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

Optimal. Leaf size=144 \[ -\frac{a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{5/2}}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{x}{b^3} \]

[Out]

x/b^3 - (a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^3*(a^2 - b^2)^(5/2)) + (a^
2*Cos[x]*Sin[x])/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2) + (a^2*(2*a^2 - 5*b^2)*Cos[x])/(2*b^2*(a^2 - b^2)^2*(a + b
*Sin[x]))

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Rubi [A]  time = 0.244593, antiderivative size = 144, 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 = {2792, 3021, 2735, 2660, 618, 204} \[ -\frac{a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{5/2}}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + b*Sin[x])^3,x]

[Out]

x/b^3 - (a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^3*(a^2 - b^2)^(5/2)) + (a^
2*Cos[x]*Sin[x])/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2) + (a^2*(2*a^2 - 5*b^2)*Cos[x])/(2*b^2*(a^2 - b^2)^2*(a + b
*Sin[x]))

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3021

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

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 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 ^3(x)}{(a+b \sin (x))^3} \, dx &=\frac{a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{\int \frac{a^2-2 a b \sin (x)-2 \left (a^2-b^2\right ) \sin ^2(x)}{(a+b \sin (x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\int \frac{a b \left (a^2-4 b^2\right )+2 \left (a^2-b^2\right )^2 \sin (x)}{a+b \sin (x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{x}{b^3}+\frac{a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\right )\right ) \int \frac{1}{a+b \sin (x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{x}{b^3}+\frac{a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^3 \left (a^2-b^2\right )^2}\\ &=\frac{x}{b^3}+\frac{a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\left (2 a \left (2 a^4-5 a^2 b^2+6 b^4\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 )}{b^3 \left (a^2-b^2\right )^2}\\ &=\frac{x}{b^3}-\frac{a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{5/2}}+\frac{a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.513841, size = 136, normalized size = 0.94 \[ \frac{-\frac{2 a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{3 a^2 b \left (a^2-2 b^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}-\frac{a^3 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}+2 x}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x - (2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - (a^3*b*C
os[x])/((a - b)*(a + b)*(a + b*Sin[x])^2) + (3*a^2*b*(a^2 - 2*b^2)*Cos[x])/((a - b)^2*(a + b)^2*(a + b*Sin[x])
))/(2*b^3)

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Maple [B]  time = 0.057, size = 612, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^3*arctan(tan(1/2*x))+a^4/b/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-4*b*a^2/(t
an(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3+2*a^5/b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)
^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-a^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-1
0*b^2*a/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2+7*a^4/b/(tan(1/2*x)^2*a+2*tan(1/2
*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-16*b*a^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan
(1/2*x)+2*a^5/b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)-5*a^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b
+a)^2/(a^4-2*a^2*b^2+b^4)-2*a^5/b^3/(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))+5*a^3/b/(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))-6*b*a/
(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))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.16055, size = 1742, normalized size = 12.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*x*cos(x)^2 + (2*a^7 - 3*a^5*b^2 + a^3*b^4 + 6*a*b^6 - (2*a^5*
b^2 - 5*a^3*b^4 + 6*a*b^6)*cos(x)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*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)) - 4*(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*x - 2*(2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*c
os(x) - 2*(4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*x + 3*(a^6*b^2 - 3*a^4*b^4 + 2*a^2*b^6)*cos(x))*sin(x))/(
a^8*b^3 - 2*a^6*b^5 + 2*a^2*b^9 - b^11 - (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*cos(x)^2 + 2*(a^7*b^4 - 3*a^
5*b^6 + 3*a^3*b^8 - a*b^10)*sin(x)), -1/2*(2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*x*cos(x)^2 - (2*a^7 - 3*a
^5*b^2 + a^3*b^4 + 6*a*b^6 - (2*a^5*b^2 - 5*a^3*b^4 + 6*a*b^6)*cos(x)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*
sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - 2*(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^
8)*x - (2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*cos(x) - (4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*x + 3*(a^6*b^2 -
3*a^4*b^4 + 2*a^2*b^6)*cos(x))*sin(x))/(a^8*b^3 - 2*a^6*b^5 + 2*a^2*b^9 - b^11 - (a^6*b^5 - 3*a^4*b^7 + 3*a^2*
b^9 - b^11)*cos(x)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*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)**3/(a+b*sin(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.64813, size = 316, normalized size = 2.19 \begin{align*} -\frac{{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a b^{4}\right )}{\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^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt{a^{2} - b^{2}}} + \frac{a^{4} b \tan \left (\frac{1}{2} \, x\right )^{3} - 4 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, a^{5} \tan \left (\frac{1}{2} \, x\right )^{2} - a^{3} b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 10 \, a b^{4} \tan \left (\frac{1}{2} \, x\right )^{2} + 7 \, a^{4} b \tan \left (\frac{1}{2} \, x\right ) - 16 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right ) + 2 \, a^{5} - 5 \, a^{3} b^{2}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}^{2}} + \frac{x}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^5 - 5*a^3*b^2 + 6*a*b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/
((a^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(a^2 - b^2)) + (a^4*b*tan(1/2*x)^3 - 4*a^2*b^3*tan(1/2*x)^3 + 2*a^5*tan(1/2*x
)^2 - a^3*b^2*tan(1/2*x)^2 - 10*a*b^4*tan(1/2*x)^2 + 7*a^4*b*tan(1/2*x) - 16*a^2*b^3*tan(1/2*x) + 2*a^5 - 5*a^
3*b^2)/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2) + x/b^3