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

Optimal. Leaf size=33 \[ -\frac{a^2 c^2 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5} \]

[Out]

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

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Rubi [A]  time = 0.0858378, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2736, 2671} \[ -\frac{a^2 c^2 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2671

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

Rubi steps

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

Mathematica [B]  time = 0.388886, size = 81, normalized size = 2.45 \[ \frac{c^2 \left (10 \sin \left (\frac{1}{2} (e+f x)\right )+5 \sin \left (\frac{3}{2} (e+f x)\right )-\sin \left (\frac{5}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{10 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(10*Sin[(e + f*x)/2] + 5*Sin[(3*(e + f*x))/2] - Sin[(5*(e + f*x))/2
]))/(10*a^3*f*(1 + Sin[e + f*x])^3)

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Maple [B]  time = 0.083, size = 88, normalized size = 2.7 \begin{align*} 2\,{\frac{{c}^{2}}{f{a}^{3}} \left ( -8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}-{\frac{16}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{5}}}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-1}+8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}+4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2} \right ) } \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))^3,x)

[Out]

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

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Maxima [B]  time = 1.19826, size = 748, normalized size = 22.67 \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))^2/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(c^2*(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/(c
os(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*c^2*(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)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B]  time = 1.28623, size = 400, normalized size = 12.12 \begin{align*} -\frac{c^{2} \cos \left (f x + e\right )^{3} + 3 \, c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2} -{\left (c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2}\right )} \sin \left (f x + e\right )}{5 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 49.998, size = 362, normalized size = 10.97 \begin{align*} \begin{cases} \frac{2 c^{2} \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 a^{3} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 5 a^{3} f} + \frac{20 c^{2} \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 a^{3} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 5 a^{3} f} + \frac{10 c^{2} \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 25 a^{3} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 5 a^{3} 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 )^{3}} & \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))**3,x)

[Out]

Piecewise((2*c**2*tan(e/2 + f*x/2)**5/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*
f*tan(e/2 + f*x/2)**3 + 50*a**3*f*tan(e/2 + f*x/2)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f) + 20*c**2*tan(e
/2 + f*x/2)**3/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*f*tan(e/2 + f*x/2)**3 +
50*a**3*f*tan(e/2 + f*x/2)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f) + 10*c**2*tan(e/2 + f*x/2)/(5*a**3*f*t
an(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*f*tan(e/2 + f*x/2)**3 + 50*a**3*f*tan(e/2 + f*x/2
)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f), Ne(f, 0)), (x*(-c*sin(e) + c)**2/(a*sin(e) + a)**3, True))

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Giac [A]  time = 1.89197, size = 81, normalized size = 2.45 \begin{align*} -\frac{2 \,{\left (5 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 10 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c^{2}\right )}}{5 \, a^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} \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))^3,x, algorithm="giac")

[Out]

-2/5*(5*c^2*tan(1/2*f*x + 1/2*e)^4 + 10*c^2*tan(1/2*f*x + 1/2*e)^2 + c^2)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)