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

Optimal. Leaf size=35 \[ \frac{A \tan (e+f x)}{a c f}+\frac{B \sec (e+f x)}{a c f} \]

[Out]

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

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Rubi [A]  time = 0.135558, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2967, 2669, 3767, 8} \[ \frac{A \tan (e+f x)}{a c f}+\frac{B \sec (e+f x)}{a c f} \]

Antiderivative was successfully verified.

[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 2967

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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 3767

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]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx &=\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 \operatorname{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]  time = 0.0283891, size = 35, normalized size = 1. \[ \frac{A \tan (e+f x)}{a c f}+\frac{B \sec (e+f x)}{a c f} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.056, size = 57, normalized size = 1.6 \begin{align*} 2\,{\frac{1}{acf} \left ( -{\frac{A/2-B/2}{\tan \left ( 1/2\,fx+e/2 \right ) +1}}-{\frac{A/2+B/2}{\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))/(c-c*sin(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 0.9652, size = 47, normalized size = 1.34 \begin{align*} \frac{\frac{A \tan \left (f x + e\right )}{a c} + \frac{B}{a c \cos \left (f x + e\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))/(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

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Fricas [A]  time = 1.35681, size = 58, normalized size = 1.66 \begin{align*} \frac{A \sin \left (f x + e\right ) + B}{a c f \cos \left (f x + e\right )} \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))/(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.16627, size = 83, normalized size = 2.37 \begin{align*} \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} \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))/(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))

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Giac [A]  time = 1.20056, size = 55, normalized size = 1.57 \begin{align*} -\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} \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))/(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)