\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx\) [59]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 36, antiderivative size = 142 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {(A+B) \sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {(4 A-3 B) \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(4 A-3 B) \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {2 (4 A-3 B) \tan (e+f x)}{35 a c^4 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2938, 2751, 3852, 8} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {2 (4 A-3 B) \tan (e+f x)}{35 a c^4 f}+\frac {(4 A-3 B) \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {(4 A-3 B) \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(A+B) \sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3} \]

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

((A + B)*Sec[e + f*x])/(7*a*c*f*(c - c*Sin[e + f*x])^3) + ((4*A - 3*B)*Sec[e + f*x])/(35*a*f*(c^2 - c^2*Sin[e
+ f*x])^2) + ((4*A - 3*B)*Sec[e + f*x])/(35*a*f*(c^4 - c^4*Sin[e + f*x])) + (2*(4*A - 3*B)*Tan[e + f*x])/(35*a
*c^4*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[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[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2938

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] + Dist[(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]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[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] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx}{a c} \\ & = \frac {(A+B) \sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac {(4 A-3 B) \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{7 a c^2} \\ & = \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 {(3 (4 A-3 B)) \int \frac {\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{35 a c^3} \\ & = \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)) \int \sec ^2(e+f x) \, dx}{35 a c^4} \\ & = \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)) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{35 a c^4 f} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.02 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.69 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (560 B+(-406 A+182 B) \cos (e+f x)+224 (4 A-3 B) \cos (2 (e+f x))+174 A \cos (3 (e+f x))-78 B \cos (3 (e+f x))-64 A \cos (4 (e+f x))+48 B \cos (4 (e+f x))+896 A \sin (e+f x)-672 B \sin (e+f x)+406 A \sin (2 (e+f x))-182 B \sin (2 (e+f x))-384 A \sin (3 (e+f x))+288 B \sin (3 (e+f x))-29 A \sin (4 (e+f x))+13 B \sin (4 (e+f x)))}{2240 a c^4 f (-1+\sin (e+f x))^4 (1+\sin (e+f x))} \]

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

((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)]))/
(2240*a*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.96 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96

method result size
risch \(\frac {4 i \left (56 i A \,{\mathrm e}^{3 i \left (f x +e \right )}-42 i B \,{\mathrm e}^{3 i \left (f x +e \right )}+35 B \,{\mathrm e}^{4 i \left (f x +e \right )}-24 i A \,{\mathrm e}^{i \left (f x +e \right )}+56 A \,{\mathrm e}^{2 i \left (f x +e \right )}+18 i B \,{\mathrm e}^{i \left (f x +e \right )}-42 B \,{\mathrm e}^{2 i \left (f x +e \right )}-4 A +3 B \right )}{35 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a \,c^{4} f}\) \(136\)
parallelrisch \(\frac {-70 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (210 A -70 B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-350 A +140 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (210 A -210 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (14 A +112 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-154 A -42 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (86 A -12 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-26 A +2 B}{35 f \,c^{4} a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) \(171\)
derivativedivides \(\frac {-\frac {2 \left (\frac {A}{16}-\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (4 A +4 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {12 A +12 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {18 A +14 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (19 A +17 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {15 A}{16}+\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {17 A}{4}+\frac {7 B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {45 A}{4}+\frac {27 B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}}{a \,c^{4} f}\) \(189\)
default \(\frac {-\frac {2 \left (\frac {A}{16}-\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (4 A +4 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {12 A +12 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {18 A +14 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (19 A +17 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {15 A}{16}+\frac {B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {17 A}{4}+\frac {7 B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {45 A}{4}+\frac {27 B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}}{a \,c^{4} f}\) \(189\)
norman \(\frac {\frac {\left (6 A -2 B \right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}-\frac {26 A -2 B}{35 a f c}-\frac {12 \left (4 A -3 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f c}-\frac {2 A \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}+\frac {2 \left (4 A -18 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f c}-\frac {4 \left (3 A -B \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}+\frac {20 \left (A +B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7 a f c}+\frac {2 \left (6 A -4 B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}-\frac {\left (36 A +8 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{7 a f c}+\frac {2 \left (43 A -6 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{35 a f c}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) \(313\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

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)/a/c^4
/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {2 \, {\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{4} - 9 \, {\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (6 \, {\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} - 20 \, A + 15 \, B\right )} \sin \left (f x + e\right ) + 15 \, A - 20 \, B}{35 \, {\left (3 \, a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right ) - {\left (a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2468 vs. \(2 (122) = 244\).

Time = 9.50 (sec) , antiderivative size = 2468, normalized size of antiderivative = 17.38 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)

[Out]

Piecewise((-70*A*tan(e/2 + f*x/2)**7/(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) + 210*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*tan(e/2 + f*x/2)**7 + 490*a*c**4*f*tan(e/2 + f*x/2)**6 - 490*a*c**4*f*ta
n(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 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) + 210*A*tan(e/2 + f*x/
2)**4/(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) + 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/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) + 86*A*tan(e/2 + f*x/2)/(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) - 26*A/(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) - 70*B*tan(e/2 + f*x/2)**6/(35*a*c**4*f*tan(e/2 + f*x/2)**8 - 2
10*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) +
 140*B*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 + 210*a*c**4*f*tan(e/2 + f*x/2) - 35*a*c**4*f) - 210*B*tan(e/2 + f*x/2)**4/(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) + 112*B*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 - 49
0*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) - 42*B*tan(e/2 + f*x/2)**2/(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*ta
n(e/2 + f*x/2) - 35*a*c**4*f) - 12*B*tan(e/2 + f*x/2)/(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) + 2*B/(35*a*c**4*f*ta
n(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), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e) + a)*(-c*sin(e) + c)**4), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (137) = 274\).

Time = 0.24 (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=-\frac {2 \, {\left (\frac {A {\left (\frac {43 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {77 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {7 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {175 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {35 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - 13\right )}}{a c^{4} - \frac {6 \, a c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {14 \, a c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {14 \, a c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {14 \, a c^{4} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {14 \, a c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {6 \, a c^{4} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a c^{4} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}} - \frac {B {\left (\frac {6 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {56 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {70 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 1\right )}}{a c^{4} - \frac {6 \, a c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {14 \, a c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {14 \, a c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {14 \, a c^{4} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {14 \, a c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {6 \, a c^{4} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a c^{4} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}}\right )}}{35 \, f} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

-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 - 13)/(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) - 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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.57 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=-\frac {\frac {35 \, {\left (A - B\right )}}{a c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {525 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 35 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1960 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 280 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 4025 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 665 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 4480 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 1120 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3143 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 791 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1176 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 392 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 243 \, A - 51 \, B}{a c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}}}{280 \, f} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

-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*t
an(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

Mupad [B] (verification not implemented)

Time = 13.14 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.68 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx=\frac {2\,\left (\frac {35\,B}{4}+\frac {91\,A\,\cos \left (e+f\,x\right )}{4}-\frac {7\,B\,\cos \left (e+f\,x\right )}{4}+14\,A\,\sin \left (e+f\,x\right )-\frac {21\,B\,\sin \left (e+f\,x\right )}{2}+14\,A\,\cos \left (2\,e+2\,f\,x\right )-\frac {39\,A\,\cos \left (3\,e+3\,f\,x\right )}{4}-A\,\cos \left (4\,e+4\,f\,x\right )-\frac {21\,B\,\cos \left (2\,e+2\,f\,x\right )}{2}+\frac {3\,B\,\cos \left (3\,e+3\,f\,x\right )}{4}+\frac {3\,B\,\cos \left (4\,e+4\,f\,x\right )}{4}-\frac {91\,A\,\sin \left (2\,e+2\,f\,x\right )}{4}-6\,A\,\sin \left (3\,e+3\,f\,x\right )+\frac {13\,A\,\sin \left (4\,e+4\,f\,x\right )}{8}+\frac {7\,B\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {9\,B\,\sin \left (3\,e+3\,f\,x\right )}{2}-\frac {B\,\sin \left (4\,e+4\,f\,x\right )}{8}\right )}{35\,a\,c^4\,f\,\left (\frac {7\,\cos \left (e+f\,x\right )}{2}-\frac {3\,\cos \left (3\,e+3\,f\,x\right )}{2}-\frac {7\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {\sin \left (4\,e+4\,f\,x\right )}{4}\right )} \]

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^4),x)

[Out]

(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))