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

Optimal. Leaf size=55 \[ -\frac{\cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac{\cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

-Cos[e + f*x]/(3*f*(a + a*Sin[e + f*x])^2) - Cos[e + f*x]/(3*f*(a^2 + a^2*Sin[e + f*x]))

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Rubi [A]  time = 0.0278865, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2650, 2648} \[ -\frac{\cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac{\cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(-2),x]

[Out]

-Cos[e + f*x]/(3*f*(a + a*Sin[e + f*x])^2) - Cos[e + f*x]/(3*f*(a^2 + a^2*Sin[e + f*x]))

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0995097, size = 54, normalized size = 0.98 \[ -\frac{-4 \sin (e+f x)+\sin (2 (e+f x))+4 \cos (e+f x)+\cos (2 (e+f x))-3}{6 a^2 f (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-3 + 4*Cos[e + f*x] + Cos[2*(e + f*x)] - 4*Sin[e + f*x] + Sin[2*(e + f*x)])/(6*a^2*f*(1 + Sin[e + f*x])^2)

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Maple [A]  time = 0.039, size = 53, normalized size = 1. \begin{align*} 2\,{\frac{1}{{a}^{2}f} \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-1}-2/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.18346, size = 158, normalized size = 2.87 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{3 \,{\left (a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.45571, size = 239, normalized size = 4.35 \begin{align*} \frac{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right ) + 2 \, \cos \left (f x + e\right ) + 1}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.99466, size = 146, normalized size = 2.65 \begin{align*} \begin{cases} \frac{2 \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} - \frac{2}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} & \text{for}\: f \neq 0 \\\frac{x}{\left (a \sin{\left (e \right )} + a\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,x)

[Out]

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

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Giac [A]  time = 1.28746, size = 68, normalized size = 1.24 \begin{align*} -\frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2\right )}}{3 \, a^{2} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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