\(\int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx\) [231]

Optimal result

Integrand size = 24, antiderivative size = 33 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=-\frac {a x}{c}+\frac {2 a \cos (e+f x)}{f (c-c \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, 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 = {2814, 2727} \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=\frac {2 a \cos (e+f x)}{f (c-c \sin (e+f x))}-\frac {a x}{c} \]

[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 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 {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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(83\) vs. \(2(33)=66\).

Time = 1.74 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.52 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=\frac {a \left (-f x \cos \left (\frac {f x}{2}\right )+4 \sin \left (\frac {f x}{2}\right )+f x \sin \left (e+\frac {f x}{2}\right )\right )}{c f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[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]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {a x}{c}+\frac {4 a}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}\)	\(32\)
derivativedivides	\(\frac {2 a \left (-\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f c}\)	\(38\)
default	\(\frac {2 a \left (-\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}\right )}{f c}\)	\(38\)
parallelrisch	\(-\frac {a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) f x -f x +4\right )}{f c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(41\)
norman	\(\frac {\frac {a x}{c}+\frac {a x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {4 a}{c f}-\frac {a x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {a x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {4 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(117\)

[In]

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

[Out]

-a*x/c+4*a/f/c/(exp(I*(f*x+e))-I)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=-\frac {a f x + {\left (a f x - 2 \, a\right )} \cos \left (f x + e\right ) - {\left (a f x + 2 \, a\right )} \sin \left (f x + e\right ) - 2 \, a}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (26) = 52\).

Time = 0.58 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.67 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=\begin {cases} - \frac {a f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} + \frac {a f x}{c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} - \frac {4 a}{c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )}{- c \sin {\left (e \right )} + c} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (34) = 68\).

Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.48 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (a {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {a}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=-\frac {\frac {{\left (f x + e\right )} a}{c} + \frac {4 \, a}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}}{f} \]

[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

Mupad [B] (verification not implemented)

Time = 6.42 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {a+a \sin (e+f x)}{c-c \sin (e+f x)} \, dx=-\frac {a\,x}{c}-\frac {a\,\left (e+f\,x\right )-a\,\left (e+f\,x-4\right )}{c\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )} \]

[In]

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

[Out]

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