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(f*x+e)/a/c/f

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Rubi [A]  time = 0.07, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 8

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

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]

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*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \frac {\tan (e+f x)}{a c f} \]

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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fricas [A]  time = 0.43, size = 24, normalized size = 1.50 \[ \frac {\sin \left (f x + e\right )}{a c f \cos \left (f x + e\right )} \]

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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giac [A]  time = 0.16, size = 17, normalized size = 1.06 \[ \frac {\tan \left (f x + e\right )}{a c f} \]

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)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +a \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )}\, dx \]

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 = 0.31, size = 16, normalized size = 1.00 \[ \frac {\tan \left (f x + e\right )}{a c f} \]

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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mupad [B]  time = 6.85, size = 35, normalized size = 2.19 \[ -\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,c\,f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.65, size = 49, normalized size = 3.06 \[ \begin {cases} - \frac {2 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - a c f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\relax (e )} + a\right ) \left (- c \sin {\relax (e )} + c\right )} & \text {otherwise} \end {cases} \]

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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