\(\int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx\) [262]

Optimal result

Integrand size = 26, antiderivative size = 92 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {15 c^3 x}{2 a}-\frac {15 c^3 \cos (e+f x)}{2 a f}-\frac {2 a^2 c^3 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^3}-\frac {5 c^3 \cos ^3(e+f x)}{2 f (a+a \sin (e+f x))} \]

[Out]

-15/2*c^3*x/a-15/2*c^3*cos(f*x+e)/a/f-2*a^2*c^3*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^3-5/2*c^3*cos(f*x+e)^3/f/(a+a*
sin(f*x+e))

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 = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2815, 2759, 2758, 2761, 8} \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {2 a^2 c^3 \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^3}-\frac {15 c^3 \cos (e+f x)}{2 a f}-\frac {5 c^3 \cos ^3(e+f x)}{2 f (a \sin (e+f x)+a)}-\frac {15 c^3 x}{2 a} \]

[In]

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

[Out]

(-15*c^3*x)/(2*a) - (15*c^3*Cos[e + f*x])/(2*a*f) - (2*a^2*c^3*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*x])^3) - (5
*c^3*Cos[e + f*x]^3)/(2*f*(a + a*Sin[e + f*x]))

Rule 8

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

Rule 2758

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

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

Mathematica [A] (verified)

Time = 11.77 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.68 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^3 \left (\sin \left (\frac {1}{2} (e+f x)\right ) (-64+30 e+30 f x+16 \cos (e+f x)-\sin (2 (e+f x)))+\cos \left (\frac {1}{2} (e+f x)\right ) (30 (e+f x)+16 \cos (e+f x)-\sin (2 (e+f x)))\right )}{4 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (1+\sin (e+f x))} \]

[In]

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

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^3*(Sin[(e + f*x)/2]*(-64 + 30*e + 30*f*x + 16*C
os[e + f*x] - Sin[2*(e + f*x)]) + Cos[(e + f*x)/2]*(30*(e + f*x) + 16*Cos[e + f*x] - Sin[2*(e + f*x)])))/(4*a*
f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(1 + Sin[e + f*x]))

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(-\frac {c^{3} \left (60 \cos \left (f x +e \right ) f x +16 \cos \left (2 f x +2 e \right )-\sin \left (3 f x +3 e \right )-65 \sin \left (f x +e \right )+96 \cos \left (f x +e \right )+80\right )}{8 a f \cos \left (f x +e \right )}\)	\(70\)
derivativedivides	\(\frac {2 c^{3} \left (-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+4 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+4}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f a}\)	\(96\)
default	\(\frac {2 c^{3} \left (-\frac {8}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+4 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+4}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {15 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f a}\)	\(96\)
risch	\(-\frac {15 c^{3} x}{2 a}-\frac {2 c^{3} {\mathrm e}^{i \left (f x +e \right )}}{f a}-\frac {2 c^{3} {\mathrm e}^{-i \left (f x +e \right )}}{f a}-\frac {16 c^{3}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {c^{3} \sin \left (2 f x +2 e \right )}{4 f a}\)	\(96\)
norman	\(\frac {-\frac {7 c^{3}}{f a}+\frac {10 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f a}+\frac {17 c^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {5 c^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {35 c^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {15 c^{3} x}{2 a}-\frac {15 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a}-\frac {45 c^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {45 c^{3} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {45 c^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {45 c^{3} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {15 c^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {15 c^{3} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 a}-\frac {12 c^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {42 c^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(319\)

[In]

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

[Out]

-1/8*c^3/a/f*(60*cos(f*x+e)*f*x+16*cos(2*f*x+2*e)-sin(3*f*x+3*e)-65*sin(f*x+e)+96*cos(f*x+e)+80)/cos(f*x+e)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.39 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {c^{3} \cos \left (f x + e\right )^{3} + 15 \, c^{3} f x + 8 \, c^{3} \cos \left (f x + e\right )^{2} + 16 \, c^{3} + {\left (15 \, c^{3} f x + 23 \, c^{3}\right )} \cos \left (f x + e\right ) + {\left (15 \, c^{3} f x - c^{3} \cos \left (f x + e\right )^{2} + 7 \, c^{3} \cos \left (f x + e\right ) - 16 \, c^{3}\right )} \sin \left (f x + e\right )}{2 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \]

[In]

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

[Out]

-1/2*(c^3*cos(f*x + e)^3 + 15*c^3*f*x + 8*c^3*cos(f*x + e)^2 + 16*c^3 + (15*c^3*f*x + 23*c^3)*cos(f*x + e) + (
15*c^3*f*x - c^3*cos(f*x + e)^2 + 7*c^3*cos(f*x + e) - 16*c^3)*sin(f*x + e))/(a*f*cos(f*x + e) + a*f*sin(f*x +
e) + a*f)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1170 vs. \(2 (85) = 170\).

Time = 1.93 (sec) , antiderivative size = 1170, normalized size of antiderivative = 12.72 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-15*c**3*f*x*tan(e/2 + f*x/2)**5/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan
(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 15*c**3*f*x*tan(e/2 + f*x/2)*
*4/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)
**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 30*c**3*f*x*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan
(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 3
0*c**3*f*x*tan(e/2 + f*x/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)
**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 15*c**3*f*x*tan(e/2 + f*x/2)/(2*a*f*tan(e/
2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(
e/2 + f*x/2) + 2*a*f) - 15*c**3*f*x/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f
*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 34*c**3*tan(e/2 + f*x/2)**4/(2*a*f*ta
n(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*
tan(e/2 + f*x/2) + 2*a*f) - 18*c**3*tan(e/2 + f*x/2)**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4
+ 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 78*c**3*tan(e/2 +
f*x/2)**2/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2
+ f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) - 14*c**3*tan(e/2 + f*x/2)/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*ta
n(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f) -
48*c**3/(2*a*f*tan(e/2 + f*x/2)**5 + 2*a*f*tan(e/2 + f*x/2)**4 + 4*a*f*tan(e/2 + f*x/2)**3 + 4*a*f*tan(e/2 + f
*x/2)**2 + 2*a*f*tan(e/2 + f*x/2) + 2*a*f), Ne(f, 0)), (x*(-c*sin(e) + c)**3/(a*sin(e) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (86) = 172\).

Time = 0.28 (sec) , antiderivative size = 424, normalized size of antiderivative = 4.61 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {c^{3} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 4}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} + 6 \, c^{3} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} + 6 \, c^{3} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac {2 \, c^{3}}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{f} \]

[In]

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

[Out]

-(c^3*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 +
a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 6*c^3*((sin(f*x + e)/(
cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1
))/a) + 6*c^3*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) + 2*c^3/
(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.21 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )} c^{3}}{a} + \frac {32 \, c^{3}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 8 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 \, c^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} a}}{2 \, f} \]

[In]

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

[Out]

-1/2*(15*(f*x + e)*c^3/a + 32*c^3/(a*(tan(1/2*f*x + 1/2*e) + 1)) + 2*(c^3*tan(1/2*f*x + 1/2*e)^3 + 8*c^3*tan(1
/2*f*x + 1/2*e)^2 - c^3*tan(1/2*f*x + 1/2*e) + 8*c^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*a))/f

Mupad [B] (verification not implemented)

Time = 8.96 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.35 \[ \int \frac {(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx=\frac {\frac {15\,c^3\,\left (e+f\,x\right )}{2}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {15\,c^3\,\left (e+f\,x\right )}{2}-\frac {c^3\,\left (15\,e+15\,f\,x+14\right )}{2}\right )-\frac {c^3\,\left (15\,e+15\,f\,x+48\right )}{2}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {15\,c^3\,\left (e+f\,x\right )}{2}-\frac {c^3\,\left (15\,e+15\,f\,x+34\right )}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (15\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (30\,e+30\,f\,x+18\right )}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (15\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (30\,e+30\,f\,x+78\right )}{2}\right )}{a\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^2}-\frac {15\,c^3\,x}{2\,a} \]

[In]

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

[Out]

((15*c^3*(e + f*x))/2 + tan(e/2 + (f*x)/2)*((15*c^3*(e + f*x))/2 - (c^3*(15*e + 15*f*x + 14))/2) - (c^3*(15*e
+ 15*f*x + 48))/2 + tan(e/2 + (f*x)/2)^4*((15*c^3*(e + f*x))/2 - (c^3*(15*e + 15*f*x + 34))/2) + tan(e/2 + (f*
x)/2)^3*(15*c^3*(e + f*x) - (c^3*(30*e + 30*f*x + 18))/2) + tan(e/2 + (f*x)/2)^2*(15*c^3*(e + f*x) - (c^3*(30*
e + 30*f*x + 78))/2))/(a*f*(tan(e/2 + (f*x)/2) + 1)*(tan(e/2 + (f*x)/2)^2 + 1)^2) - (15*c^3*x)/(2*a)