\(\int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx\) [232]

Optimal result

Integrand size = 24, antiderivative size = 30 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=\frac {a c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, 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} \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=\frac {a c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[In]

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

[Out]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(30)=60\).

Time = 1.62 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.47 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=-\frac {a \left (-3 \cos \left (e+\frac {f x}{2}\right )+\cos \left (e+\frac {3 f x}{2}\right )\right )}{3 c^2 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27

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

[In]

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

[Out]

2/3*a*(-3*tan(1/2*f*x+1/2*e)^2-1)/f/c^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. \(2 (29) = 58\).

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.47 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=\frac {a \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) + 2 \, a\right )} \sin \left (f x + e\right ) - 2 \, a}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

1/3*(a*cos(f*x + e)^2 - a*cos(f*x + e) - (a*cos(f*x + e) + 2*a)*sin(f*x + e) - 2*a)/(c^2*f*cos(f*x + e)^2 - c^
2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^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. \(2 (26) = 52\).

Time = 1.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 5.27 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=\begin {cases} - \frac {6 a \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {2 a}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )}{\left (- c \sin {\left (e \right )} + c\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-6*a*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*ta
n(e/2 + f*x/2) - 3*c**2*f) - 2*a/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e
/2 + f*x/2) - 3*c**2*f), Ne(f, 0)), (x*(a*sin(e) + a)/(-c*sin(e) + c)**2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (29) = 58\).

Time = 0.20 (sec) , antiderivative size = 217, normalized size of antiderivative = 7.23 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {a {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac {a {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a\right )}}{3 \, c^{2} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.73 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx=-\frac {2\,a\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3\right )}{3\,c^2\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^3} \]

[In]

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

[Out]

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