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

Optimal. Leaf size=92 \[ -\frac{2 a^2 c^3 \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^3}-\frac{15 c^3 \cos (e+f x)}{2 a f}-\frac{5 c^3 \cos ^3(e+f x)}{2 f (a \sin (e+f x)+a)}-\frac{15 c^3 x}{2 a} \]

[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]))

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Rubi [A]  time = 0.17626, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2736, 2680, 2679, 2682, 8} \[ -\frac{2 a^2 c^3 \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^3}-\frac{15 c^3 \cos (e+f x)}{2 a f}-\frac{5 c^3 \cos ^3(e+f x)}{2 f (a \sin (e+f x)+a)}-\frac{15 c^3 x}{2 a} \]

Antiderivative was successfully verified.

[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 2736

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]))

Rule 2680

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[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 2682

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 8

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

Rubi steps

\begin{align*} \int \frac{(c-c \sin (e+f x))^3}{a+a \sin (e+f x)} \, dx &=\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]  time = 0.507868, size = 155, normalized size = 1.68 \[ \frac{c^3 (\sin (e+f x)-1)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right ) (-\sin (2 (e+f x))+16 \cos (e+f x)+30 e+30 f x-64)+\cos \left (\frac{1}{2} (e+f x)\right ) (30 (e+f x)-\sin (2 (e+f x))+16 \cos (e+f x))\right )}{4 a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[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]))

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Maple [B]  time = 0.078, size = 181, normalized size = 2. \begin{align*} -{\frac{{c}^{3}}{af} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-8\,{\frac{{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{c}^{3}}{af}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-8\,{\frac{{c}^{3}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-15\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{af}}-16\,{\frac{{c}^{3}}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*c^3/a/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3-8/f*c^3/a/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/
2*e)^2+1/f*c^3/a/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)-8/f*c^3/a/(1+tan(1/2*f*x+1/2*e)^2)^2-15/f*c^3/a
*arctan(tan(1/2*f*x+1/2*e))-16/f*c^3/a/(tan(1/2*f*x+1/2*e)+1)

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Maxima [B]  time = 2.02787, size = 572, normalized size = 6.22 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.39535, size = 313, normalized size = 3.4 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 16.8178, size = 1170, normalized size = 12.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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) + 14*c**3*tan(e/2 + f*x/2)**5/(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) - 20*c**3*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) + 10*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) - 50*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)
- 34*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))

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Giac [A]  time = 2.07312, size = 158, normalized size = 1.72 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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