3.265 \(\int \frac{1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx\)

Optimal. Leaf size=16 \[ \frac{\tan (e+f x)}{a c f} \]

[Out]

Tan[e + f*x]/(a*c*f)

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Rubi [A]  time = 0.0674709, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 3767, 8} \[ \frac{\tan (e+f x)}{a c f} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])),x]

[Out]

Tan[e + f*x]/(a*c*f)

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 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{1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))} \, dx &=\frac{\int \sec ^2(e+f x) \, dx}{a c}\\ &=-\frac{\operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a c f}\\ &=\frac{\tan (e+f x)}{a c f}\\ \end{align*}

Mathematica [A]  time = 0.0123707, size = 16, normalized size = 1. \[ \frac{\tan (e+f x)}{a c f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])),x]

[Out]

Tan[e + f*x]/(a*c*f)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+a\sin \left ( fx+e \right ) \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x)

[Out]

int(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x)

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Maxima [A]  time = 1.75087, size = 22, normalized size = 1.38 \begin{align*} \frac{\tan \left (f x + e\right )}{a c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

tan(f*x + e)/(a*c*f)

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Fricas [A]  time = 1.27716, size = 47, normalized size = 2.94 \begin{align*} \frac{\sin \left (f x + e\right )}{a c f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

sin(f*x + e)/(a*c*f*cos(f*x + e))

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Sympy [A]  time = 2.81456, size = 49, normalized size = 3.06 \begin{align*} \begin{cases} - \frac{2 \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} & \text{for}\: f \neq 0 \\\frac{x}{\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(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x)

[Out]

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

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Giac [A]  time = 2.03161, size = 23, normalized size = 1.44 \begin{align*} \frac{\tan \left (f x + e\right )}{a c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

tan(f*x + e)/(a*c*f)