Integrand size = 34, antiderivative size = 142 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac {a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac {a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac {a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )} \] Output:
2/7*a*(A+B)*cos(f*x+e)/f/(c-c*sin(f*x+e))^4-1/35*a*(A+15*B)*cos(f*x+e)/c/f /(c-c*sin(f*x+e))^3-1/105*a*(2*A-5*B)*cos(f*x+e)/f/(c^2-c^2*sin(f*x+e))^2- 1/105*a*(2*A-5*B)*cos(f*x+e)/f/(c^4-c^4*sin(f*x+e))
Time = 6.84 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.23 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {a \left (35 (4 A-B) \cos \left (e+\frac {f x}{2}\right )-42 A \cos \left (e+\frac {3 f x}{2}\right )+2 A \cos \left (3 e+\frac {7 f x}{2}\right )-5 B \cos \left (3 e+\frac {7 f x}{2}\right )+70 A \sin \left (\frac {f x}{2}\right )+140 B \sin \left (\frac {f x}{2}\right )+105 B \sin \left (2 e+\frac {3 f x}{2}\right )+14 A \sin \left (2 e+\frac {5 f x}{2}\right )-35 B \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{420 c^4 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \] Input:
Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x]) ^4,x]
Output:
(a*(35*(4*A - B)*Cos[e + (f*x)/2] - 42*A*Cos[e + (3*f*x)/2] + 2*A*Cos[3*e + (7*f*x)/2] - 5*B*Cos[3*e + (7*f*x)/2] + 70*A*Sin[(f*x)/2] + 140*B*Sin[(f *x)/2] + 105*B*Sin[2*e + (3*f*x)/2] + 14*A*Sin[2*e + (5*f*x)/2] - 35*B*Sin [2*e + (5*f*x)/2]))/(420*c^4*f*(Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - S in[(e + f*x)/2])^7)
Time = 0.76 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 3446, 3042, 3336, 25, 3042, 3229, 3042, 3129, 3042, 3127}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a c \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \frac {\cos (e+f x)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5}dx\) |
| \(\Big \downarrow \) 3336 |
| \(\displaystyle a c \left (\frac {\int -\frac {(A+8 B) c+7 B \sin (e+f x) c}{(c-c \sin (e+f x))^3}dx}{7 c^3}+\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\int \frac {(A+8 B) c+7 B \sin (e+f x) c}{(c-c \sin (e+f x))^3}dx}{7 c^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\int \frac {(A+8 B) c+7 B \sin (e+f x) c}{(c-c \sin (e+f x))^3}dx}{7 c^3}\right )\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\frac {1}{5} (2 A-5 B) \int \frac {1}{(c-c \sin (e+f x))^2}dx+\frac {c (A+15 B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}}{7 c^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\frac {1}{5} (2 A-5 B) \int \frac {1}{(c-c \sin (e+f x))^2}dx+\frac {c (A+15 B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}}{7 c^3}\right )\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\frac {1}{5} (2 A-5 B) \left (\frac {\int \frac {1}{c-c \sin (e+f x)}dx}{3 c}+\frac {\cos (e+f x)}{3 f (c-c \sin (e+f x))^2}\right )+\frac {c (A+15 B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}}{7 c^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\frac {1}{5} (2 A-5 B) \left (\frac {\int \frac {1}{c-c \sin (e+f x)}dx}{3 c}+\frac {\cos (e+f x)}{3 f (c-c \sin (e+f x))^2}\right )+\frac {c (A+15 B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}}{7 c^3}\right )\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle a c \left (\frac {2 (A+B) \cos (e+f x)}{7 c f (c-c \sin (e+f x))^4}-\frac {\frac {c (A+15 B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}+\frac {1}{5} (2 A-5 B) \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))}+\frac {\cos (e+f x)}{3 f (c-c \sin (e+f x))^2}\right )}{7 c^3}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^4,x]
Output:
a*c*((2*(A + B)*Cos[e + f*x])/(7*c*f*(c - c*Sin[e + f*x])^4) - (((A + 15*B )*c*Cos[e + f*x])/(5*f*(c - c*Sin[e + f*x])^3) + ((2*A - 5*B)*(Cos[e + f*x ]/(3*f*(c - c*Sin[e + f*x])^2) + Cos[e + f*x]/(3*c*f*(c - c*Sin[e + f*x])) ))/5)/(7*c^3))
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*( (c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[2*(b*c - a*d)*Cos [e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(2*m + 3))), x] + Simp[1/(b^ 3*(2*m + 3)) Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d* (2*m + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]
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 = 1.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.92
| method | result | size |
| parallelrisch | \(-\frac {2 \left (A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (-2 A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\frac {\left (13 A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3}+\frac {2 \left (-5 A +2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\frac {13 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5}+\frac {\left (-\frac {8 A}{5}+B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}+\frac {23 A}{105}-\frac {B}{21}\right ) a}{f \,c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(130\) |
| risch | \(-\frac {2 i a \left (140 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-35 i B \,{\mathrm e}^{4 i \left (f x +e \right )}+105 B \,{\mathrm e}^{5 i \left (f x +e \right )}-42 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-70 A \,{\mathrm e}^{3 i \left (f x +e \right )}-140 B \,{\mathrm e}^{3 i \left (f x +e \right )}+2 i A -14 A \,{\mathrm e}^{i \left (f x +e \right )}-5 i B +35 B \,{\mathrm e}^{i \left (f x +e \right )}\right )}{105 f \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7}}\) | \(133\) |
| derivativedivides | \(\frac {2 a \left (-\frac {28 A +14 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {68 A +60 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {56 A +40 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {8 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {48 A +48 B}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {16 A +16 B}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\right )}{f \,c^{4}}\) | \(159\) |
| default | \(\frac {2 a \left (-\frac {28 A +14 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {68 A +60 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {56 A +40 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {8 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {48 A +48 B}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {16 A +16 B}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\right )}{f \,c^{4}}\) | \(159\) |
| norman | \(\frac {\frac {\left (4 A a -2 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f c}-\frac {46 A a -10 B a}{105 f c}-\frac {2 A a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{f c}+\frac {\left (16 A a -10 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 f c}+\frac {2 \left (22 A a -10 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 f c}-\frac {\left (38 A a -2 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 f c}+\frac {\left (44 A a -20 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{5 f c}+\frac {2 \left (46 A a -20 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5 f c}-\frac {2 \left (184 A a -10 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 f c}-\frac {\left (638 A a -20 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{105 f c}-\frac {2 \left (1024 A a -40 B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{105 f c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(321\) |
Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x,method=_RETURNV ERBOSE)
Output:
-2*(A*tan(1/2*f*x+1/2*e)^6+(-2*A+B)*tan(1/2*f*x+1/2*e)^5+1/3*(13*A-B)*tan( 1/2*f*x+1/2*e)^4+2/3*(-5*A+2*B)*tan(1/2*f*x+1/2*e)^3+13/5*A*tan(1/2*f*x+1/ 2*e)^2+1/3*(-8/5*A+B)*tan(1/2*f*x+1/2*e)+23/105*A-1/21*B)*a/f/c^4/(tan(1/2 *f*x+1/2*e)-1)^7
Time = 0.08 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.77 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{4} + 4 \, {\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, A + 10 \, B\right )} a \cos \left (f x + e\right )^{2} + 15 \, {\left (A + B\right )} a \cos \left (f x + e\right ) + 30 \, {\left (A + B\right )} a - {\left ({\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{2} - 15 \, {\left (A + B\right )} a \cos \left (f x + e\right ) - 30 \, {\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorith m="fricas")
Output:
1/105*((2*A - 5*B)*a*cos(f*x + e)^4 + 4*(2*A - 5*B)*a*cos(f*x + e)^3 - 3*( 3*A + 10*B)*a*cos(f*x + e)^2 + 15*(A + B)*a*cos(f*x + e) + 30*(A + B)*a - ((2*A - 5*B)*a*cos(f*x + e)^3 - 3*(2*A - 5*B)*a*cos(f*x + e)^2 - 15*(A + B )*a*cos(f*x + e) - 30*(A + B)*a)*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. 1831 vs. \(2 (124) = 248\).
Time = 9.54 (sec) , antiderivative size = 1831, normalized size of antiderivative = 12.89 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)
Output:
Piecewise((-210*A*a*tan(e/2 + f*x/2)**6/(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*t an(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) + 420*A*a*t an(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*a*tan(e/2 + f*x/2)**4/(1 05*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*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 - 22 05*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*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*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 ...
Leaf count of result is larger than twice the leaf count of optimal. 1080 vs. \(2 (138) = 276\).
Time = 0.07 (sec) , antiderivative size = 1080, normalized size of antiderivative = 7.61 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorith m="maxima")
Output:
2/105*(A*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) + B*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*s in(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*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*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^...
Time = 0.23 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.24 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=-\frac {2 \, {\left (105 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 210 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 105 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 455 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 35 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 350 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 140 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 273 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 56 \, A a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, B a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 23 \, A a - 5 \, B a\right )}}{105 \, c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorith m="giac")
Output:
-2/105*(105*A*a*tan(1/2*f*x + 1/2*e)^6 - 210*A*a*tan(1/2*f*x + 1/2*e)^5 + 105*B*a*tan(1/2*f*x + 1/2*e)^5 + 455*A*a*tan(1/2*f*x + 1/2*e)^4 - 35*B*a*t an(1/2*f*x + 1/2*e)^4 - 350*A*a*tan(1/2*f*x + 1/2*e)^3 + 140*B*a*tan(1/2*f *x + 1/2*e)^3 + 273*A*a*tan(1/2*f*x + 1/2*e)^2 - 56*A*a*tan(1/2*f*x + 1/2* e) + 35*B*a*tan(1/2*f*x + 1/2*e) + 23*A*a - 5*B*a)/(c^4*f*(tan(1/2*f*x + 1 /2*e) - 1)^7)
Time = 35.77 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.61 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=-\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {15\,B\,a}{4}-\frac {171\,A\,a}{2}+\frac {353\,A\,a\,\cos \left (e+f\,x\right )}{8}+\frac {5\,B\,a\,\cos \left (e+f\,x\right )}{4}+\frac {595\,A\,a\,\sin \left (e+f\,x\right )}{8}-35\,B\,a\,\sin \left (e+f\,x\right )+\frac {43\,A\,a\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {25\,A\,a\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {5\,B\,a\,\cos \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,a\,\cos \left (3\,e+3\,f\,x\right )}{4}-\frac {77\,A\,a\,\sin \left (2\,e+2\,f\,x\right )}{4}-\frac {21\,A\,a\,\sin \left (3\,e+3\,f\,x\right )}{8}+\frac {35\,B\,a\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{105\,c^4\,f\,\left (\frac {35\,\sqrt {2}\,\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}{8}-\frac {21\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}-\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{8}-\frac {7\,\sqrt {2}\,\cos \left (\frac {5\,e}{2}+\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{8}+\frac {\sqrt {2}\,\cos \left (\frac {7\,e}{2}-\frac {\pi }{4}+\frac {7\,f\,x}{2}\right )}{8}\right )} \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x)))/(c - c*sin(e + f*x))^4,x)
Output:
-(2*cos(e/2 + (f*x)/2)*((15*B*a)/4 - (171*A*a)/2 + (353*A*a*cos(e + f*x))/ 8 + (5*B*a*cos(e + f*x))/4 + (595*A*a*sin(e + f*x))/8 - 35*B*a*sin(e + f*x ) + (43*A*a*cos(2*e + 2*f*x))/2 - (25*A*a*cos(3*e + 3*f*x))/8 - (5*B*a*cos (2*e + 2*f*x))/4 + (5*B*a*cos(3*e + 3*f*x))/4 - (77*A*a*sin(2*e + 2*f*x))/ 4 - (21*A*a*sin(3*e + 3*f*x))/8 + (35*B*a*sin(2*e + 2*f*x))/4))/(105*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 = 0.17 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.71 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx=\frac {2 a \left (-15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7} a -105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a -105 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b +70 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} a +35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} b -175 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a -140 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b +42 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a -49 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a -35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b -8 a +5 b \right )}{105 c^{4} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-21 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )} \] Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)
Output:
(2*a*( - 15*tan((e + f*x)/2)**7*a - 105*tan((e + f*x)/2)**5*a - 105*tan((e + f*x)/2)**5*b + 70*tan((e + f*x)/2)**4*a + 35*tan((e + f*x)/2)**4*b - 17 5*tan((e + f*x)/2)**3*a - 140*tan((e + f*x)/2)**3*b + 42*tan((e + f*x)/2)* *2*a - 49*tan((e + f*x)/2)*a - 35*tan((e + f*x)/2)*b - 8*a + 5*b))/(105*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 - 21*tan((e + f*x)/2)** 2 + 7*tan((e + f*x)/2) - 1))