Integrand size = 36, antiderivative size = 135 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac {4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f} \] Output:
1/7*(A+B)*sec(f*x+e)^3/a^2/f/(c^2-c^2*sin(f*x+e))^2+1/35*(5*A-2*B)*sec(f*x +e)^3/a^2/f/(c^4-c^4*sin(f*x+e))+4/35*(5*A-2*B)*tan(f*x+e)/a^2/c^4/f+4/105 *(5*A-2*B)*tan(f*x+e)^3/a^2/c^4/f
Leaf count is larger than twice the leaf count of optimal. \(285\) vs. \(2(135)=270\).
Time = 5.74 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.11 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-2688 B+42 (25 A+4 B) \cos (e+f x)-512 (5 A-2 B) \cos (2 (e+f x))+225 A \cos (3 (e+f x))+36 B \cos (3 (e+f x))-1280 A \cos (4 (e+f x))+512 B \cos (4 (e+f x))-75 A \cos (5 (e+f x))-12 B \cos (5 (e+f x))-4480 A \sin (e+f x)+1792 B \sin (e+f x)-600 A \sin (2 (e+f x))-96 B \sin (2 (e+f x))-960 A \sin (3 (e+f x))+384 B \sin (3 (e+f x))-300 A \sin (4 (e+f x))-48 B \sin (4 (e+f x))+320 A \sin (5 (e+f x))-128 B \sin (5 (e+f x)))}{13440 a^2 c^4 f (-1+\sin (e+f x))^4 (1+\sin (e+f x))^2} \] Input:
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x] )^4),x]
Output:
-1/13440*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-2688*B + 42*(25*A + 4*B)*Cos[e + f*x] - 512*(5*A - 2*B)*Cos[ 2*(e + f*x)] + 225*A*Cos[3*(e + f*x)] + 36*B*Cos[3*(e + f*x)] - 1280*A*Cos [4*(e + f*x)] + 512*B*Cos[4*(e + f*x)] - 75*A*Cos[5*(e + f*x)] - 12*B*Cos[ 5*(e + f*x)] - 4480*A*Sin[e + f*x] + 1792*B*Sin[e + f*x] - 600*A*Sin[2*(e + f*x)] - 96*B*Sin[2*(e + f*x)] - 960*A*Sin[3*(e + f*x)] + 384*B*Sin[3*(e + f*x)] - 300*A*Sin[4*(e + f*x)] - 48*B*Sin[4*(e + f*x)] + 320*A*Sin[5*(e + f*x)] - 128*B*Sin[5*(e + f*x)]))/(a^2*c^4*f*(-1 + Sin[e + f*x])^4*(1 + S in[e + f*x])^2)
Time = 0.69 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.84, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3446, 3042, 3338, 3042, 3151, 3042, 4254, 2009}
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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \frac {\sec ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A+B \sin (e+f x)}{\cos (e+f x)^4 (c-c \sin (e+f x))^2}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle \frac {\frac {(5 A-2 B) \int \frac {\sec ^4(e+f x)}{c-c \sin (e+f x)}dx}{7 c}+\frac {(A+B) \sec ^3(e+f x)}{7 f (c-c \sin (e+f x))^2}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(5 A-2 B) \int \frac {1}{\cos (e+f x)^4 (c-c \sin (e+f x))}dx}{7 c}+\frac {(A+B) \sec ^3(e+f x)}{7 f (c-c \sin (e+f x))^2}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {\frac {(5 A-2 B) \left (\frac {4 \int \sec ^4(e+f x)dx}{5 c}+\frac {\sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}\right )}{7 c}+\frac {(A+B) \sec ^3(e+f x)}{7 f (c-c \sin (e+f x))^2}}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(5 A-2 B) \left (\frac {4 \int \csc \left (e+f x+\frac {\pi }{2}\right )^4dx}{5 c}+\frac {\sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}\right )}{7 c}+\frac {(A+B) \sec ^3(e+f x)}{7 f (c-c \sin (e+f x))^2}}{a^2 c^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {(5 A-2 B) \left (\frac {\sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}-\frac {4 \int \left (\tan ^2(e+f x)+1\right )d(-\tan (e+f x))}{5 c f}\right )}{7 c}+\frac {(A+B) \sec ^3(e+f x)}{7 f (c-c \sin (e+f x))^2}}{a^2 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {(A+B) \sec ^3(e+f x)}{7 f (c-c \sin (e+f x))^2}+\frac {(5 A-2 B) \left (\frac {\sec ^3(e+f x)}{5 f (c-c \sin (e+f x))}-\frac {4 \left (-\frac {1}{3} \tan ^3(e+f x)-\tan (e+f x)\right )}{5 c f}\right )}{7 c}}{a^2 c^2}\) |
Input:
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4),x ]
Output:
(((A + B)*Sec[e + f*x]^3)/(7*f*(c - c*Sin[e + f*x])^2) + ((5*A - 2*B)*(Sec [e + f*x]^3/(5*f*(c - c*Sin[e + f*x])) - (4*(-Tan[e + f*x] - Tan[e + f*x]^ 3/3))/(5*c*f)))/(7*c))/(a^2*c^2)
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 = 1.54 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.19
| method | result | size |
| risch | \(-\frac {16 \left (-8 B \,{\mathrm e}^{i \left (f x +e \right )}-28 i B \,{\mathrm e}^{4 i \left (f x +e \right )}+70 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-5 i A +2 i B +15 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-16 B \,{\mathrm e}^{3 i \left (f x +e \right )}+20 A \,{\mathrm e}^{i \left (f x +e \right )}-6 i B \,{\mathrm e}^{2 i \left (f x +e \right )}+40 A \,{\mathrm e}^{3 i \left (f x +e \right )}+42 B \,{\mathrm e}^{5 i \left (f x +e \right )}\right )}{105 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} f \,a^{2} c^{4}}\) | \(161\) |
| parallelrisch | \(\frac {-210 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}+\left (420 A -210 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+\left (-280 A +280 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+\left (-560 A -280 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (420 A -168 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+\left (280 A -28 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-760 A +136 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (240 A -264 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (30 A +72 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-60 A -18 B}{105 f \,a^{2} c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(209\) |
| derivativedivides | \(\frac {-\frac {-\frac {A}{8}+\frac {B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{8}-\frac {B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {3 A}{16}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (2 A +2 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {6 A +6 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {10 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (10 A +9 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {13 A}{16}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {23 A}{8}+\frac {11 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {55 A}{8}+\frac {35 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}}{f \,a^{2} c^{4}}\) | \(233\) |
| default | \(\frac {-\frac {-\frac {A}{8}+\frac {B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{8}-\frac {B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {3 A}{16}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (2 A +2 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {6 A +6 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {10 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (10 A +9 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {13 A}{16}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {23 A}{8}+\frac {11 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {55 A}{8}+\frac {35 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}}{f \,a^{2} c^{4}}\) | \(233\) |
| norman | \(\frac {-\frac {20 A +6 B}{35 a f c}-\frac {4 \left (10 A +11 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 a f c}-\frac {2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{a f c}+\frac {2 \left (30 A -47 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{35 a f c}-\frac {2 \left (7 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 a f c}+\frac {2 \left (2 A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a f c}+\frac {4 \left (5 A +4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 a f c}-\frac {\left (4 A +14 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 a f c}+\frac {2 \left (5 A +12 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{35 a f c}-\frac {4 \left (85 A +8 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{105 a f c}-\frac {2 \left (365 A -104 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{105 a f c}+\frac {\left (520 A -292 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{105 a f c}}{a \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(379\) |
Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x,method=_RETUR NVERBOSE)
Output:
-16/105*(-8*B*exp(I*(f*x+e))-28*I*B*exp(4*I*(f*x+e))+70*I*A*exp(4*I*(f*x+e ))-5*I*A+2*I*B+15*I*A*exp(2*I*(f*x+e))-16*B*exp(3*I*(f*x+e))+20*A*exp(I*(f *x+e))-6*I*B*exp(2*I*(f*x+e))+40*A*exp(3*I*(f*x+e))+42*B*exp(5*I*(f*x+e))) /(exp(I*(f*x+e))-I)^7/(exp(I*(f*x+e))+I)^3/f/a^2/c^4
Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=-\frac {16 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - {\left (8 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 12 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - 25 \, A + 10 \, B\right )} \sin \left (f x + e\right ) - 10 \, A + 25 \, B}{105 \, {\left (a^{2} c^{4} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3}\right )}} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algori thm="fricas")
Output:
-1/105*(16*(5*A - 2*B)*cos(f*x + e)^4 - 8*(5*A - 2*B)*cos(f*x + e)^2 - (8* (5*A - 2*B)*cos(f*x + e)^4 - 12*(5*A - 2*B)*cos(f*x + e)^2 - 25*A + 10*B)* sin(f*x + e) - 10*A + 25*B)/(a^2*c^4*f*cos(f*x + e)^5 + 2*a^2*c^4*f*cos(f* x + e)^3*sin(f*x + e) - 2*a^2*c^4*f*cos(f*x + e)^3)
Leaf count of result is larger than twice the leaf count of optimal. 4228 vs. \(2 (122) = 244\).
Time = 18.28 (sec) , antiderivative size = 4228, normalized size of antiderivative = 31.32 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**4,x)
Output:
Piecewise((-210*A*tan(e/2 + f*x/2)**9/(105*a**2*c**4*f*tan(e/2 + f*x/2)**1 0 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2) **8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x /2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 420*A*tan(e/2 + f*x/2)**8/(105*a**2*c**4*f*t an(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4* f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c* *4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2 *c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a* *2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) - 280*A*tan(e/2 + f*x/2)**7/ (105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)** 9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2) **7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f* x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) - 560*A*ta n(e/2 + f*x/2)**6/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f* tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4* f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c **4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a...
Leaf count of result is larger than twice the leaf count of optimal. 835 vs. \(2 (129) = 258\).
Time = 0.07 (sec) , antiderivative size = 835, normalized size of antiderivative = 6.19 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algori thm="maxima")
Output:
-2/105*(B*(36*sin(f*x + e)/(cos(f*x + e) + 1) - 132*sin(f*x + e)^2/(cos(f* x + e) + 1)^2 + 68*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 - 84*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 140*sin(f *x + e)^6/(cos(f*x + e) + 1)^6 + 140*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 105*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 9)/(a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 + 14*a^2*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8 *a^2*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3*a^2*c^4*sin(f*x + e)^8/(c os(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - a^2*c ^4*sin(f*x + e)^10/(cos(f*x + e) + 1)^10) + 5*A*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 24*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 76*sin(f*x + e)^3/(cos (f*x + e) + 1)^3 + 28*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 42*sin(f*x + e )^5/(cos(f*x + e) + 1)^5 - 56*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 28*sin (f*x + e)^7/(cos(f*x + e) + 1)^7 + 42*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 21*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 6)/(a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 + 14*a^2*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8 *a^2*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3*a^2*c^4*sin(f*x + e)^8...
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (129) = 258\).
Time = 0.23 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.05 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=-\frac {\frac {35 \, {\left (9 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 \, A - 5 \, B\right )}}{a^{2} c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {1365 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 210 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 5775 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 105 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 12250 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 175 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 14350 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 910 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 10185 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 756 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3955 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 427 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 760 \, A - 31 \, B}{a^{2} c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}}}{840 \, f} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algori thm="giac")
Output:
-1/840*(35*(9*A*tan(1/2*f*x + 1/2*e)^2 - 6*B*tan(1/2*f*x + 1/2*e)^2 + 15*A *tan(1/2*f*x + 1/2*e) - 9*B*tan(1/2*f*x + 1/2*e) + 8*A - 5*B)/(a^2*c^4*(ta n(1/2*f*x + 1/2*e) + 1)^3) + (1365*A*tan(1/2*f*x + 1/2*e)^6 + 210*B*tan(1/ 2*f*x + 1/2*e)^6 - 5775*A*tan(1/2*f*x + 1/2*e)^5 - 105*B*tan(1/2*f*x + 1/2 *e)^5 + 12250*A*tan(1/2*f*x + 1/2*e)^4 - 175*B*tan(1/2*f*x + 1/2*e)^4 - 14 350*A*tan(1/2*f*x + 1/2*e)^3 + 910*B*tan(1/2*f*x + 1/2*e)^3 + 10185*A*tan( 1/2*f*x + 1/2*e)^2 - 756*B*tan(1/2*f*x + 1/2*e)^2 - 3955*A*tan(1/2*f*x + 1 /2*e) + 427*B*tan(1/2*f*x + 1/2*e) + 760*A - 31*B)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f
Time = 36.98 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.46 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=-\frac {\left (\frac {32\,A}{21}-\frac {64\,B}{105}-\frac {16\,A\,\sin \left (e+f\,x\right )}{21}+\frac {32\,B\,\sin \left (e+f\,x\right )}{105}\right )\,{\cos \left (e+f\,x\right )}^4+\left (\frac {8\,A}{7}+\frac {12\,B}{35}-\frac {8\,A\,\sin \left (e+f\,x\right )}{7}-\frac {12\,B\,\sin \left (e+f\,x\right )}{35}+\frac {\left (4\,\sin \left (e+f\,x\right )-4\right )\,\left (\frac {4\,A}{7}+\frac {6\,B}{35}\right )}{2}\right )\,{\cos \left (e+f\,x\right )}^3+\left (\frac {32\,B}{105}-\frac {16\,A}{21}+\frac {8\,A\,\sin \left (e+f\,x\right )}{7}-\frac {16\,B\,\sin \left (e+f\,x\right )}{35}\right )\,{\cos \left (e+f\,x\right )}^2-\frac {4\,A}{21}+\frac {10\,B}{21}+\frac {10\,A\,\sin \left (e+f\,x\right )}{21}-\frac {4\,B\,\sin \left (e+f\,x\right )}{21}}{a^2\,c^4\,f\,\left (4\,{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )-4\,{\cos \left (e+f\,x\right )}^3+2\,{\cos \left (e+f\,x\right )}^5\right )} \] Input:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^4),x )
Output:
-((10*B)/21 - (4*A)/21 + (10*A*sin(e + f*x))/21 - (4*B*sin(e + f*x))/21 + cos(e + f*x)^3*((8*A)/7 + (12*B)/35 - (8*A*sin(e + f*x))/7 - (12*B*sin(e + f*x))/35 + ((4*sin(e + f*x) - 4)*((4*A)/7 + (6*B)/35))/2) - cos(e + f*x)^ 2*((16*A)/21 - (32*B)/105 - (8*A*sin(e + f*x))/7 + (16*B*sin(e + f*x))/35) + cos(e + f*x)^4*((32*A)/21 - (64*B)/105 - (16*A*sin(e + f*x))/21 + (32*B *sin(e + f*x))/105))/(a^2*c^4*f*(4*cos(e + f*x)^3*sin(e + f*x) - 4*cos(e + f*x)^3 + 2*cos(e + f*x)^5))
Time = 0.18 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.06 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx=\frac {-45 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a +18 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +90 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a -36 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -90 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +36 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +45 \cos \left (f x +e \right ) a -18 \cos \left (f x +e \right ) b +80 \sin \left (f x +e \right )^{5} a -32 \sin \left (f x +e \right )^{5} b -160 \sin \left (f x +e \right )^{4} a +64 \sin \left (f x +e \right )^{4} b -40 \sin \left (f x +e \right )^{3} a +16 \sin \left (f x +e \right )^{3} b +240 \sin \left (f x +e \right )^{2} a -96 \sin \left (f x +e \right )^{2} b -90 a \sin \left (f x +e \right )+36 \sin \left (f x +e \right ) b -60 a -18 b}{210 \cos \left (f x +e \right ) a^{2} c^{4} f \left (\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{3}+2 \sin \left (f x +e \right )-1\right )} \] Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x)
Output:
( - 45*cos(e + f*x)*sin(e + f*x)**4*a + 18*cos(e + f*x)*sin(e + f*x)**4*b + 90*cos(e + f*x)*sin(e + f*x)**3*a - 36*cos(e + f*x)*sin(e + f*x)**3*b - 90*cos(e + f*x)*sin(e + f*x)*a + 36*cos(e + f*x)*sin(e + f*x)*b + 45*cos(e + f*x)*a - 18*cos(e + f*x)*b + 80*sin(e + f*x)**5*a - 32*sin(e + f*x)**5* b - 160*sin(e + f*x)**4*a + 64*sin(e + f*x)**4*b - 40*sin(e + f*x)**3*a + 16*sin(e + f*x)**3*b + 240*sin(e + f*x)**2*a - 96*sin(e + f*x)**2*b - 90*s in(e + f*x)*a + 36*sin(e + f*x)*b - 60*a - 18*b)/(210*cos(e + f*x)*a**2*c* *4*f*(sin(e + f*x)**4 - 2*sin(e + f*x)**3 + 2*sin(e + f*x) - 1))