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

Optimal. Leaf size=32 \[ -\frac{2 c \cos (e+f x)}{f (a \sin (e+f x)+a)}-\frac{c x}{a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0439464, antiderivative size = 32, 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 = {2735, 2648} \[ -\frac{2 c \cos (e+f x)}{f (a \sin (e+f x)+a)}-\frac{c x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{c-c \sin (e+f x)}{a+a \sin (e+f x)} \, dx &=-\frac{c x}{a}+(2 c) \int \frac{1}{a+a \sin (e+f x)} \, dx\\ &=-\frac{c x}{a}-\frac{2 c \cos (e+f x)}{f (a+a \sin (e+f x))}\\ \end{align*}

Mathematica [B]  time = 0.180096, size = 79, normalized size = 2.47 \[ -\frac{c \left (f x \sin \left (e+\frac{f x}{2}\right )-4 \sin \left (\frac{f x}{2}\right )+f x \cos \left (\frac{f x}{2}\right )\right )}{a f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(f*x*Cos[(f*x)/2] - 4*Sin[(f*x)/2] + f*x*Sin[e + (f*x)/2]))/(a*f*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])))

________________________________________________________________________________________

Maple [A]  time = 0.056, size = 43, normalized size = 1.3 \begin{align*} -2\,{\frac{c\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{af}}-4\,{\frac{c}{af \left ( \tan \left ( 1/2\,fx+e/2 \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)),x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.88392, size = 104, normalized size = 3.25 \begin{align*} -\frac{2 \,{\left (c{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac{c}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.31992, size = 159, normalized size = 4.97 \begin{align*} -\frac{c f x +{\left (c f x + 2 \, c\right )} \cos \left (f x + e\right ) +{\left (c f x - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a 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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 3.36027, size = 90, normalized size = 2.81 \begin{align*} \begin{cases} - \frac{c f x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{c f x}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{4 c}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} & \text{for}\: f \neq 0 \\\frac{x \left (- c \sin{\left (e \right )} + c\right )}{a \sin{\left (e \right )} + a} & \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)),x)

[Out]

Piecewise((-c*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2) + a*f) - c*f*x/(a*f*tan(e/2 + f*x/2) + a*f) - 4*c/(a*
f*tan(e/2 + f*x/2) + a*f), Ne(f, 0)), (x*(-c*sin(e) + c)/(a*sin(e) + a), True))

________________________________________________________________________________________

Giac [A]  time = 2.05113, size = 50, normalized size = 1.56 \begin{align*} -\frac{\frac{{\left (f x + e\right )} c}{a} + \frac{4 \, c}{a{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}}}{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)),x, algorithm="giac")

[Out]

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