\(\int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx\) [273]

Optimal result

Integrand size = 24, antiderivative size = 29 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=-\frac {a c \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2815, 2750} \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=-\frac {a c \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

[In]

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

[Out]

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

Rule 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2815

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*} \text {integral}& = (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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(29)=58\).

Time = 1.40 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=\frac {c \left (-3 \cos \left (e+\frac {f x}{2}\right )+\cos \left (e+\frac {3 f x}{2}\right )\right )}{3 a^2 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

[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 [A] (verified)

Time = 0.72 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31

method	result	size
parallelrisch	\(\frac {2 c \left (-3 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-1\right )}{3 f \,a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\)	\(38\)
risch	\(\frac {2 c \,{\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 c}{3}}{f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\)	\(39\)
derivativedivides	\(\frac {2 c \left (\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {4}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\right )}{f \,a^{2}}\)	\(56\)
default	\(\frac {2 c \left (\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {4}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\right )}{f \,a^{2}}\)	\(56\)
norman	\(\frac {-\frac {2 c \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {2 c}{3 f a}-\frac {8 c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}}{a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\)	\(83\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (27) = 54\).

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.59 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=-\frac {c \cos \left (f x + e\right )^{2} - c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) + 2 \, c\right )} \sin \left (f x + e\right ) - 2 \, c}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

[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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (27) = 54\).

Time = 1.15 (sec) , antiderivative size = 158, normalized size of antiderivative = 5.45 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=\begin {cases} - \frac {6 c \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {2 c}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\left (e \right )} + c\right )}{\left (a \sin {\left (e \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (27) = 54\).

Time = 0.20 (sec) , antiderivative size = 215, normalized size of antiderivative = 7.41 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac {c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c\right )}}{3 \, a^{2} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 6.43 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx=\frac {2\,c\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3\right )}{3\,a^2\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^3} \]

[In]

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

[Out]

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