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

Optimal. Leaf size=29 \[ -\frac{a c \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0668245, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2736, 2671} \[ -\frac{a c \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 0.264063, size = 70, normalized size = 2.41 \[ \frac{c \left (\cos \left (e+\frac{3 f x}{2}\right )-3 \cos \left (e+\frac{f x}{2}\right )\right )}{3 a^2 f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.069, size = 56, normalized size = 1.9 \begin{align*} 2\,{\frac{c}{{a}^{2}f} \left ( -4/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}+2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.77858, size = 290, normalized size = 10. \begin{align*} -\frac{2 \,{\left (\frac{c{\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{c{\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))/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*(c*(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)
- c*(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

________________________________________________________________________________________

Fricas [B]  time = 1.29729, size = 251, normalized size = 8.66 \begin{align*} -\frac{c \cos \left (f x + e\right )^{2} - c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) + 2 \, c\right )} \sin \left (f x + e\right ) - 2 \, c}{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))/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 6.88263, size = 158, normalized size = 5.45 \begin{align*} \begin{cases} - \frac{6 c \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{2 c}{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 )}{\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))/(a+a*sin(f*x+e))**2,x)

[Out]

Piecewise((-6*c*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*ta
n(e/2 + f*x/2) + 3*a**2*f) - 2*c/(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)/(a*sin(e) + a)**2, True))

________________________________________________________________________________________

Giac [A]  time = 2.04768, size = 53, normalized size = 1.83 \begin{align*} -\frac{2 \,{\left (3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c\right )}}{3 \, a^{2} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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