\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx\) [76]

Optimal result

Integrand size = 36, antiderivative size = 90 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}+\frac {(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f} \]

[Out]

-1/5*(A-B)*sec(f*x+e)^3/c^2/f/(a^3+a^3*sin(f*x+e))+1/5*(4*A+B)*tan(f*x+e)/a^3/c^2/f+1/15*(4*A+B)*tan(f*x+e)^3/
a^3/c^2/f

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3046, 2938, 3852} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f}+\frac {(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}-\frac {(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3 \sin (e+f x)+a^3\right )} \]

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^2),x]

[Out]

-1/5*((A - B)*Sec[e + f*x]^3)/(c^2*f*(a^3 + a^3*Sin[e + f*x])) + ((4*A + B)*Tan[e + f*x])/(5*a^3*c^2*f) + ((4*
A + B)*Tan[e + f*x]^3)/(15*a^3*c^2*f)

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 ^4(e+f x) (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{a^2 c^2} \\ & = -\frac {(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {(4 A+B) \int \sec ^4(e+f x) \, dx}{5 a^3 c^2} \\ & = -\frac {(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(4 A+B) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 a^3 c^2 f} \\ & = -\frac {(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}+\frac {(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(237\) vs. \(2(90)=180\).

Time = 1.99 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.63 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (240 B+54 (A-B) \cos (e+f x)-32 (4 A+B) \cos (2 (e+f x))+18 A \cos (3 (e+f x))-18 B \cos (3 (e+f x))-64 A \cos (4 (e+f x))-16 B \cos (4 (e+f x))+384 A \sin (e+f x)+96 B \sin (e+f x)+18 A \sin (2 (e+f x))-18 B \sin (2 (e+f x))+128 A \sin (3 (e+f x))+32 B \sin (3 (e+f x))+9 A \sin (4 (e+f x))-9 B \sin (4 (e+f x)))}{960 a^3 c^2 f (-1+\sin (e+f x))^2 (1+\sin (e+f x))^3} \]

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^2),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(240*B + 54*(A - B)*Cos[e + f*x]
- 32*(4*A + B)*Cos[2*(e + f*x)] + 18*A*Cos[3*(e + f*x)] - 18*B*Cos[3*(e + f*x)] - 64*A*Cos[4*(e + f*x)] - 16*B
*Cos[4*(e + f*x)] + 384*A*Sin[e + f*x] + 96*B*Sin[e + f*x] + 18*A*Sin[2*(e + f*x)] - 18*B*Sin[2*(e + f*x)] + 1
28*A*Sin[3*(e + f*x)] + 32*B*Sin[3*(e + f*x)] + 9*A*Sin[4*(e + f*x)] - 9*B*Sin[4*(e + f*x)]))/(960*a^3*c^2*f*(
-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.51

method	result	size
risch	\(\frac {4 i \left (24 i A \,{\mathrm e}^{3 i \left (f x +e \right )}+6 i B \,{\mathrm e}^{3 i \left (f x +e \right )}+15 B \,{\mathrm e}^{4 i \left (f x +e \right )}+8 i A \,{\mathrm e}^{i \left (f x +e \right )}-8 A \,{\mathrm e}^{2 i \left (f x +e \right )}+2 i B \,{\mathrm e}^{i \left (f x +e \right )}-2 B \,{\mathrm e}^{2 i \left (f x +e \right )}-4 A -B \right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2} a^{3}}\)	\(136\)
parallelrisch	\(\frac {-30 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-30 A -30 B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (10 A -20 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (50 A -10 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-26 A +16 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-42 A -18 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-18 A -12 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+6 A -6 B}{15 f \,c^{2} a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\)	\(171\)
derivativedivides	\(\frac {-\frac {2 \left (\frac {B}{4}+\frac {A}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {\frac {B}{4}+\frac {A}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (A -B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 B -2 A}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}-2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{3} c^{2} f}\)	\(185\)
default	\(\frac {-\frac {2 \left (\frac {B}{4}+\frac {A}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {\frac {B}{4}+\frac {A}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (A -B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 B -2 A}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}-2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{3} c^{2} f}\)	\(185\)
norman	\(\frac {-\frac {6 A +4 B}{10 c f a}-\frac {4 \left (4 A +B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f a}+\frac {A \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}-\frac {\left (14 A +16 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{10 c f a}-\frac {\left (6 A +4 B \right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c f a}-\frac {\left (2 A +8 B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f a}-\frac {2 \left (8 A +2 B \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f a}-\frac {\left (16 A +4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 c f a}+\frac {2 \left (8 A +2 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f a}+\frac {\left (38 A -28 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(318\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

4/15*I*(24*I*A*exp(3*I*(f*x+e))+6*I*B*exp(3*I*(f*x+e))+15*B*exp(4*I*(f*x+e))+8*I*A*exp(I*(f*x+e))-8*A*exp(2*I*
(f*x+e))+2*I*B*exp(I*(f*x+e))-2*B*exp(2*I*(f*x+e))-4*A-B)/(exp(I*(f*x+e))+I)^5/(exp(I*(f*x+e))-I)^3/f/c^2/a^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (4 \, A + B\right )} \cos \left (f x + e\right )^{4} - {\left (4 \, A + B\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, {\left (4 \, A + B\right )} \cos \left (f x + e\right )^{2} + 4 \, A + B\right )} \sin \left (f x + e\right ) - A - 4 \, B}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/15*(2*(4*A + B)*cos(f*x + e)^4 - (4*A + B)*cos(f*x + e)^2 - (2*(4*A + B)*cos(f*x + e)^2 + 4*A + B)*sin(f*x
+ e) - A - 4*B)/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*f*cos(f*x + e)^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2674 vs. \(2 (82) = 164\).

Time = 9.00 (sec) , antiderivative size = 2674, normalized size of antiderivative = 29.71 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((-30*A*tan(e/2 + f*x/2)**7/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7
- 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3
+ 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 30*A*tan(e/2 + f*x
/2)**6/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x
/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*
x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) + 10*A*tan(e/2 + f*x/2)**5/(15*a**3*c**2*f*tan(e/2
+ f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/
2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e
/2 + f*x/2) - 15*a**3*c**2*f) + 50*A*tan(e/2 + f*x/2)**4/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*
tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f
*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f)
- 26*A*tan(e/2 + f*x/2)**3/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*
c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3
*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 42*A*tan(e/2 + f*x/2)**2/(15
*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 9
0*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 -
30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 18*A*tan(e/2 + f*x/2)/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8
+ 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5
+ 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) -
15*a**3*c**2*f) + 6*A/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2
*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**
2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 30*B*tan(e/2 + f*x/2)**6/(15*a**
3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a*
*3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a
**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 20*B*tan(e/2 + f*x/2)**5/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 +
30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5
+ 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) -
15*a**3*c**2*f) - 10*B*tan(e/2 + f*x/2)**4/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/
2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x
/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) + 16*B*tan(e/2
+ f*x/2)**3/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2
+ f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/
2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 18*B*tan(e/2 + f*x/2)**2/(15*a**3*c**2*f*t
an(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*
tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f
*tan(e/2 + f*x/2) - 15*a**3*c**2*f) - 12*B*tan(e/2 + f*x/2)/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2
*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**
2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*
f) - 6*B/(15*a**3*c**2*f*tan(e/2 + f*x/2)**8 + 30*a**3*c**2*f*tan(e/2 + f*x/2)**7 - 30*a**3*c**2*f*tan(e/2 + f
*x/2)**6 - 90*a**3*c**2*f*tan(e/2 + f*x/2)**5 + 90*a**3*c**2*f*tan(e/2 + f*x/2)**3 + 30*a**3*c**2*f*tan(e/2 +
f*x/2)**2 - 30*a**3*c**2*f*tan(e/2 + f*x/2) - 15*a**3*c**2*f), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e) + a)**3
*(-c*sin(e) + c)**2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 650 vs. \(2 (84) = 168\).

Time = 0.22 (sec) , antiderivative size = 650, normalized size of antiderivative = 7.22 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {A {\left (\frac {9 \, \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 {13 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {25 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {15 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - 3\right )}}{a^{3} c^{2} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{3} c^{2} \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 {9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {8 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {10 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + 3\right )}}{a^{3} c^{2} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {6 \, a^{3} c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {2 \, a^{3} c^{2} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{3} c^{2} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}}\right )}}{15 \, f} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

2/15*(A*(9*sin(f*x + e)/(cos(f*x + e) + 1) + 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 - 25*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3)/(a^3*c^2 + 2*a^3*c^2*sin(f*x + e)/(co
s(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 6*a^3*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^
3 + 6*a^3*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 2*a^3*c^2*
sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3*c^2*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) + B*(6*sin(f*x + e)/(cos(f*
x + e) + 1) + 9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 8*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 + 10*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 3)/(
a^3*c^2 + 2*a^3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 6*a^3*c^
2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6*a^3*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)
^6/(cos(f*x + e) + 1)^6 - 2*a^3*c^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3*c^2*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (84) = 168\).

Time = 0.47 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.46 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=-\frac {\frac {5 \, {\left (15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 24 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, A + 7 \, B\right )}}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}} + \frac {165 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 480 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 60 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 70 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 400 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 20 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113 \, A - 13 \, B}{a^{3} c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{120 \, f} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/120*(5*(15*A*tan(1/2*f*x + 1/2*e)^2 + 9*B*tan(1/2*f*x + 1/2*e)^2 - 24*A*tan(1/2*f*x + 1/2*e) - 12*B*tan(1/2
*f*x + 1/2*e) + 13*A + 7*B)/(a^3*c^2*(tan(1/2*f*x + 1/2*e) - 1)^3) + (165*A*tan(1/2*f*x + 1/2*e)^4 - 45*B*tan(
1/2*f*x + 1/2*e)^4 + 480*A*tan(1/2*f*x + 1/2*e)^3 - 60*B*tan(1/2*f*x + 1/2*e)^3 + 650*A*tan(1/2*f*x + 1/2*e)^2
- 70*B*tan(1/2*f*x + 1/2*e)^2 + 400*A*tan(1/2*f*x + 1/2*e) - 20*B*tan(1/2*f*x + 1/2*e) + 113*A - 13*B)/(a^3*c
^2*(tan(1/2*f*x + 1/2*e) + 1)^5))/f

Mupad [B] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.03 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx=\frac {\left (\frac {8\,A}{15}+\frac {2\,B}{15}+\frac {16\,A\,\sin \left (e+f\,x\right )}{15}+\frac {4\,B\,\sin \left (e+f\,x\right )}{15}\right )\,{\cos \left (e+f\,x\right )}^2+\frac {2\,A}{15}+\frac {8\,B}{15}+\frac {8\,A\,\sin \left (e+f\,x\right )}{15}+\frac {2\,B\,\sin \left (e+f\,x\right )}{15}}{a^3\,c^2\,f\,\left (2\,{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )+2\,{\cos \left (e+f\,x\right )}^3\right )}-\frac {\frac {2\,A}{5}-\frac {2\,B}{5}+\frac {2\,A\,\sin \left (e+f\,x\right )}{5}-\frac {2\,B\,\sin \left (e+f\,x\right )}{5}}{a^3\,c^2\,f\,\left (2\,\sin \left (e+f\,x\right )+2\right )}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {16\,A}{15}+\frac {4\,B}{15}\right )}{a^3\,c^2\,f\,\left (2\,\sin \left (e+f\,x\right )+2\right )} \]

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^2),x)

[Out]

((2*A)/15 + (8*B)/15 + (8*A*sin(e + f*x))/15 + (2*B*sin(e + f*x))/15 + cos(e + f*x)^2*((8*A)/15 + (2*B)/15 + (
16*A*sin(e + f*x))/15 + (4*B*sin(e + f*x))/15))/(a^3*c^2*f*(2*cos(e + f*x)^3*sin(e + f*x) + 2*cos(e + f*x)^3))
- ((2*A)/5 - (2*B)/5 + (2*A*sin(e + f*x))/5 - (2*B*sin(e + f*x))/5)/(a^3*c^2*f*(2*sin(e + f*x) + 2)) - (cos(e
+ f*x)*((16*A)/15 + (4*B)/15))/(a^3*c^2*f*(2*sin(e + f*x) + 2))