\(\int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx\) [268]

Optimal result

Integrand size = 23, antiderivative size = 35 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {B x}{a}-\frac {(A-B) \cos (e+f x)}{f (a+a \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2814, 2727} \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {B x}{a}-\frac {(A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)} \]

[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 2727

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]

Rule 2814

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(35)=70\).

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.26 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (B (e+f x) \cos \left (\frac {1}{2} (e+f x)\right )+(2 A+B (-2+e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{a f (1+\sin (e+f x))} \]

[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]))

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {2 B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 \left (A -B \right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a f}\)	\(42\)
default	\(\frac {2 B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 \left (A -B \right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a f}\)	\(42\)
parallelrisch	\(\frac {f x B +\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (f x B +2 A -2 B \right )}{f a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(47\)
risch	\(\frac {B x}{a}-\frac {2 A}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 B}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}\)	\(54\)
norman	\(\frac {\frac {B x}{a}+\frac {B x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {B x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {B x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {\left (-2 B +2 A \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {\left (-2 B +2 A \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(134\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {B f x + {\left (B f x - A + B\right )} \cos \left (f x + e\right ) + {\left (B f x + A - B\right )} \sin \left (f x + e\right ) - A + B}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (26) = 52\).

Time = 0.56 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.11 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\begin {cases} - \frac {2 A}{a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} + \frac {B f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} + \frac {B f x}{a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} + \frac {2 B}{a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} & \text {for}\: f \neq 0 \\\frac {x \left (A + B \sin {\left (e \right )}\right )}{a \sin {\left (e \right )} + a} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (35) = 70\).

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.23 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (B {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {A}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {\frac {{\left (f x + e\right )} B}{a} - \frac {2 \, {\left (A - B\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}}{f} \]

[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

Mupad [B] (verification not implemented)

Time = 13.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sin (e+f x)}{a+a \sin (e+f x)} \, dx=\frac {B\,x}{a}-\frac {2\,A-2\,B}{a\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )} \]

[In]

int((A + B*sin(e + f*x))/(a + a*sin(e + f*x)),x)

[Out]

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