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

Optimal. Leaf size=59 \[ \frac{x}{a^3}+\frac{29 \cos (x)}{15 \left (a^3 \sin (x)+a^3\right )}+\frac{\sin ^2(x) \cos (x)}{5 (a \sin (x)+a)^3}-\frac{7 \cos (x)}{15 a (a \sin (x)+a)^2} \]

[Out]

x/a^3 + (Cos[x]*Sin[x]^2)/(5*(a + a*Sin[x])^3) - (7*Cos[x])/(15*a*(a + a*Sin[x])^2) + (29*Cos[x])/(15*(a^3 + a
^3*Sin[x]))

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Rubi [A]  time = 0.156979, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2765, 2968, 3019, 2735, 2648} \[ \frac{x}{a^3}+\frac{29 \cos (x)}{15 \left (a^3 \sin (x)+a^3\right )}+\frac{\sin ^2(x) \cos (x)}{5 (a \sin (x)+a)^3}-\frac{7 \cos (x)}{15 a (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + a*Sin[x])^3,x]

[Out]

x/a^3 + (Cos[x]*Sin[x]^2)/(5*(a + a*Sin[x])^3) - (7*Cos[x])/(15*a*(a + a*Sin[x])^2) + (29*Cos[x])/(15*(a^3 + a
^3*Sin[x]))

Rule 2765

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

Rule 2968

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

Rule 3019

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 - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[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 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 ^3(x)}{(a+a \sin (x))^3} \, dx &=\frac{\cos (x) \sin ^2(x)}{5 (a+a \sin (x))^3}-\frac{\int \frac{\sin (x) (2 a-5 a \sin (x))}{(a+a \sin (x))^2} \, dx}{5 a^2}\\ &=\frac{\cos (x) \sin ^2(x)}{5 (a+a \sin (x))^3}-\frac{\int \frac{2 a \sin (x)-5 a \sin ^2(x)}{(a+a \sin (x))^2} \, dx}{5 a^2}\\ &=\frac{\cos (x) \sin ^2(x)}{5 (a+a \sin (x))^3}-\frac{7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac{\int \frac{-14 a^2+15 a^2 \sin (x)}{a+a \sin (x)} \, dx}{15 a^4}\\ &=\frac{x}{a^3}+\frac{\cos (x) \sin ^2(x)}{5 (a+a \sin (x))^3}-\frac{7 \cos (x)}{15 a (a+a \sin (x))^2}-\frac{29 \int \frac{1}{a+a \sin (x)} \, dx}{15 a^2}\\ &=\frac{x}{a^3}+\frac{\cos (x) \sin ^2(x)}{5 (a+a \sin (x))^3}-\frac{7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac{29 \cos (x)}{15 \left (a^3+a^3 \sin (x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.176702, size = 112, normalized size = 1.9 \[ \frac{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \left (150 x \sin \left (\frac{x}{2}\right )-370 \sin \left (\frac{x}{2}\right )+75 x \sin \left (\frac{3 x}{2}\right )-90 \sin \left (\frac{3 x}{2}\right )-15 x \sin \left (\frac{5 x}{2}\right )+64 \sin \left (\frac{5 x}{2}\right )+30 (5 x-9) \cos \left (\frac{x}{2}\right )+(230-75 x) \cos \left (\frac{3 x}{2}\right )-15 x \cos \left (\frac{5 x}{2}\right )\right )}{60 a^3 (\sin (x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[x/2] + Sin[x/2])*(30*(-9 + 5*x)*Cos[x/2] + (230 - 75*x)*Cos[(3*x)/2] - 15*x*Cos[(5*x)/2] - 370*Sin[x/2]
+ 150*x*Sin[x/2] - 90*Sin[(3*x)/2] + 75*x*Sin[(3*x)/2] + 64*Sin[(5*x)/2] - 15*x*Sin[(5*x)/2]))/(60*a^3*(1 + Si
n[x])^3)

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Maple [A]  time = 0.043, size = 77, normalized size = 1.3 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{a}^{3}}}-4\,{\frac{1}{{a}^{3} \left ( \tan \left ( x/2 \right ) +1 \right ) ^{4}}}+{\frac{8}{5\,{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-5}}+{\frac{4}{3\,{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{{a}^{3} \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}+2\,{\frac{1}{{a}^{3} \left ( \tan \left ( x/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a^3*arctan(tan(1/2*x))-4/a^3/(tan(1/2*x)+1)^4+8/5/a^3/(tan(1/2*x)+1)^5+4/3/a^3/(tan(1/2*x)+1)^3+2/a^3/(tan(1
/2*x)+1)^2+2/a^3/(tan(1/2*x)+1)

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Maxima [B]  time = 2.55501, size = 194, normalized size = 3.29 \begin{align*} \frac{2 \,{\left (\frac{95 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{145 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{75 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + 22\right )}}{15 \,{\left (a^{3} + \frac{5 \, a^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{10 \, a^{3} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}\right )}} + \frac{2 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(95*sin(x)/(cos(x) + 1) + 145*sin(x)^2/(cos(x) + 1)^2 + 75*sin(x)^3/(cos(x) + 1)^3 + 15*sin(x)^4/(cos(x)
+ 1)^4 + 22)/(a^3 + 5*a^3*sin(x)/(cos(x) + 1) + 10*a^3*sin(x)^2/(cos(x) + 1)^2 + 10*a^3*sin(x)^3/(cos(x) + 1)^
3 + 5*a^3*sin(x)^4/(cos(x) + 1)^4 + a^3*sin(x)^5/(cos(x) + 1)^5) + 2*arctan(sin(x)/(cos(x) + 1))/a^3

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Fricas [B]  time = 1.42613, size = 340, normalized size = 5.76 \begin{align*} \frac{{\left (15 \, x + 32\right )} \cos \left (x\right )^{3} +{\left (45 \, x - 19\right )} \cos \left (x\right )^{2} - 6 \,{\left (5 \, x + 9\right )} \cos \left (x\right ) +{\left ({\left (15 \, x - 32\right )} \cos \left (x\right )^{2} - 3 \,{\left (10 \, x + 17\right )} \cos \left (x\right ) - 60 \, x + 3\right )} \sin \left (x\right ) - 60 \, x - 3}{15 \,{\left (a^{3} \cos \left (x\right )^{3} + 3 \, a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3} +{\left (a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*x + 32)*cos(x)^3 + (45*x - 19)*cos(x)^2 - 6*(5*x + 9)*cos(x) + ((15*x - 32)*cos(x)^2 - 3*(10*x + 17)
*cos(x) - 60*x + 3)*sin(x) - 60*x - 3)/(a^3*cos(x)^3 + 3*a^3*cos(x)^2 - 2*a^3*cos(x) - 4*a^3 + (a^3*cos(x)^2 -
2*a^3*cos(x) - 4*a^3)*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+a*sin(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.23968, size = 69, normalized size = 1.17 \begin{align*} \frac{x}{a^{3}} + \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, x\right )^{4} + 75 \, \tan \left (\frac{1}{2} \, x\right )^{3} + 145 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 95 \, \tan \left (\frac{1}{2} \, x\right ) + 22\right )}}{15 \, a^{3}{\left (\tan \left (\frac{1}{2} \, x\right ) + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a^3 + 2/15*(15*tan(1/2*x)^4 + 75*tan(1/2*x)^3 + 145*tan(1/2*x)^2 + 95*tan(1/2*x) + 22)/(a^3*(tan(1/2*x) + 1)
^5)