3.272 \(\int \frac{(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=70 \[ \frac{2 c^2 \cos (e+f x)}{f \left (a^2 \sin (e+f x)+a^2\right )}+\frac{c^2 x}{a^2}-\frac{2 a c^2 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.131786, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2680, 8} \[ \frac{2 c^2 \cos (e+f x)}{f \left (a^2 \sin (e+f x)+a^2\right )}+\frac{c^2 x}{a^2}-\frac{2 a c^2 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x)/a^2 - (2*a*c^2*Cos[e + f*x]^3)/(3*f*(a + a*Sin[e + f*x])^3) + (2*c^2*Cos[e + f*x])/(f*(a^2 + a^2*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 8

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

Rubi steps

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

Mathematica [A]  time = 0.585637, size = 119, normalized size = 1.7 \[ \frac{c^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 (3 e+3 f x-8) \cos \left (\frac{1}{2} (e+f x)\right )+(-3 e-3 f x+16) \cos \left (\frac{3}{2} (e+f x)\right )+6 \sin \left (\frac{1}{2} (e+f x)\right ) (2 (e+f x-2)+(e+f x) \cos (e+f x))\right )}{6 a^2 f (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3*(-8 + 3*e + 3*f*x)*Cos[(e + f*x)/2] + (16 - 3*e - 3*f*x)*Cos[(3*
(e + f*x))/2] + 6*(2*(-2 + e + f*x) + (e + f*x)*Cos[e + f*x])*Sin[(e + f*x)/2]))/(6*a^2*f*(1 + Sin[e + f*x])^2
)

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Maple [A]  time = 0.078, size = 71, normalized size = 1. \begin{align*} 2\,{\frac{{c}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{a}^{2}f}}-{\frac{16\,{c}^{2}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+8\,{\frac{{c}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.15464, size = 487, normalized size = 6.96 \begin{align*} \frac{2 \,{\left (c^{2}{\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}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{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 )}{a^{2}}\right )} - \frac{c^{2}{\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 )}}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{2 \, c^{2}{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.39614, size = 362, normalized size = 5.17 \begin{align*} -\frac{6 \, c^{2} f x -{\left (3 \, c^{2} f x - 8 \, c^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, c^{2} +{\left (3 \, c^{2} f x + 4 \, c^{2}\right )} \cos \left (f x + e\right ) +{\left (6 \, c^{2} f x + 4 \, c^{2} +{\left (3 \, c^{2} f x + 8 \, c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*c^2*f*x - (3*c^2*f*x - 8*c^2)*cos(f*x + e)^2 - 4*c^2 + (3*c^2*f*x + 4*c^2)*cos(f*x + e) + (6*c^2*f*x +
4*c^2 + (3*c^2*f*x + 8*c^2)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f
- (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))

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Sympy [A]  time = 12.4763, size = 486, normalized size = 6.94 \begin{align*} \begin{cases} \frac{3 c^{2} f x \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} + \frac{9 c^{2} f x \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} + \frac{9 c^{2} f x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} + \frac{3 c^{2} f x}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} - \frac{8 c^{2} \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} - \frac{24 c^{2} \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} & \text{for}\: f \neq 0 \\\frac{x \left (- c \sin{\left (e \right )} + c\right )^{2}}{\left (a \sin{\left (e \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*c**2*f*x*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**
2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 9*c**2*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan
(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 9*c**2*f*x*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x
/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 3*c**2*f*x/(3*a**2*f*tan(e/2 +
f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 8*c**2*tan(e/2 + f*x/2)**3
/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 24*c**
2*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2)
+ 3*a**2*f), Ne(f, 0)), (x*(-c*sin(e) + c)**2/(a*sin(e) + a)**2, True))

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Giac [A]  time = 2.13887, size = 78, normalized size = 1.11 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} c^{2}}{a^{2}} + \frac{8 \,{\left (3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c^{2}\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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