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

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Rubi [A]  time = 0.14, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.38, size = 94, normalized size = 0.80 \[ \frac {\left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a \cos (x) \left (3 a b-\left (a^2-4 b^2\right ) \sin (x)\right )}{2 (a-b)^2 (a+b)^2 (a+b \sin (x))^2} \]

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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fricas [B]  time = 0.52, size = 516, normalized size = 4.37 \[ \left [-\frac {2 \, {\left (a^{5} - 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \relax (x) \sin \relax (x) + {\left (a^{4} + 3 \, a^{2} b^{2} + 2 \, b^{4} - {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \sin \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} + 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \relax (x)}{4 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \relax (x)\right )}}, -\frac {{\left (a^{5} - 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \relax (x) \sin \relax (x) + {\left (a^{4} + 3 \, a^{2} b^{2} + 2 \, b^{4} - {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \sin \relax (x)\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \relax (x)}{2 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \relax (x)\right )}}\right ] \]

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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giac [A]  time = 0.15, size = 182, normalized size = 1.54 \[ \frac {{\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} + 2 \, b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 6 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, x\right ) + 10 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) + 3 \, a^{2} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} \]

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)

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maple [B]  time = 0.11, size = 265, normalized size = 2.25 \[ \frac {\frac {8 \left (a^{2}+2 b^{2}\right ) a \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{8 a^{4}-16 a^{2} b^{2}+8 b^{4}}+\frac {3 b \left (a^{2}+2 b^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {a \left (a^{2}-10 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {3 a^{2} b}{a^{4}-2 a^{2} b^{2}+b^{4}}}{\left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right )^{2}}+\frac {a^{2} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.23, size = 318, normalized size = 2.69 \[ \frac {\frac {3\,a^2\,b}{a^4-2\,a^2\,b^2+b^4}-\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2-10\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a^2+2\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {3\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^2+2\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2+a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (a^2+2\,b^2\right )\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+2\,b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^2+2\,b^2}\right )\,\left (a^2+2\,b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a + b*sin(x))^3,x)

[Out]

((3*a^2*b)/(a^4 + b^4 - 2*a^2*b^2) - (a*tan(x/2)*(a^2 - 10*b^2))/(a^4 + b^4 - 2*a^2*b^2) + (a*tan(x/2)^3*(a^2
+ 2*b^2))/(a^4 + b^4 - 2*a^2*b^2) + (3*b*tan(x/2)^2*(a^2 + 2*b^2))/(a^4 + b^4 - 2*a^2*b^2))/(tan(x/2)^2*(2*a^2
+ 4*b^2) + a^2 + a^2*tan(x/2)^4 + 4*a*b*tan(x/2) + 4*a*b*tan(x/2)^3) + (atan(((((a^2 + 2*b^2)*(2*a^4*b + 2*b^
5 - 4*a^2*b^3))/(2*(a + b)^(5/2)*(a - b)^(5/2)*(a^4 + b^4 - 2*a^2*b^2)) + (a*tan(x/2)*(a^2 + 2*b^2))/((a + b)^
(5/2)*(a - b)^(5/2)))*(a^4 + b^4 - 2*a^2*b^2))/(a^2 + 2*b^2))*(a^2 + 2*b^2))/((a + b)^(5/2)*(a - b)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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