3.275 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{\tan ^3(e+f x)}{3 a^2 c^2 f}+\frac{\tan (e+f x)}{a^2 c^2 f} \]

[Out]

Tan[e + f*x]/(a^2*c^2*f) + Tan[e + f*x]^3/(3*a^2*c^2*f)

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Rubi [A]  time = 0.0632528, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2736, 3767} \[ \frac{\tan ^3(e+f x)}{3 a^2 c^2 f}+\frac{\tan (e+f x)}{a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^2),x]

[Out]

Tan[e + f*x]/(a^2*c^2*f) + Tan[e + f*x]^3/(3*a^2*c^2*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]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) \, dx}{a^2 c^2}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{a^2 c^2 f}\\ &=\frac{\tan (e+f x)}{a^2 c^2 f}+\frac{\tan ^3(e+f x)}{3 a^2 c^2 f}\\ \end{align*}

Mathematica [A]  time = 0.0537988, size = 29, normalized size = 0.76 \[ \frac{\frac{1}{3} \tan ^3(e+f x)+\tan (e+f x)}{a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^2),x]

[Out]

(Tan[e + f*x] + Tan[e + f*x]^3/3)/(a^2*c^2*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 ) ^{2} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^2,x)

[Out]

int(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^2,x)

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Maxima [A]  time = 1.43591, size = 38, normalized size = 1. \begin{align*} \frac{\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )}{3 \, a^{2} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/3*(tan(f*x + e)^3 + 3*tan(f*x + e))/(a^2*c^2*f)

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Fricas [A]  time = 1.20854, size = 92, normalized size = 2.42 \begin{align*} \frac{{\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right )}{3 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/3*(2*cos(f*x + e)^2 + 1)*sin(f*x + e)/(a^2*c^2*f*cos(f*x + e)^3)

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Sympy [A]  time = 13.2746, size = 286, normalized size = 7.53 \begin{align*} \begin{cases} - \frac{6 \tan ^{5}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a^{2} c^{2} f} + \frac{4 \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a^{2} c^{2} f} - \frac{6 \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} - 3 a^{2} c^{2} f} & \text{for}\: f \neq 0 \\\frac{x}{\left (a \sin{\left (e \right )} + a\right )^{2} \left (- c \sin{\left (e \right )} + c\right )^{2}} & \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))**2/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((-6*tan(e/2 + f*x/2)**5/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a
**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) + 4*tan(e/2 + f*x/2)**3/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9
*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) - 6*tan(e/2 + f*x/2)/(3*
a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a*
*2*c**2*f), Ne(f, 0)), (x/((a*sin(e) + a)**2*(-c*sin(e) + c)**2), True))

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Giac [A]  time = 2.23937, size = 41, normalized size = 1.08 \begin{align*} \frac{\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )}{3 \, a^{2} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

1/3*(tan(f*x + e)^3 + 3*tan(f*x + e))/(a^2*c^2*f)