3.14 \(\int \frac{\sin ^2(x)}{(a+a \sin (x))^2} \, dx\)

Optimal. Leaf size=35 \[ \frac{x}{a^2}+\frac{5 \cos (x)}{3 a^2 (\sin (x)+1)}-\frac{\cos (x)}{3 (a \sin (x)+a)^2} \]

[Out]

x/a^2 + (5*Cos[x])/(3*a^2*(1 + Sin[x])) - Cos[x]/(3*(a + a*Sin[x])^2)

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Rubi [A]  time = 0.0725763, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2758, 2735, 2648} \[ \frac{x}{a^2}+\frac{5 \cos (x)}{3 a^2 (\sin (x)+1)}-\frac{\cos (x)}{3 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + a*Sin[x])^2,x]

[Out]

x/a^2 + (5*Cos[x])/(3*a^2*(1 + Sin[x])) - Cos[x]/(3*(a + a*Sin[x])^2)

Rule 2758

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b
*(2*m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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{\sin ^2(x)}{(a+a \sin (x))^2} \, dx &=-\frac{\cos (x)}{3 (a+a \sin (x))^2}+\frac{\int \frac{-2 a+3 a \sin (x)}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac{x}{a^2}-\frac{\cos (x)}{3 (a+a \sin (x))^2}-\frac{5 \int \frac{1}{a+a \sin (x)} \, dx}{3 a}\\ &=\frac{x}{a^2}-\frac{\cos (x)}{3 (a+a \sin (x))^2}+\frac{5 \cos (x)}{3 \left (a^2+a^2 \sin (x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.126773, size = 69, normalized size = 1.97 \[ \frac{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \left (3 (3 x-4) \cos \left (\frac{x}{2}\right )+(10-3 x) \cos \left (\frac{3 x}{2}\right )+6 \sin \left (\frac{x}{2}\right ) (2 x+x \cos (x)-3)\right )}{6 a^2 (\sin (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + a*Sin[x])^2,x]

[Out]

((Cos[x/2] + Sin[x/2])*(3*(-4 + 3*x)*Cos[x/2] + (10 - 3*x)*Cos[(3*x)/2] + 6*(-3 + 2*x + x*Cos[x])*Sin[x/2]))/(
6*a^2*(1 + Sin[x])^2)

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Maple [A]  time = 0.037, size = 51, normalized size = 1.5 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{a}^{2}}}-{\frac{4}{3\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{{a}^{2} \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}+2\,{\frac{1}{{a}^{2} \left ( \tan \left ( x/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a^2*arctan(tan(1/2*x))-4/3/a^2/(tan(1/2*x)+1)^3+2/a^2/(tan(1/2*x)+1)^2+2/a^2/(tan(1/2*x)+1)

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Maxima [B]  time = 2.68941, size = 122, normalized size = 3.49 \begin{align*} \frac{2 \,{\left (\frac{9 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{3 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 4\right )}}{3 \,{\left (a^{2} + \frac{3 \, a^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{3 \, a^{2} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}} + \frac{2 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(9*sin(x)/(cos(x) + 1) + 3*sin(x)^2/(cos(x) + 1)^2 + 4)/(a^2 + 3*a^2*sin(x)/(cos(x) + 1) + 3*a^2*sin(x)^2/
(cos(x) + 1)^2 + a^2*sin(x)^3/(cos(x) + 1)^3) + 2*arctan(sin(x)/(cos(x) + 1))/a^2

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Fricas [B]  time = 1.45629, size = 215, normalized size = 6.14 \begin{align*} \frac{{\left (3 \, x - 5\right )} \cos \left (x\right )^{2} -{\left (3 \, x + 4\right )} \cos \left (x\right ) -{\left ({\left (3 \, x + 5\right )} \cos \left (x\right ) + 6 \, x + 1\right )} \sin \left (x\right ) - 6 \, x + 1}{3 \,{\left (a^{2} \cos \left (x\right )^{2} - a^{2} \cos \left (x\right ) - 2 \, a^{2} -{\left (a^{2} \cos \left (x\right ) + 2 \, a^{2}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((3*x - 5)*cos(x)^2 - (3*x + 4)*cos(x) - ((3*x + 5)*cos(x) + 6*x + 1)*sin(x) - 6*x + 1)/(a^2*cos(x)^2 - a^
2*cos(x) - 2*a^2 - (a^2*cos(x) + 2*a^2)*sin(x))

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Sympy [B]  time = 7.36811, size = 369, normalized size = 10.54 \begin{align*} \frac{15 x \tan ^{3}{\left (\frac{x}{2} \right )}}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} + \frac{45 x \tan ^{2}{\left (\frac{x}{2} \right )}}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} + \frac{45 x \tan{\left (\frac{x}{2} \right )}}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} + \frac{15 x}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} - \frac{22 \tan ^{3}{\left (\frac{x}{2} \right )}}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} - \frac{36 \tan ^{2}{\left (\frac{x}{2} \right )}}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} + \frac{24 \tan{\left (\frac{x}{2} \right )}}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} + \frac{18}{15 a^{2} \tan ^{3}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan ^{2}{\left (\frac{x}{2} \right )} + 45 a^{2} \tan{\left (\frac{x}{2} \right )} + 15 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**2/(a+a*sin(x))**2,x)

[Out]

15*x*tan(x/2)**3/(15*a**2*tan(x/2)**3 + 45*a**2*tan(x/2)**2 + 45*a**2*tan(x/2) + 15*a**2) + 45*x*tan(x/2)**2/(
15*a**2*tan(x/2)**3 + 45*a**2*tan(x/2)**2 + 45*a**2*tan(x/2) + 15*a**2) + 45*x*tan(x/2)/(15*a**2*tan(x/2)**3 +
45*a**2*tan(x/2)**2 + 45*a**2*tan(x/2) + 15*a**2) + 15*x/(15*a**2*tan(x/2)**3 + 45*a**2*tan(x/2)**2 + 45*a**2
*tan(x/2) + 15*a**2) - 22*tan(x/2)**3/(15*a**2*tan(x/2)**3 + 45*a**2*tan(x/2)**2 + 45*a**2*tan(x/2) + 15*a**2)
- 36*tan(x/2)**2/(15*a**2*tan(x/2)**3 + 45*a**2*tan(x/2)**2 + 45*a**2*tan(x/2) + 15*a**2) + 24*tan(x/2)/(15*a
**2*tan(x/2)**3 + 45*a**2*tan(x/2)**2 + 45*a**2*tan(x/2) + 15*a**2) + 18/(15*a**2*tan(x/2)**3 + 45*a**2*tan(x/
2)**2 + 45*a**2*tan(x/2) + 15*a**2)

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Giac [A]  time = 1.63975, size = 47, normalized size = 1.34 \begin{align*} \frac{x}{a^{2}} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 9 \, \tan \left (\frac{1}{2} \, x\right ) + 4\right )}}{3 \, a^{2}{\left (\tan \left (\frac{1}{2} \, x\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a^2 + 2/3*(3*tan(1/2*x)^2 + 9*tan(1/2*x) + 4)/(a^2*(tan(1/2*x) + 1)^3)