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]

-1/5*a*c*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^4-1/15*c*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^3

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 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]

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 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]))

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 = 0.34, 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)

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 154, normalized size = 2.66 \[ \frac {c \cos \left (f x + e\right )^{3} - 2 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) - {\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + 6 \, c\right )} \sin \left (f x + e\right ) + 6 \, c}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

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))

________________________________________________________________________________________

giac [A]  time = 0.70, size = 84, normalized size = 1.45 \[ -\frac {2 \, {\left (15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, c\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.26, size = 86, normalized size = 1.48 \[ \frac {2 c \left (\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\right )}{f \,a^{3}} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.98, size = 387, normalized size = 6.67 \[ -\frac {2 \, {\left (\frac {c {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]

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

________________________________________________________________________________________

mupad [B]  time = 7.24, size = 134, normalized size = 2.31 \[ -\frac {2\,c\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+25\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+15\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{15\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 8.43, size = 573, normalized size = 9.88 \[ \begin {cases} - \frac {30 c \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {30 c \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {50 c \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {10 c \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {8 c}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\relax (e )} + c\right )}{\left (a \sin {\relax (e )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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((-30*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) - 30*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) - 50*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) - 10*c*tan(e/2 + f*x/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*t
an(e/2 + f*x/2) + 15*a**3*f) - 8*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))

________________________________________________________________________________________