\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx\) [77]

Optimal result

Integrand size = 36, antiderivative size = 84 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=\frac {B \sec ^5(e+f x)}{5 a^3 c^3 f}+\frac {A \tan (e+f x)}{a^3 c^3 f}+\frac {2 A \tan ^3(e+f x)}{3 a^3 c^3 f}+\frac {A \tan ^5(e+f x)}{5 a^3 c^3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 84, 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, 2748, 3852} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=\frac {A \tan ^5(e+f x)}{5 a^3 c^3 f}+\frac {2 A \tan ^3(e+f x)}{3 a^3 c^3 f}+\frac {A \tan (e+f x)}{a^3 c^3 f}+\frac {B \sec ^5(e+f x)}{5 a^3 c^3 f} \]

[In]

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

[Out]

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

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 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 \sec ^6(e+f x) (A+B \sin (e+f x)) \, dx}{a^3 c^3} \\ & = \frac {B \sec ^5(e+f x)}{5 a^3 c^3 f}+\frac {A \int \sec ^6(e+f x) \, dx}{a^3 c^3} \\ & = \frac {B \sec ^5(e+f x)}{5 a^3 c^3 f}-\frac {A \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{a^3 c^3 f} \\ & = \frac {B \sec ^5(e+f x)}{5 a^3 c^3 f}+\frac {A \tan (e+f x)}{a^3 c^3 f}+\frac {2 A \tan ^3(e+f x)}{3 a^3 c^3 f}+\frac {A \tan ^5(e+f x)}{5 a^3 c^3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=\frac {B \sec ^5(e+f x)}{5 a^3 c^3 f}+\frac {A \left (\tan (e+f x)+\frac {2}{3} \tan ^3(e+f x)+\frac {1}{5} \tan ^5(e+f x)\right )}{a^3 c^3 f} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99

method	result	size
risch	\(\frac {\frac {16 i A \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}+\frac {32 B \,{\mathrm e}^{5 i \left (f x +e \right )}}{5}+\frac {32 i A \,{\mathrm e}^{4 i \left (f x +e \right )}}{3}+\frac {16 i A}{15}}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3} a^{3}}\)	\(83\)
parallelrisch	\(\frac {-2 A \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-2 B \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {8 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-\frac {116 A \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15}-4 B \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {8 A \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {2 B}{5}}{f \,c^{3} a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\)	\(138\)
derivativedivides	\(\frac {-\frac {A +B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {\frac {7 A}{8}+\frac {5 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {A}{2}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (\frac {11 A}{8}+\frac {9 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {-A +B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-\frac {7 A}{8}+\frac {5 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{2}-\frac {B}{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (\frac {A}{2}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {11 A}{8}-\frac {9 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} c^{3} f}\)	\(227\)
default	\(\frac {-\frac {A +B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {\frac {7 A}{8}+\frac {5 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {A}{2}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (\frac {11 A}{8}+\frac {9 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {-A +B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-\frac {7 A}{8}+\frac {5 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{2}-\frac {B}{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (\frac {A}{2}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {11 A}{8}-\frac {9 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} c^{3} f}\)	\(227\)
norman	\(\frac {-\frac {2 B}{5 a c f}-\frac {2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a c f}+\frac {2 A \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {76 A \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a c f}+\frac {2 A \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a c f}-\frac {2 A \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f c}-\frac {76 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f a}-\frac {4 B \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {2 B \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f a}-\frac {2 B \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f a}-\frac {2 B \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f a}-\frac {4 B \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(313\)

[In]

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

[Out]

16/15*(5*I*A*exp(2*I*(f*x+e))+6*B*exp(5*I*(f*x+e))+10*I*A*exp(4*I*(f*x+e))+I*A)/(exp(I*(f*x+e))+I)^5/(exp(I*(f
*x+e))-I)^5/f/c^3/a^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=\frac {{\left (8 \, A \cos \left (f x + e\right )^{4} + 4 \, A \cos \left (f x + e\right )^{2} + 3 \, A\right )} \sin \left (f x + e\right ) + 3 \, B}{15 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \]

[In]

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

[Out]

1/15*((8*A*cos(f*x + e)^4 + 4*A*cos(f*x + e)^2 + 3*A)*sin(f*x + e) + 3*B)/(a^3*c^3*f*cos(f*x + e)^5)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1098 vs. \(2 (78) = 156\).

Time = 6.10 (sec) , antiderivative size = 1098, normalized size of antiderivative = 13.07 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-30*A*tan(e/2 + f*x/2)**9/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f*x/2)**8
+ 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)
**2 - 15*a**3*c**3*f) + 40*A*tan(e/2 + f*x/2)**7/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2
+ f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(
e/2 + f*x/2)**2 - 15*a**3*c**3*f) - 116*A*tan(e/2 + f*x/2)**5/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c
**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**
3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) + 40*A*tan(e/2 + f*x/2)**3/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10
- 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)
**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) - 30*A*tan(e/2 + f*x/2)/(15*a**3*c**3*f*tan(e/2 + f
*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2
+ f*x/2)**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) - 30*B*tan(e/2 + f*x/2)**8/(15*a**3*c**3*f
*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c*
*3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) - 60*B*tan(e/2 + f*x/2)**4/(15
*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 -
150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) - 6*B/(15*a**3*c**
3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3
*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f), Ne(f, 0)), (x*(A + B*sin(e
))/((a*sin(e) + a)**3*(-c*sin(e) + c)**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=\frac {\frac {{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} A}{a^{3} c^{3}} + \frac {3 \, B}{a^{3} c^{3} \cos \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

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

[Out]

1/15*((3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*A/(a^3*c^3) + 3*B/(a^3*c^3*cos(f*x + e)^5))/f

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 20 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 58 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 30 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 20 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} a^{3} c^{3} f} \]

[In]

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

[Out]

-2/15*(15*A*tan(1/2*f*x + 1/2*e)^9 + 15*B*tan(1/2*f*x + 1/2*e)^8 - 20*A*tan(1/2*f*x + 1/2*e)^7 + 58*A*tan(1/2*
f*x + 1/2*e)^5 + 30*B*tan(1/2*f*x + 1/2*e)^4 - 20*A*tan(1/2*f*x + 1/2*e)^3 + 15*A*tan(1/2*f*x + 1/2*e) + 3*B)/
((tan(1/2*f*x + 1/2*e)^2 - 1)^5*a^3*c^3*f)

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3} \, dx=-\frac {2\,\left (15\,A\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9+15\,B\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-20\,A\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+58\,A\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+30\,B\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-20\,A\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,A\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+3\,B\right )}{15\,a^3\,c^3\,f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^5} \]

[In]

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

[Out]

-(2*(3*B + 15*A*tan(e/2 + (f*x)/2) - 20*A*tan(e/2 + (f*x)/2)^3 + 58*A*tan(e/2 + (f*x)/2)^5 - 20*A*tan(e/2 + (f
*x)/2)^7 + 15*A*tan(e/2 + (f*x)/2)^9 + 30*B*tan(e/2 + (f*x)/2)^4 + 15*B*tan(e/2 + (f*x)/2)^8))/(15*a^3*c^3*f*(
tan(e/2 + (f*x)/2)^2 - 1)^5)