3.196 \(\int \frac{\sin ^2(x)}{(a+b \sin (x))^3} \, dx\)

Optimal. Leaf size=118 \[ \frac{\left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))} \]

[Out]

((a^2 + 2*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (a^2*Cos[x])/(2*b*(a^2 - b^2)*(a
+ b*Sin[x])^2) - (a*(a^2 - 4*b^2)*Cos[x])/(2*b*(a^2 - b^2)^2*(a + b*Sin[x]))

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Rubi [A]  time = 0.143604, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2790, 2754, 12, 2660, 618, 204} \[ \frac{\left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Sin[x])^3,x]

[Out]

((a^2 + 2*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (a^2*Cos[x])/(2*b*(a^2 - b^2)*(a
+ b*Sin[x])^2) - (a*(a^2 - 4*b^2)*Cos[x])/(2*b*(a^2 - b^2)^2*(a + b*Sin[x]))

Rule 2790

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

Rule 2754

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^2(x)}{(a+b \sin (x))^3} \, dx &=\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{\int \frac{2 a b+\left (a^2-2 b^2\right ) \sin (x)}{(a+b \sin (x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\int \frac{b \left (a^2+2 b^2\right )}{a+b \sin (x)} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\left (a^2+2 b^2\right ) \int \frac{1}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\left (a^2+2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\left (2 \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac{\left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a^2 \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{a \left (a^2-4 b^2\right ) \cos (x)}{2 b \left (a^2-b^2\right )^2 (a+b \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.351667, size = 94, normalized size = 0.8 \[ \frac{\left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a \cos (x) \left (3 a b-\left (a^2-4 b^2\right ) \sin (x)\right )}{2 (a-b)^2 (a+b)^2 (a+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Sin[x])^3,x]

[Out]

((a^2 + 2*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (a*Cos[x]*(3*a*b - (a^2 - 4*b^2)*
Sin[x]))/(2*(a - b)^2*(a + b)^2*(a + b*Sin[x])^2)

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Maple [B]  time = 0.051, size = 265, normalized size = 2.3 \begin{align*} 8\,{\frac{1}{ \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) ^{2}} \left ( 1/8\,{\frac{ \left ({a}^{2}+2\,{b}^{2} \right ) a \left ( \tan \left ( x/2 \right ) \right ) ^{3}}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}}+3/8\,{\frac{b \left ({a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}}-1/8\,{\frac{a \left ({a}^{2}-10\,{b}^{2} \right ) \tan \left ( x/2 \right ) }{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}}+3/8\,{\frac{{a}^{2}b}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+{\frac{{a}^{2}}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}\arctan \left ({\frac{1}{2} \left ( 2\,a\tan \left ( x/2 \right ) +2\,b \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}+2\,{\frac{{b}^{2}}{ \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a+b*sin(x))^3,x)

[Out]

8*(1/8*(a^2+2*b^2)*a/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3+3/8*b*(a^2+2*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-1/8*a
*(a^2-10*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)+3/8*a^2*b/(a^4-2*a^2*b^2+b^4))/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^
2+a^2/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+2/(a^4-2*a^2*b^2+b^
4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a+b*sin(x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.10686, size = 1138, normalized size = 9.64 \begin{align*} \left [-\frac{2 \,{\left (a^{5} - 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{4} + 3 \, a^{2} b^{2} + 2 \, b^{4} -{\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \sin \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 6 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )}{4 \,{\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} -{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}, -\frac{{\left (a^{5} - 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{4} + 3 \, a^{2} b^{2} + 2 \, b^{4} -{\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \sin \left (x\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} -{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a+b*sin(x))^3,x, algorithm="fricas")

[Out]

[-1/4*(2*(a^5 - 5*a^3*b^2 + 4*a*b^4)*cos(x)*sin(x) + (a^4 + 3*a^2*b^2 + 2*b^4 - (a^2*b^2 + 2*b^4)*cos(x)^2 + 2
*(a^3*b + 2*a*b^3)*sin(x))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(a*cos(
x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 6*(a^4*b - a^2*b^3)*cos(x
))/(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^
3 + 3*a^3*b^5 - a*b^7)*sin(x)), -1/2*((a^5 - 5*a^3*b^2 + 4*a*b^4)*cos(x)*sin(x) + (a^4 + 3*a^2*b^2 + 2*b^4 - (
a^2*b^2 + 2*b^4)*cos(x)^2 + 2*(a^3*b + 2*a*b^3)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2
)*cos(x))) - 3*(a^4*b - a^2*b^3)*cos(x))/(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6
- b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(x))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**2/(a+b*sin(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.80612, size = 246, normalized size = 2.08 \begin{align*} \frac{{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (a^{2} + 2 \, b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{a^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 3 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{2} + 6 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{2} - a^{3} \tan \left (\frac{1}{2} \, x\right ) + 10 \, a b^{2} \tan \left (\frac{1}{2} \, x\right ) + 3 \, a^{2} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a+b*sin(x))^3,x, algorithm="giac")

[Out]

(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*(a^2 + 2*b^2)/((a^4 - 2*a^2*b^2
+ b^4)*sqrt(a^2 - b^2)) + (a^3*tan(1/2*x)^3 + 2*a*b^2*tan(1/2*x)^3 + 3*a^2*b*tan(1/2*x)^2 + 6*b^3*tan(1/2*x)^
2 - a^3*tan(1/2*x) + 10*a*b^2*tan(1/2*x) + 3*a^2*b)/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x)
+ a)^2)