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+1/5*cos(x)*sin(x)^2/(a+a*sin(x))^3-7/15*cos(x)/a/(a+a*sin(x))^2+29/15*cos(x)/(a^3+a^3*sin(x))

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Rubi [A]  time = 0.16, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.18, size = 112, normalized size = 1.90 \[ \frac {\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \left (150 x \sin \left (\frac {x}{2}\right )-370 \sin \left (\frac {x}{2}\right )+75 x \sin \left (\frac {3 x}{2}\right )-90 \sin \left (\frac {3 x}{2}\right )-15 x \sin \left (\frac {5 x}{2}\right )+64 \sin \left (\frac {5 x}{2}\right )+30 (5 x-9) \cos \left (\frac {x}{2}\right )+(230-75 x) \cos \left (\frac {3 x}{2}\right )-15 x \cos \left (\frac {5 x}{2}\right )\right )}{60 a^3 (\sin (x)+1)^3} \]

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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fricas [B]  time = 0.44, size = 119, normalized size = 2.02 \[ \frac {{\left (15 \, x + 32\right )} \cos \relax (x)^{3} + {\left (45 \, x - 19\right )} \cos \relax (x)^{2} - 6 \, {\left (5 \, x + 9\right )} \cos \relax (x) + {\left ({\left (15 \, x - 32\right )} \cos \relax (x)^{2} - 3 \, {\left (10 \, x + 17\right )} \cos \relax (x) - 60 \, x + 3\right )} \sin \relax (x) - 60 \, x - 3}{15 \, {\left (a^{3} \cos \relax (x)^{3} + 3 \, a^{3} \cos \relax (x)^{2} - 2 \, a^{3} \cos \relax (x) - 4 \, a^{3} + {\left (a^{3} \cos \relax (x)^{2} - 2 \, a^{3} \cos \relax (x) - 4 \, a^{3}\right )} \sin \relax (x)\right )}} \]

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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giac [A]  time = 0.33, size = 51, normalized size = 0.86 \[ \frac {x}{a^{3}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 75 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 145 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 95 \, \tan \left (\frac {1}{2} \, x\right ) + 22\right )}}{15 \, a^{3} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{5}} \]

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)

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maple [A]  time = 0.10, size = 77, normalized size = 1.31 \[ \frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}-\frac {4}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {8}{5 a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {4}{3 a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

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 = 1.02, size = 144, normalized size = 2.44 \[ \frac {2 \, {\left (\frac {95 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {145 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {75 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {15 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + 22\right )}}{15 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {10 \, a^{3} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {a^{3} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}}\right )}} + \frac {2 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{3}} \]

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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mupad [B]  time = 6.74, size = 50, normalized size = 0.85 \[ \frac {x}{a^3}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {58\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{3}+\frac {38\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+\frac {44}{15}}{a^3\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a + a*sin(x))^3,x)

[Out]

x/a^3 + ((38*tan(x/2))/3 + (58*tan(x/2)^2)/3 + 10*tan(x/2)^3 + 2*tan(x/2)^4 + 44/15)/(a^3*(tan(x/2) + 1)^5)

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sympy [B]  time = 11.82, size = 777, normalized size = 13.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*x*tan(x/2)**5/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 75
*a**3*tan(x/2) + 15*a**3) + 75*x*tan(x/2)**4/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3
+ 150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3) + 150*x*tan(x/2)**3/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x
/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3) + 150*x*tan(x/2)**2/(15*a**
3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3
) + 75*x*tan(x/2)/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 7
5*a**3*tan(x/2) + 15*a**3) + 15*x/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3
*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3) + 30*tan(x/2)**4/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a
**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3) + 150*tan(x/2)**3/(15*a**3*tan(x/2)**5 +
75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3) + 290*tan(x/2)
**2/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2
) + 15*a**3) + 190*tan(x/2)/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 + 150*a**3*tan(x
/2)**2 + 75*a**3*tan(x/2) + 15*a**3) + 44/(15*a**3*tan(x/2)**5 + 75*a**3*tan(x/2)**4 + 150*a**3*tan(x/2)**3 +
150*a**3*tan(x/2)**2 + 75*a**3*tan(x/2) + 15*a**3)

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