3.268 \(\int \frac{A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=35 \[ \frac{B x}{a}-\frac{(A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)} \]

[Out]

(B*x)/a - ((A - B)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x]))

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Rubi [A]  time = 0.0485771, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2735, 2648} \[ \frac{B x}{a}-\frac{(A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x]),x]

[Out]

(B*x)/a - ((A - B)*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{A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx &=\frac{B x}{a}-(-A+B) \int \frac{1}{a+a \sin (e+f x)} \, dx\\ &=\frac{B x}{a}-\frac{(A-B) \cos (e+f x)}{f (a+a \sin (e+f x))}\\ \end{align*}

Mathematica [B]  time = 0.155121, size = 79, normalized size = 2.26 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right ) (2 A+B (e+f x-2))+B (e+f x) \cos \left (\frac{1}{2} (e+f x)\right )\right )}{a f (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x]),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(B*(e + f*x)*Cos[(e + f*x)/2] + (2*A + B*(-2 + e + f*x))*Sin[(e + f*x)/
2]))/(a*f*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.042, size = 65, normalized size = 1.9 \begin{align*} 2\,{\frac{B\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{af}}-2\,{\frac{A}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{B}{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((A+B*sin(f*x+e))/(a+a*sin(f*x+e)),x)

[Out]

2/a/f*B*arctan(tan(1/2*f*x+1/2*e))-2/a/f/(tan(1/2*f*x+1/2*e)+1)*A+2/a/f/(tan(1/2*f*x+1/2*e)+1)*B

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Maxima [B]  time = 1.43761, size = 105, normalized size = 3. \begin{align*} \frac{2 \,{\left (B{\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{A}{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((A+B*sin(f*x+e))/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.83652, size = 166, normalized size = 4.74 \begin{align*} \frac{B f x +{\left (B f x - A + B\right )} \cos \left (f x + e\right ) +{\left (B f x + A - B\right )} \sin \left (f x + e\right ) - A + B}{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((A+B*sin(f*x+e))/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

(B*f*x + (B*f*x - A + B)*cos(f*x + e) + (B*f*x + A - B)*sin(f*x + e) - A + B)/(a*f*cos(f*x + e) + a*f*sin(f*x
+ e) + a*f)

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Sympy [A]  time = 2.03555, size = 109, normalized size = 3.11 \begin{align*} \begin{cases} - \frac{2 A}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} + \frac{B 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{B f x}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} + \frac{2 B}{a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} & \text{for}\: f \neq 0 \\\frac{x \left (A + B \sin{\left (e \right )}\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((A+B*sin(f*x+e))/(a+a*sin(f*x+e)),x)

[Out]

Piecewise((-2*A/(a*f*tan(e/2 + f*x/2) + a*f) + B*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2) + a*f) + B*f*x/(a*
f*tan(e/2 + f*x/2) + a*f) + 2*B/(a*f*tan(e/2 + f*x/2) + a*f), Ne(f, 0)), (x*(A + B*sin(e))/(a*sin(e) + a), Tru
e))

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Giac [A]  time = 1.22262, size = 54, normalized size = 1.54 \begin{align*} \frac{\frac{{\left (f x + e\right )} B}{a} - \frac{2 \,{\left (A - B\right )}}{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((A+B*sin(f*x+e))/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

((f*x + e)*B/a - 2*(A - B)/(a*(tan(1/2*f*x + 1/2*e) + 1)))/f