Integrand size = 36, antiderivative size = 142 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {(A+B) \sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {(4 A-3 B) \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(4 A-3 B) \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {2 (4 A-3 B) \tan (e+f x)}{35 a c^4 f} \] Output:
1/7*(A+B)*sec(f*x+e)/a/c/f/(c-c*sin(f*x+e))^3+1/35*(4*A-3*B)*sec(f*x+e)/a/ f/(c^2-c^2*sin(f*x+e))^2+1/35*(4*A-3*B)*sec(f*x+e)/a/f/(c^4-c^4*sin(f*x+e) )+2/35*(4*A-3*B)*tan(f*x+e)/a/c^4/f
Time = 4.99 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.69 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (560 B+(-406 A+182 B) \cos (e+f x)+224 (4 A-3 B) \cos (2 (e+f x))+174 A \cos (3 (e+f x))-78 B \cos (3 (e+f x))-64 A \cos (4 (e+f x))+48 B \cos (4 (e+f x))+896 A \sin (e+f x)-672 B \sin (e+f x)+406 A \sin (2 (e+f x))-182 B \sin (2 (e+f x))-384 A \sin (3 (e+f x))+288 B \sin (3 (e+f x))-29 A \sin (4 (e+f x))+13 B \sin (4 (e+f x)))}{2240 a c^4 f (-1+\sin (e+f x))^4 (1+\sin (e+f x))} \] Input:
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^ 4),x]
Output:
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(560*B + (-406*A + 182*B)*Cos[e + f*x] + 224*(4*A - 3*B)*Cos[2*(e + f*x )] + 174*A*Cos[3*(e + f*x)] - 78*B*Cos[3*(e + f*x)] - 64*A*Cos[4*(e + f*x) ] + 48*B*Cos[4*(e + f*x)] + 896*A*Sin[e + f*x] - 672*B*Sin[e + f*x] + 406* A*Sin[2*(e + f*x)] - 182*B*Sin[2*(e + f*x)] - 384*A*Sin[3*(e + f*x)] + 288 *B*Sin[3*(e + f*x)] - 29*A*Sin[4*(e + f*x)] + 13*B*Sin[4*(e + f*x)]))/(224 0*a*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x]))
Time = 0.81 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.90, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {3042, 3446, 3042, 3338, 3042, 3151, 3042, 3151, 3042, 4254, 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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a) (c-c \sin (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a) (c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A+B \sin (e+f x)}{\cos (e+f x)^2 (c-c \sin (e+f x))^3}dx}{a c}\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^2}dx}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))^2}dx}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \left (\frac {3 \int \frac {\sec ^2(e+f x)}{c-c \sin (e+f x)}dx}{5 c}+\frac {\sec (e+f x)}{5 f (c-c \sin (e+f x))^2}\right )}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \left (\frac {3 \int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))}dx}{5 c}+\frac {\sec (e+f x)}{5 f (c-c \sin (e+f x))^2}\right )}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \left (\frac {3 \left (\frac {2 \int \sec ^2(e+f x)dx}{3 c}+\frac {\sec (e+f x)}{3 f (c-c \sin (e+f x))}\right )}{5 c}+\frac {\sec (e+f x)}{5 f (c-c \sin (e+f x))^2}\right )}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \left (\frac {3 \left (\frac {2 \int \csc \left (e+f x+\frac {\pi }{2}\right )^2dx}{3 c}+\frac {\sec (e+f x)}{3 f (c-c \sin (e+f x))}\right )}{5 c}+\frac {\sec (e+f x)}{5 f (c-c \sin (e+f x))^2}\right )}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {(4 A-3 B) \left (\frac {3 \left (\frac {\sec (e+f x)}{3 f (c-c \sin (e+f x))}-\frac {2 \int 1d(-\tan (e+f x))}{3 c f}\right )}{5 c}+\frac {\sec (e+f x)}{5 f (c-c \sin (e+f x))^2}\right )}{7 c}+\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}}{a c}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {(A+B) \sec (e+f x)}{7 f (c-c \sin (e+f x))^3}+\frac {(4 A-3 B) \left (\frac {\sec (e+f x)}{5 f (c-c \sin (e+f x))^2}+\frac {3 \left (\frac {2 \tan (e+f x)}{3 c f}+\frac {\sec (e+f x)}{3 f (c-c \sin (e+f x))}\right )}{5 c}\right )}{7 c}}{a c}\) |
Input:
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]
Output:
(((A + B)*Sec[e + f*x])/(7*f*(c - c*Sin[e + f*x])^3) + ((4*A - 3*B)*(Sec[e + f*x]/(5*f*(c - c*Sin[e + f*x])^2) + (3*(Sec[e + f*x]/(3*f*(c - c*Sin[e + f*x])) + (2*Tan[e + f*x])/(3*c*f)))/(5*c)))/(7*c))/(a*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*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[(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]))
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Result contains complex when optimal does not.
Time = 0.86 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(\frac {4 i \left (56 i A \,{\mathrm e}^{3 i \left (f x +e \right )}-42 i B \,{\mathrm e}^{3 i \left (f x +e \right )}+35 B \,{\mathrm e}^{4 i \left (f x +e \right )}-24 i A \,{\mathrm e}^{i \left (f x +e \right )}+56 A \,{\mathrm e}^{2 i \left (f x +e \right )}+18 i B \,{\mathrm e}^{i \left (f x +e \right )}-42 B \,{\mathrm e}^{2 i \left (f x +e \right )}-4 A +3 B \right )}{35 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) f a \,c^{4}}\) | \(136\) |
| parallelrisch | \(\frac {-70 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+\left (210 A -70 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (-350 A +140 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (210 A -210 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (14 A +112 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (-154 A -42 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (86 A -12 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-26 A +2 B}{35 f a \,c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(171\) |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {A}{16}-\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (4 A +4 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {12 A +12 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {18 A +14 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (19 A +17 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {15 A}{16}+\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {17 A}{4}+\frac {7 B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {45 A}{4}+\frac {27 B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}}{f a \,c^{4}}\) | \(189\) |
| default | \(\frac {-\frac {2 \left (\frac {A}{16}-\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (4 A +4 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {12 A +12 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {18 A +14 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (19 A +17 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {15 A}{16}+\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {17 A}{4}+\frac {7 B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {45 A}{4}+\frac {27 B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}}{f a \,c^{4}}\) | \(189\) |
| norman | \(\frac {\frac {\left (6 A -2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f c}-\frac {26 A -2 B}{35 a f c}-\frac {12 \left (4 A -3 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5 a f c}-\frac {2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a f c}+\frac {2 \left (4 A -18 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 a f c}-\frac {4 \left (3 A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f c}+\frac {20 \left (A +B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{7 a f c}+\frac {2 \left (6 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f c}-\frac {\left (36 A +8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{7 a f c}+\frac {2 \left (43 A -6 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{35 a f c}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(313\) |
Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x,method=_RETURNV ERBOSE)
Output:
4/35*I*(56*I*A*exp(3*I*(f*x+e))-42*I*B*exp(3*I*(f*x+e))+35*B*exp(4*I*(f*x+ e))-24*I*A*exp(I*(f*x+e))+56*A*exp(2*I*(f*x+e))+18*I*B*exp(I*(f*x+e))-42*B *exp(2*I*(f*x+e))-4*A+3*B)/(exp(I*(f*x+e))-I)^7/(exp(I*(f*x+e))+I)/f/a/c^4
Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {2 \, {\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{4} - 9 \, {\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (6 \, {\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} - 20 \, A + 15 \, B\right )} \sin \left (f x + e\right ) + 15 \, A - 20 \, B}{35 \, {\left (3 \, a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right ) - {\left (a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorith m="fricas")
Output:
1/35*(2*(4*A - 3*B)*cos(f*x + e)^4 - 9*(4*A - 3*B)*cos(f*x + e)^2 + (6*(4* A - 3*B)*cos(f*x + e)^2 - 20*A + 15*B)*sin(f*x + e) + 15*A - 20*B)/(3*a*c^ 4*f*cos(f*x + e)^3 - 4*a*c^4*f*cos(f*x + e) - (a*c^4*f*cos(f*x + e)^3 - 4* a*c^4*f*cos(f*x + e))*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 2468 vs. \(2 (122) = 244\).
Time = 10.03 (sec) , antiderivative size = 2468, normalized size of antiderivative = 17.38 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)
Output:
Piecewise((-70*A*tan(e/2 + f*x/2)**7/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 21 0*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a* c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4 *f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) + 21 0*A*tan(e/2 + f*x/2)**6/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*ta n(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/ 2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) - 350*A*tan(e/2 + f*x/2)**5/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2 )**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 2 10*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) + 210*A*tan(e/2 + f*x/2)**4/(3 5*a*c**4*f*tan(e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a* c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4 *f*tan(e/2 + f*x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*t an(e/2 + f*x/2) - 35*a*c**4*f) + 14*A*tan(e/2 + f*x/2)**3/(35*a*c**4*f*tan (e/2 + f*x/2)**8 - 210*a*c**4*f*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*tan(e/2 + f*x/2)**5 + 490*a*c**4*f*tan(e/2 + f *x/2)**3 - 490*a*c**4*f*tan(e/2 + f*x/2)**2 + 210*a*c**4*f*tan(e/2 + f*x/2 ) - 35*a*c**4*f) - 154*A*tan(e/2 + f*x/2)**2/(35*a*c**4*f*tan(e/2 + f*x...
Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (137) = 274\).
Time = 0.06 (sec) , antiderivative size = 619, normalized size of antiderivative = 4.36 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorith m="maxima")
Output:
-2/35*(A*(43*sin(f*x + e)/(cos(f*x + e) + 1) - 77*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 105*sin(f*x + e)^4/( cos(f*x + e) + 1)^4 - 175*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 105*sin(f* x + e)^6/(cos(f*x + e) + 1)^6 - 35*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 1 3)/(a*c^4 - 6*a*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 14*a*c^4*sin(f*x + e )^2/(cos(f*x + e) + 1)^2 - 14*a*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 14*a*c^4*sin(f*x + e)^6/(co s(f*x + e) + 1)^6 + 6*a*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a*c^4*si n(f*x + e)^8/(cos(f*x + e) + 1)^8) - B*(6*sin(f*x + e)/(cos(f*x + e) + 1) + 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 56*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 70*sin(f*x + e)^5/(cos (f*x + e) + 1)^5 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 1)/(a*c^4 - 6* a*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 14*a*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 14*a*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a*c^4*sin(f *x + e)^5/(cos(f*x + e) + 1)^5 - 14*a*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1 )^6 + 6*a*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a*c^4*sin(f*x + e)^8/( cos(f*x + e) + 1)^8))/f
Time = 0.24 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.57 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=-\frac {\frac {35 \, {\left (A - B\right )}}{a c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {525 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 35 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1960 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 280 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 4025 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 665 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 4480 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 1120 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3143 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 791 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1176 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 392 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 243 \, A - 51 \, B}{a c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}}}{280 \, f} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorith m="giac")
Output:
-1/280*(35*(A - B)/(a*c^4*(tan(1/2*f*x + 1/2*e) + 1)) + (525*A*tan(1/2*f*x + 1/2*e)^6 + 35*B*tan(1/2*f*x + 1/2*e)^6 - 1960*A*tan(1/2*f*x + 1/2*e)^5 + 280*B*tan(1/2*f*x + 1/2*e)^5 + 4025*A*tan(1/2*f*x + 1/2*e)^4 - 665*B*tan (1/2*f*x + 1/2*e)^4 - 4480*A*tan(1/2*f*x + 1/2*e)^3 + 1120*B*tan(1/2*f*x + 1/2*e)^3 + 3143*A*tan(1/2*f*x + 1/2*e)^2 - 791*B*tan(1/2*f*x + 1/2*e)^2 - 1176*A*tan(1/2*f*x + 1/2*e) + 392*B*tan(1/2*f*x + 1/2*e) + 243*A - 51*B)/ (a*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f
Time = 37.69 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.68 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {2\,\left (\frac {35\,B}{4}+\frac {91\,A\,\cos \left (e+f\,x\right )}{4}-\frac {7\,B\,\cos \left (e+f\,x\right )}{4}+14\,A\,\sin \left (e+f\,x\right )-\frac {21\,B\,\sin \left (e+f\,x\right )}{2}+14\,A\,\cos \left (2\,e+2\,f\,x\right )-\frac {39\,A\,\cos \left (3\,e+3\,f\,x\right )}{4}-A\,\cos \left (4\,e+4\,f\,x\right )-\frac {21\,B\,\cos \left (2\,e+2\,f\,x\right )}{2}+\frac {3\,B\,\cos \left (3\,e+3\,f\,x\right )}{4}+\frac {3\,B\,\cos \left (4\,e+4\,f\,x\right )}{4}-\frac {91\,A\,\sin \left (2\,e+2\,f\,x\right )}{4}-6\,A\,\sin \left (3\,e+3\,f\,x\right )+\frac {13\,A\,\sin \left (4\,e+4\,f\,x\right )}{8}+\frac {7\,B\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {9\,B\,\sin \left (3\,e+3\,f\,x\right )}{2}-\frac {B\,\sin \left (4\,e+4\,f\,x\right )}{8}\right )}{35\,a\,c^4\,f\,\left (\frac {7\,\cos \left (e+f\,x\right )}{2}-\frac {3\,\cos \left (3\,e+3\,f\,x\right )}{2}-\frac {7\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {\sin \left (4\,e+4\,f\,x\right )}{4}\right )} \] Input:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^4),x)
Output:
(2*((35*B)/4 + (91*A*cos(e + f*x))/4 - (7*B*cos(e + f*x))/4 + 14*A*sin(e + f*x) - (21*B*sin(e + f*x))/2 + 14*A*cos(2*e + 2*f*x) - (39*A*cos(3*e + 3* f*x))/4 - A*cos(4*e + 4*f*x) - (21*B*cos(2*e + 2*f*x))/2 + (3*B*cos(3*e + 3*f*x))/4 + (3*B*cos(4*e + 4*f*x))/4 - (91*A*sin(2*e + 2*f*x))/4 - 6*A*sin (3*e + 3*f*x) + (13*A*sin(4*e + 4*f*x))/8 + (7*B*sin(2*e + 2*f*x))/4 + (9* B*sin(3*e + 3*f*x))/2 - (B*sin(4*e + 4*f*x))/8))/(35*a*c^4*f*((7*cos(e + f *x))/2 - (3*cos(3*e + 3*f*x))/2 - (7*sin(2*e + 2*f*x))/2 + sin(4*e + 4*f*x )/4))
Time = 0.17 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.80 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {4 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a -3 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +9 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +12 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -9 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -4 \cos \left (f x +e \right ) a +3 \cos \left (f x +e \right ) b +24 \sin \left (f x +e \right )^{4} a -18 \sin \left (f x +e \right )^{4} b -72 \sin \left (f x +e \right )^{3} a +54 \sin \left (f x +e \right )^{3} b +60 \sin \left (f x +e \right )^{2} a -45 \sin \left (f x +e \right )^{2} b +12 a \sin \left (f x +e \right )-9 \sin \left (f x +e \right ) b -39 a +3 b}{105 \cos \left (f x +e \right ) a \,c^{4} f \left (\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1\right )} \] Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)
Output:
(4*cos(e + f*x)*sin(e + f*x)**3*a - 3*cos(e + f*x)*sin(e + f*x)**3*b - 12* cos(e + f*x)*sin(e + f*x)**2*a + 9*cos(e + f*x)*sin(e + f*x)**2*b + 12*cos (e + f*x)*sin(e + f*x)*a - 9*cos(e + f*x)*sin(e + f*x)*b - 4*cos(e + f*x)* a + 3*cos(e + f*x)*b + 24*sin(e + f*x)**4*a - 18*sin(e + f*x)**4*b - 72*si n(e + f*x)**3*a + 54*sin(e + f*x)**3*b + 60*sin(e + f*x)**2*a - 45*sin(e + f*x)**2*b + 12*sin(e + f*x)*a - 9*sin(e + f*x)*b - 39*a + 3*b)/(105*cos(e + f*x)*a*c**4*f*(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1 ))