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

Optimal. Leaf size=58 \[ -\frac{c \cos ^3(e+f x)}{15 f (a \sin (e+f x)+a)^3}-\frac{a c \cos ^3(e+f x)}{5 f (a \sin (e+f x)+a)^4} \]

[Out]

-(a*c*Cos[e + f*x]^3)/(5*f*(a + a*Sin[e + f*x])^4) - (c*Cos[e + f*x]^3)/(15*f*(a + a*Sin[e + f*x])^3)

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Rubi [A]  time = 0.106667, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2736, 2672, 2671} \[ -\frac{c \cos ^3(e+f x)}{15 f (a \sin (e+f x)+a)^3}-\frac{a c \cos ^3(e+f x)}{5 f (a \sin (e+f x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c*Cos[e + f*x]^3)/(5*f*(a + a*Sin[e + f*x])^4) - (c*Cos[e + f*x]^3)/(15*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 2672

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*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 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))^3} \, dx &=(a c) \int \frac{\cos ^2(e+f x)}{(a+a \sin (e+f x))^4} \, dx\\ &=-\frac{a c \cos ^3(e+f x)}{5 f (a+a \sin (e+f x))^4}+\frac{1}{5} c \int \frac{\cos ^2(e+f x)}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac{a c \cos ^3(e+f x)}{5 f (a+a \sin (e+f x))^4}-\frac{c \cos ^3(e+f x)}{15 f (a+a \sin (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.329819, size = 92, normalized size = 1.59 \[ \frac{c \left (\sin \left (2 e+\frac{5 f x}{2}\right )-15 \cos \left (e+\frac{f x}{2}\right )+5 \cos \left (e+\frac{3 f x}{2}\right )+5 \sin \left (\frac{f x}{2}\right )\right )}{30 a^3 f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-15*Cos[e + (f*x)/2] + 5*Cos[e + (3*f*x)/2] + 5*Sin[(f*x)/2] + Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(Cos[e/2]
+ Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [A]  time = 0.074, size = 86, normalized size = 1.5 \begin{align*} 2\,{\frac{c}{f{a}^{3}} \left ( -14/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}-8/5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-5}+4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-1}+3\, \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))/(a+a*sin(f*x+e))^3,x)

[Out]

2/f*c/a^3*(-14/3/(tan(1/2*f*x+1/2*e)+1)^3-8/5/(tan(1/2*f*x+1/2*e)+1)^5+4/(tan(1/2*f*x+1/2*e)+1)^4-1/(tan(1/2*f
*x+1/2*e)+1)+3/(tan(1/2*f*x+1/2*e)+1)^2)

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Maxima [B]  time = 1.20779, size = 522, normalized size = 9. \begin{align*} -\frac{2 \,{\left (\frac{c{\left (\frac{20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac{5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac{3 \, c{\left (\frac{5 \, \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{5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac{5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, 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))^3,x, algorithm="maxima")

[Out]

-2/15*(c*(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/(cos
(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) - 3*c*(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)/(co
s(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 = 1.28202, size = 381, normalized size = 6.57 \begin{align*} \frac{c \cos \left (f x + e\right )^{3} - 2 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) -{\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + 6 \, c\right )} \sin \left (f x + e\right ) + 6 \, c}{15 \,{\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))/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(c*cos(f*x + e)^3 - 2*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) - (c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + 6*c)*s
in(f*x + e) + 6*c)/(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*co
s(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 = 19.1246, size = 573, normalized size = 9.88 \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))/(a+a*sin(f*x+e))**3,x)

[Out]

Piecewise((2*c*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f
*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 20*c*tan(e/2
+ f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 +
150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 10*c*tan(e/2 + f*x/2)**3/(15*a**3*
f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 +
f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*c*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5
+ 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*
tan(e/2 + f*x/2) + 15*a**3*f) - 6*c/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*
f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (
x*(-c*sin(e) + c)/(a*sin(e) + a)**3, True))

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Giac [A]  time = 2.04911, size = 113, normalized size = 1.95 \begin{align*} -\frac{2 \,{\left (15 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 15 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 25 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, c\right )}}{15 \, 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))/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*(15*c*tan(1/2*f*x + 1/2*e)^4 + 15*c*tan(1/2*f*x + 1/2*e)^3 + 25*c*tan(1/2*f*x + 1/2*e)^2 + 5*c*tan(1/2*f
*x + 1/2*e) + 4*c)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)