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

Optimal. Leaf size=33 \[ \frac{2 a \cos (e+f x)}{f (c-c \sin (e+f x))}-\frac{a x}{c} \]

[Out]

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

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Rubi [A]  time = 0.0484159, antiderivative size = 33, 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} \[ \frac{2 a \cos (e+f x)}{f (c-c \sin (e+f x))}-\frac{a x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.182072, size = 83, normalized size = 2.52 \[ \frac{a \left (f x \sin \left (e+\frac{f x}{2}\right )+4 \sin \left (\frac{f x}{2}\right )-f x \cos \left (\frac{f x}{2}\right )\right )}{c f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.057, size = 43, normalized size = 1.3 \begin{align*} -2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{cf}}-4\,{\frac{a}{cf \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((a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 2.23758, size = 111, normalized size = 3.36 \begin{align*} -\frac{2 \,{\left (a{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac{1}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac{a}{c - \frac{c \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((a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63003, size = 159, normalized size = 4.82 \begin{align*} -\frac{a f x +{\left (a f x - 2 \, a\right )} \cos \left (f x + e\right ) -{\left (a f x + 2 \, a\right )} \sin \left (f x + e\right ) - 2 \, a}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.91624, size = 88, normalized size = 2.67 \begin{align*} \begin{cases} - \frac{a f x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{c f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - c f} + \frac{a f x}{c f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - c f} - \frac{4 a}{c f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - c f} & \text{for}\: f \neq 0 \\\frac{x \left (a \sin{\left (e \right )} + a\right )}{- c \sin{\left (e \right )} + c} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.672, size = 50, normalized size = 1.52 \begin{align*} -\frac{\frac{{\left (f x + e\right )} a}{c} + \frac{4 \, a}{c{\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((a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

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