\(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx\) [254]

Optimal result

Integrand size = 26, antiderivative size = 92 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {5 a^3 x}{c^2}-\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]

[Out]

5*a^3*x/c^2-5*a^3*cos(f*x+e)/c^2/f+2/3*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^4-10/3*a^3*cos(f*x+e)^3/f/(c-c*
sin(f*x+e))^2

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2759, 2761, 8} \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=-\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac {5 a^3 x}{c^2}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]

[In]

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

[Out]

(5*a^3*x)/c^2 - (5*a^3*Cos[e + f*x])/(c^2*f) + (2*a^3*c^2*Cos[e + f*x]^5)/(3*f*(c - c*Sin[e + f*x])^4) - (10*a
^3*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^2)

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 2761

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*((g*Cos[e
+ f*x])^(p - 1)/(b*f*(p - 1))), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[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))^5} \, dx \\ & = \frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {1}{3} \left (5 a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^3} \, dx \\ & = \frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac {\left (5 a^3\right ) \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{c} \\ & = -\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac {\left (5 a^3\right ) \int 1 \, dx}{c^2} \\ & = \frac {5 a^3 x}{c^2}-\frac {5 a^3 \cos (e+f x)}{c^2 f}+\frac {2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac {10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.11 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.62 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (23+15 e+15 f x) \cos \left (\frac {1}{2} (e+f x)\right )-(121+30 e+30 f x) \cos \left (\frac {3}{2} (e+f x)\right )+3 \cos \left (\frac {5}{2} (e+f x)\right )-6 (31+20 e+20 f x+2 (-2+5 e+5 f x) \cos (e+f x)-\cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{12 c^2 f (-1+\sin (e+f x))^2} \]

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*(23 + 15*e + 15*f*x)*Cos[(e + f*x)/2] - (121 + 30*e + 30*f*x)*Co
s[(3*(e + f*x))/2] + 3*Cos[(5*(e + f*x))/2] - 6*(31 + 20*e + 20*f*x + 2*(-2 + 5*e + 5*f*x)*Cos[e + f*x] - Cos[
2*(e + f*x)])*Sin[(e + f*x)/2]))/(12*c^2*f*(-1 + Sin[e + f*x])^2)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {1}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+5 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\)	\(87\)
default	\(\frac {2 a^{3} \left (-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {1}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+5 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\)	\(87\)
risch	\(\frac {5 a^{3} x}{c^{2}}-\frac {a^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 c^{2} f}-\frac {a^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 c^{2} f}-\frac {8 \left (-12 i a^{3} {\mathrm e}^{i \left (f x +e \right )}+9 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-7 a^{3}\right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\)	\(108\)
parallelrisch	\(-\frac {a^{3} \left (-90 \cos \left (f x +e \right ) f x -30 f x \cos \left (3 f x +3 e \right )+56 \sin \left (3 f x +3 e \right )+156 \cos \left (2 f x +2 e \right )-72 \sin \left (f x +e \right )+138 \cos \left (f x +e \right )+3 \cos \left (4 f x +4 e \right )+46 \cos \left (3 f x +3 e \right )+25\right )}{6 f \,c^{2} \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\)	\(117\)
norman	\(\frac {\frac {8 a^{3} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {5 a^{3} x}{c}+\frac {46 a^{3}}{3 c f}+\frac {15 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {30 a^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {50 a^{3} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {60 a^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {60 a^{3} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {50 a^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {30 a^{3} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {15 a^{3} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {5 a^{3} x \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {110 a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {130 a^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {78 a^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {34 a^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {58 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {38 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}-\frac {106 a^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(406\)

[In]

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

[Out]

2/f*a^3/c^2*(-16/3/(tan(1/2*f*x+1/2*e)-1)^3-8/(tan(1/2*f*x+1/2*e)-1)^2+4/(tan(1/2*f*x+1/2*e)-1)-1/(1+tan(1/2*f
*x+1/2*e)^2)+5*arctan(tan(1/2*f*x+1/2*e)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (90) = 180\).

Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.00 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=-\frac {3 \, a^{3} \cos \left (f x + e\right )^{3} + 30 \, a^{3} f x + 8 \, a^{3} - {\left (15 \, a^{3} f x + 31 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (15 \, a^{3} f x - 26 \, a^{3}\right )} \cos \left (f x + e\right ) - {\left (30 \, a^{3} f x - 3 \, a^{3} \cos \left (f x + e\right )^{2} - 8 \, a^{3} + {\left (15 \, a^{3} f x - 34 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

-1/3*(3*a^3*cos(f*x + e)^3 + 30*a^3*f*x + 8*a^3 - (15*a^3*f*x + 31*a^3)*cos(f*x + e)^2 + (15*a^3*f*x - 26*a^3)
*cos(f*x + e) - (30*a^3*f*x - 3*a^3*cos(f*x + e)^2 - 8*a^3 + (15*a^3*f*x - 34*a^3)*cos(f*x + e))*sin(f*x + e))
/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*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. \(2 (87) = 174\).

Time = 3.80 (sec) , antiderivative size = 1282, normalized size of antiderivative = 13.93 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((15*a**3*f*x*tan(e/2 + f*x/2)**5/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c
**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 45*a**3*f*
x*tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2
)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 60*a**3*f*x*tan(e/2 + f*x/2)**3
/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(
e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 60*a**3*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 +
f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c
**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 45*a**3*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan
(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) -
3*c**2*f) - 15*a**3*f*x/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*
x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 24*a**3*tan(e/2 + f*x/2)**4/
(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e
/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 102*a**3*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/
2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*
f*tan(e/2 + f*x/2) - 3*c**2*f) + 82*a**3*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2
+ f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c*
*2*f) - 114*a**3*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan
(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 46*a**3/(3*c**2*f*t
an(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)
**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f), Ne(f, 0)), (x*(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. 594 vs. \(2 (90) = 180\).

Time = 0.30 (sec) , antiderivative size = 594, normalized size of antiderivative = 6.46 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {2 \, {\left (2 \, a^{3} {\left (\frac {\frac {12 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {11 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {9 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 5}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {4 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} + 3 \, a^{3} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 4}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} - \frac {a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

[In]

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

[Out]

2/3*(2*a^3*((12*sin(f*x + e)/(cos(f*x + e) + 1) - 11*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 - 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) +
4*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^2*sin(f*x + e)^4/
(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^
2) + 3*a^3*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f
*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)
^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^
2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3))/f

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{2}} - \frac {6 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} c^{2}} + \frac {8 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, a^{3}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \]

[In]

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

[Out]

1/3*(15*(f*x + e)*a^3/c^2 - 6*a^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*c^2) + 8*(3*a^3*tan(1/2*f*x + 1/2*e)^2 - 12*a^
3*tan(1/2*f*x + 1/2*e) + 5*a^3)/(c^2*(tan(1/2*f*x + 1/2*e) - 1)^3))/f

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.37 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx=\frac {5\,a^3\,x}{c^2}+\frac {5\,a^3\,\left (e+f\,x\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (45\,e+45\,f\,x-114\right )}{3}\right )-\frac {a^3\,\left (15\,e+15\,f\,x-46\right )}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (15\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (45\,e+45\,f\,x-24\right )}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (20\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (60\,e+60\,f\,x-82\right )}{3}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (20\,a^3\,\left (e+f\,x\right )-\frac {a^3\,\left (60\,e+60\,f\,x-102\right )}{3}\right )}{c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(5*a^3*x)/c^2 + (5*a^3*(e + f*x) - tan(e/2 + (f*x)/2)*(15*a^3*(e + f*x) - (a^3*(45*e + 45*f*x - 114))/3) - (a^
3*(15*e + 15*f*x - 46))/3 + tan(e/2 + (f*x)/2)^4*(15*a^3*(e + f*x) - (a^3*(45*e + 45*f*x - 24))/3) + tan(e/2 +
(f*x)/2)^2*(20*a^3*(e + f*x) - (a^3*(60*e + 60*f*x - 82))/3) - tan(e/2 + (f*x)/2)^3*(20*a^3*(e + f*x) - (a^3*
(60*e + 60*f*x - 102))/3))/(c^2*f*(tan(e/2 + (f*x)/2) - 1)^3*(tan(e/2 + (f*x)/2)^2 + 1))