\(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx\) [255]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=-\frac {a^3 x}{c^3}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \]

[Out]

-a^3*x/c^3+2/5*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^5-2/3*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^3+2*a^3*cos(f
*x+e)/f/(c^3-c^3*sin(f*x+e))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2815, 2759, 8} \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\frac {2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {a^3 x}{c^3}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[In]

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

[Out]

-((a^3*x)/c^3) + (2*a^3*c^2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^5) - (2*a^3*Cos[e + f*x]^3)/(3*f*(c - c*
Sin[e + f*x])^3) + (2*a^3*Cos[e + f*x])/(f*(c^3 - c^3*Sin[e + f*x]))

Rule 8

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

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, 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]))

Rubi steps \begin{align*} \text {integral}& = \left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^6} \, dx \\ & = \frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\left (a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^4} \, dx \\ & = \frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {a^3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{c} \\ & = \frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {a^3 \int 1 \, dx}{c^3} \\ & = -\frac {a^3 x}{c^3}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(106)=212\).

Time = 6.53 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.35 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-44 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-15 (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+48 \sin \left (\frac {1}{2} (e+f x)\right )-88 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+92 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (a+a \sin (e+f x))^3}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^3} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(24*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 44*(Cos[(e + f*x)/2] - Sin[
(e + f*x)/2])^3 - 15*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 48*Sin[(e + f*x)/2] - 88*(Cos[(e + f*
x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 92*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(a
+ a*Sin[e + f*x])^3)/(15*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.95 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {a^{3} x}{c^{3}}+\frac {-24 i a^{3} {\mathrm e}^{3 i \left (f x +e \right )}+12 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+\frac {56 i a^{3} {\mathrm e}^{i \left (f x +e \right )}}{3}-\frac {112 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{3}+\frac {92 a^{3}}{15}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\)	\(97\)
derivativedivides	\(\frac {2 a^{3} \left (-\frac {32}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {40}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\)	\(100\)
default	\(\frac {2 a^{3} \left (-\frac {32}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {40}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{3}}\)	\(100\)
parallelrisch	\(-\frac {\left (\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) f x +\left (-5 f x +4\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (10 f x -8\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-10 f x +\frac {80}{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (5 f x -\frac {40}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-f x +\frac {52}{15}\right ) a^{3}}{f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(115\)
norman	\(\frac {\frac {a^{3} x}{c}+\frac {8 a^{3} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {48 a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {52 a^{3}}{15 c f}-\frac {5 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}+\frac {13 a^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {25 a^{3} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {38 a^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {46 a^{3} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {46 a^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {38 a^{3} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {25 a^{3} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {13 a^{3} x \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {5 a^{3} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {a^{3} x \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {4 a^{3} \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {40 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}+\frac {64 a^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {112 a^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {116 a^{3} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {472 a^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {556 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}-\frac {1432 a^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(489\)

[In]

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

[Out]

-a^3*x/c^3+4/15*(-90*I*a^3*exp(3*I*(f*x+e))+45*a^3*exp(4*I*(f*x+e))+70*I*a^3*exp(I*(f*x+e))-140*a^3*exp(2*I*(f
*x+e))+23*a^3)/(exp(I*(f*x+e))-I)^5/f/c^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (105) = 210\).

Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.22 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\frac {60 \, a^{3} f x - {\left (15 \, a^{3} f x - 46 \, a^{3}\right )} \cos \left (f x + e\right )^{3} - 24 \, a^{3} - {\left (45 \, a^{3} f x + 2 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, a^{3} f x - 12 \, a^{3}\right )} \cos \left (f x + e\right ) - {\left (60 \, a^{3} f x + 24 \, a^{3} - {\left (15 \, a^{3} f x + 46 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, a^{3} f x - 8 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

1/15*(60*a^3*f*x - (15*a^3*f*x - 46*a^3)*cos(f*x + e)^3 - 24*a^3 - (45*a^3*f*x + 2*a^3)*cos(f*x + e)^2 + 6*(5*
a^3*f*x - 12*a^3)*cos(f*x + e) - (60*a^3*f*x + 24*a^3 - (15*a^3*f*x + 46*a^3)*cos(f*x + e)^2 + 6*(5*a^3*f*x -
8*a^3)*cos(f*x + e))*sin(f*x + e))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c
^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1282 vs. \(2 (95) = 190\).

Time = 7.60 (sec) , antiderivative size = 1282, normalized size of antiderivative = 12.09 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-15*a**3*f*x*tan(e/2 + f*x/2)**5/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 1
50*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 75*
a**3*f*x*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e
/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 150*a**3*f*x*tan(e
/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3
- 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 150*a**3*f*x*tan(e/2 + f*x/2)**2
/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*
tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 75*a**3*f*x*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/
2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**
2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 15*a**3*f*x/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 +
f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15
*c**3*f) - 60*a**3*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c*
*3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 120*a**3
*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x
/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 400*a**3*tan(e/2 + f*x/2)*
*2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*
f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 200*a**3*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2
+ f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2
+ 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 52*a**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/
2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3
*f), Ne(f, 0)), (x*(a*sin(e) + a)**3/(-c*sin(e) + c)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (105) = 210\).

Time = 0.30 (sec) , antiderivative size = 785, normalized size of antiderivative = 7.41 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-2/15*(a^3*((95*sin(f*x + e)/(cos(f*x + e) + 1) - 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 22)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) +
1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) +
1))/c^3) + a^3*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)
^3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e)
+ 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 9*a^3*(5*sin(f*x + e)/(cos(f*x + e)
+ 1) - 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3*sin(f*x
+ e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) +
1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 6*a^3*(5*sin(f*x
+ e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e)
+ 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*
x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{3}} + \frac {4 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 30 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 100 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 50 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, a^{3}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{15 \, f} \]

[In]

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

[Out]

-1/15*(15*(f*x + e)*a^3/c^3 + 4*(15*a^3*tan(1/2*f*x + 1/2*e)^4 - 30*a^3*tan(1/2*f*x + 1/2*e)^3 + 100*a^3*tan(1
/2*f*x + 1/2*e)^2 - 50*a^3*tan(1/2*f*x + 1/2*e) + 13*a^3)/(c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f

Mupad [B] (verification not implemented)

Time = 8.26 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.92 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^3} \, dx=-\frac {a^3\,x}{c^3}-\frac {a^3\,\left (e+f\,x\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (75\,e+75\,f\,x-200\right )}{15}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-52\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (5\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (75\,e+75\,f\,x-60\right )}{15}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (10\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (150\,e+150\,f\,x-120\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (10\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (150\,e+150\,f\,x-400\right )}{15}\right )}{c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5} \]

[In]

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

[Out]

- (a^3*x)/c^3 - (a^3*(e + f*x) - tan(e/2 + (f*x)/2)*(5*a^3*(e + f*x) - (a^3*(75*e + 75*f*x - 200))/15) - (a^3*
(15*e + 15*f*x - 52))/15 + tan(e/2 + (f*x)/2)^4*(5*a^3*(e + f*x) - (a^3*(75*e + 75*f*x - 60))/15) - tan(e/2 +
(f*x)/2)^3*(10*a^3*(e + f*x) - (a^3*(150*e + 150*f*x - 120))/15) + tan(e/2 + (f*x)/2)^2*(10*a^3*(e + f*x) - (a
^3*(150*e + 150*f*x - 400))/15))/(c^3*f*(tan(e/2 + (f*x)/2) - 1)^5)