}{35 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} f \,c^{4}}\) | \(206\) |
| norman | \(\frac {\frac {\left (10 a^{2} A -10 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f c}-\frac {12 a^{2} A -2 a^{2} B}{35 f c}-\frac {2 a^{2} A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}}{f c}+\frac {2 \left (a^{2} A -a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{f c}+\frac {\left (2 a^{2} A -2 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 f c}-\frac {2 \left (7 a^{2} A +a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{f c}+\frac {2 \left (13 a^{2} A -13 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{5 f c}+\frac {4 \left (23 a^{2} A -23 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{5 f c}+\frac {\left (76 a^{2} A -76 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5 f c}-\frac {2 \left (109 a^{2} A +11 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{35 f c}-\frac {\left (176 a^{2} A +34 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{5 f c}-\frac {4 \left (367 a^{2} A +73 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{35 f c}-\frac {\left (862 a^{2} A +148 a^{2} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{35 f c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(425\) |
Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x,method=_RETUR NVERBOSE)
Output:
-2*a^2*(A*tan(1/2*f*x+1/2*e)^6+(-A+B)*tan(1/2*f*x+1/2*e)^5+(4*A+B)*tan(1/2 *f*x+1/2*e)^4+2*(-A+B)*tan(1/2*f*x+1/2*e)^3+1/5*(13*A+2*B)*tan(1/2*f*x+1/2 *e)^2+1/5*(-A+B)*tan(1/2*f*x+1/2*e)+6/35*A-1/35*B)/f/c^4/(tan(1/2*f*x+1/2* e)-1)^7
Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (73) = 146\).
Time = 0.08 (sec) , antiderivative size = 263, normalized size of antiderivative = 3.51 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=-\frac {{\left (A - 6 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + {\left (13 \, A + 27 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 10 \, {\left (A + B\right )} a^{2} \cos \left (f x + e\right ) - 20 \, {\left (A + B\right )} a^{2} - {\left ({\left (A - 6 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - {\left (3 \, A + 17 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 10 \, {\left (A + B\right )} a^{2} \cos \left (f x + e\right ) + 20 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )}{35 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algori thm="fricas")
Output:
-1/35*((A - 6*B)*a^2*cos(f*x + e)^4 + (4*A + 11*B)*a^2*cos(f*x + e)^3 + (1 3*A + 27*B)*a^2*cos(f*x + e)^2 - 10*(A + B)*a^2*cos(f*x + e) - 20*(A + B)* a^2 - ((A - 6*B)*a^2*cos(f*x + e)^3 - (3*A + 17*B)*a^2*cos(f*x + e)^2 + 10 *(A + B)*a^2*cos(f*x + e) + 20*(A + B)*a^2)*sin(f*x + e))/(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*cos(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)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 2008 vs. \(2 (66) = 132\).
Time = 15.86 (sec) , antiderivative size = 2008, normalized size of antiderivative = 26.77 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)
Output:
Piecewise((-70*A*a**2*tan(e/2 + f*x/2)**6/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c* *4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*ta n(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 70*A*a**2*t an(e/2 + f*x/2)**5/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f *x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)** 4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245 *c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) - 280*A*a**2*tan(e/2 + f*x/2)**4/(35 *c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f* tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/ 2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/ 2) - 35*c**4*f) + 140*A*a**2*tan(e/2 + f*x/2)**3/(35*c**4*f*tan(e/2 + f*x/ 2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c* *4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) - 182* A*a**2*tan(e/2 + f*x/2)**2/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan (e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)* *2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 14*A*a**2*tan(e/2 + f*x/2) /(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*...
Leaf count of result is larger than twice the leaf count of optimal. 1571 vs. \(2 (73) = 146\).
Time = 0.09 (sec) , antiderivative size = 1571, normalized size of antiderivative = 20.95 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algori thm="maxima")
Output:
2/105*(2*A*a^2*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(c os(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 - 1 3)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(c os(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*s in(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/(co s(f*x + e) + 1)^7) + B*a^2*(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/(co s(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*a^2*(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*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f...
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (73) = 146\).
Time = 0.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.89 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=-\frac {2 \, {\left (35 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 35 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 35 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 140 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 35 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 70 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 70 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 91 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 14 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, A a^{2} - B a^{2}\right )}}{35 \, c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algori thm="giac")
Output:
-2/35*(35*A*a^2*tan(1/2*f*x + 1/2*e)^6 - 35*A*a^2*tan(1/2*f*x + 1/2*e)^5 + 35*B*a^2*tan(1/2*f*x + 1/2*e)^5 + 140*A*a^2*tan(1/2*f*x + 1/2*e)^4 + 35*B *a^2*tan(1/2*f*x + 1/2*e)^4 - 70*A*a^2*tan(1/2*f*x + 1/2*e)^3 + 70*B*a^2*t an(1/2*f*x + 1/2*e)^3 + 91*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 14*B*a^2*tan(1/2 *f*x + 1/2*e)^2 - 7*A*a^2*tan(1/2*f*x + 1/2*e) + 7*B*a^2*tan(1/2*f*x + 1/2 *e) + 6*A*a^2 - B*a^2)/(c^4*f*(tan(1/2*f*x + 1/2*e) - 1)^7)
Time = 33.15 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.59 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {109\,A\,a^2}{4}+\frac {11\,B\,a^2}{4}-\frac {27\,A\,a^2\,\cos \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,A\,a^2\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {13\,B\,a^2\,\cos \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,a^2\,\cos \left (3\,e+3\,f\,x\right )}{8}+\frac {7\,A\,a^2\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {7\,A\,a^2\,\sin \left (3\,e+3\,f\,x\right )}{8}-\frac {7\,B\,a^2\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {7\,B\,a^2\,\sin \left (3\,e+3\,f\,x\right )}{8}-\frac {121\,A\,a^2\,\cos \left (e+f\,x\right )}{8}-\frac {9\,B\,a^2\,\cos \left (e+f\,x\right )}{8}-\frac {105\,A\,a^2\,\sin \left (e+f\,x\right )}{8}+\frac {105\,B\,a^2\,\sin \left (e+f\,x\right )}{8}\right )}{35\,c^4\,f\,\left (\frac {35\,\sqrt {2}\,\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}{8}-\frac {21\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}-\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{8}-\frac {7\,\sqrt {2}\,\cos \left (\frac {5\,e}{2}+\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{8}+\frac {\sqrt {2}\,\cos \left (\frac {7\,e}{2}-\frac {\pi }{4}+\frac {7\,f\,x}{2}\right )}{8}\right )} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^4,x )
Output:
(2*cos(e/2 + (f*x)/2)*((109*A*a^2)/4 + (11*B*a^2)/4 - (27*A*a^2*cos(2*e + 2*f*x))/4 + (5*A*a^2*cos(3*e + 3*f*x))/8 - (13*B*a^2*cos(2*e + 2*f*x))/4 + (5*B*a^2*cos(3*e + 3*f*x))/8 + (7*A*a^2*sin(2*e + 2*f*x))/2 + (7*A*a^2*si n(3*e + 3*f*x))/8 - (7*B*a^2*sin(2*e + 2*f*x))/2 - (7*B*a^2*sin(3*e + 3*f* x))/8 - (121*A*a^2*cos(e + f*x))/8 - (9*B*a^2*cos(e + f*x))/8 - (105*A*a^2 *sin(e + f*x))/8 + (105*B*a^2*sin(e + f*x))/8))/(35*c^4*f*((35*2^(1/2)*cos (e/2 + pi/4 + (f*x)/2))/8 - (21*2^(1/2)*cos((3*e)/2 - pi/4 + (3*f*x)/2))/8 - (7*2^(1/2)*cos((5*e)/2 + pi/4 + (5*f*x)/2))/8 + (2^(1/2)*cos((7*e)/2 - pi/4 + (7*f*x)/2))/8))
Time = 0.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.43 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {2 a^{2} \left (-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7} a -70 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a -35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b +35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} a -35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} b -105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a -70 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b +14 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -14 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b -28 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b -a +b \right )}{35 c^{4} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)
Output:
(2*a**2*( - 5*tan((e + f*x)/2)**7*a - 70*tan((e + f*x)/2)**5*a - 35*tan((e + f*x)/2)**5*b + 35*tan((e + f*x)/2)**4*a - 35*tan((e + f*x)/2)**4*b - 10 5*tan((e + f*x)/2)**3*a - 70*tan((e + f*x)/2)**3*b + 14*tan((e + f*x)/2)** 2*a - 14*tan((e + f*x)/2)**2*b - 28*tan((e + f*x)/2)*a - 7*tan((e + f*x)/2 )*b - a + b))/(35*c**4*f*(tan((e + f*x)/2)**7 - 7*tan((e + f*x)/2)**6 + 21 *tan((e + f*x)/2)**5 - 35*tan((e + f*x)/2)**4 + 35*tan((e + f*x)/2)**3 - 2 1*tan((e + f*x)/2)**2 + 7*tan((e + f*x)/2) - 1))