\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx\) [56]

Optimal result

Integrand size = 36, antiderivative size = 35 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=\frac {B \sec (e+f x)}{a c f}+\frac {A \tan (e+f x)}{a c f} \]

[Out]

B*sec(f*x+e)/a/c/f+A*tan(f*x+e)/a/c/f

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3046, 2748, 3852, 8} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=\frac {A \tan (e+f x)}{a c f}+\frac {B \sec (e+f x)}{a c f} \]

[In]

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

[Out]

(B*Sec[e + f*x])/(a*c*f) + (A*Tan[e + f*x])/(a*c*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^2(e+f x) (A+B \sin (e+f x)) \, dx}{a c} \\ & = \frac {B \sec (e+f x)}{a c f}+\frac {A \int \sec ^2(e+f x) \, dx}{a c} \\ & = \frac {B \sec (e+f x)}{a c f}-\frac {A \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a c f} \\ & = \frac {B \sec (e+f x)}{a c f}+\frac {A \tan (e+f x)}{a c f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=\frac {B \sec (e+f x)}{a c f}+\frac {A \tan (e+f x)}{a c f} \]

[In]

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

[Out]

(B*Sec[e + f*x])/(a*c*f) + (A*Tan[e + f*x])/(a*c*f)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(\frac {-2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-2 B}{f a c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(42\)
risch	\(\frac {2 i A +2 B \,{\mathrm e}^{i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a c f}\)	\(56\)
derivativedivides	\(\frac {-\frac {2 \left (\frac {A}{2}-\frac {B}{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}}{a c f}\)	\(57\)
default	\(\frac {-\frac {2 \left (\frac {A}{2}-\frac {B}{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}}{a c f}\)	\(57\)
norman	\(\frac {-\frac {2 B}{a c f}-\frac {2 A \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a c f}-\frac {2 A \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {2 B \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(123\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=\frac {A \sin \left (f x + e\right ) + B}{a c f \cos \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (26) = 52\).

Time = 0.72 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.37 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=\begin {cases} - \frac {2 A \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - a c f} - \frac {2 B}{a c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - a c f} & \text {for}\: f \neq 0 \\\frac {x \left (A + B \sin {\left (e \right )}\right )}{\left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=\frac {\frac {A \tan \left (f x + e\right )}{a c} + \frac {B}{a c \cos \left (f x + e\right )}}{f} \]

[In]

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

[Out]

(A*tan(f*x + e)/(a*c) + B/(a*c*cos(f*x + e)))/f

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=-\frac {2 \, {\left (A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a c f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.49 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx=-\frac {2\,\left (B+A\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a\,c\,f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

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