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

Optimal. Leaf size=47 \[ -\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}-\frac {2 \cos (x)}{a^2 (\sin (x)+1)}+\frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2} \]

[Out]

-2*x/a^2-4/3*cos(x)/a^2-2*cos(x)/a^2/(1+sin(x))+1/3*cos(x)*sin(x)^2/(a+a*sin(x))^2

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Rubi [A]  time = 0.14, antiderivative size = 47, 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 = {2765, 2968, 3023, 12, 2735, 2648} \[ -\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}-\frac {2 \cos (x)}{a^2 (\sin (x)+1)}+\frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + a*Sin[x])^2,x]

[Out]

(-2*x)/a^2 - (4*Cos[x])/(3*a^2) - (2*Cos[x])/(a^2*(1 + Sin[x])) + (Cos[x]*Sin[x]^2)/(3*(a + a*Sin[x])^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3023

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

Rubi steps

\begin {align*} \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx &=\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2}-\frac {\int \frac {\sin (x) (2 a-4 a \sin (x))}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2}-\frac {\int \frac {2 a \sin (x)-4 a \sin ^2(x)}{a+a \sin (x)} \, dx}{3 a^2}\\ &=-\frac {4 \cos (x)}{3 a^2}+\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2}-\frac {\int \frac {6 a^2 \sin (x)}{a+a \sin (x)} \, dx}{3 a^3}\\ &=-\frac {4 \cos (x)}{3 a^2}+\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2}-\frac {2 \int \frac {\sin (x)}{a+a \sin (x)} \, dx}{a}\\ &=-\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}+\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2}+\frac {2 \int \frac {1}{a+a \sin (x)} \, dx}{a}\\ &=-\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}+\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2}-\frac {2 \cos (x)}{a^2+a^2 \sin (x)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 84, normalized size = 1.79 \[ -\frac {\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \left (6 (6 x-5) \cos \left (\frac {x}{2}\right )+(41-12 x) \cos \left (\frac {3 x}{2}\right )-3 \cos \left (\frac {5 x}{2}\right )+6 \sin \left (\frac {x}{2}\right ) (8 x+4 (x+1) \cos (x)+\cos (2 x)-9)\right )}{12 a^2 (\sin (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((Cos[x/2] + Sin[x/2])*(6*(-5 + 6*x)*Cos[x/2] + (41 - 12*x)*Cos[(3*x)/2] - 3*Cos[(5*x)/2] + 6*(-9 + 8*x
+ 4*(1 + x)*Cos[x] + Cos[2*x])*Sin[x/2]))/(a^2*(1 + Sin[x])^2)

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fricas [B]  time = 0.46, size = 95, normalized size = 2.02 \[ -\frac {{\left (6 \, x - 11\right )} \cos \relax (x)^{2} + 3 \, \cos \relax (x)^{3} - {\left (6 \, x + 13\right )} \cos \relax (x) - {\left (2 \, {\left (3 \, x + 7\right )} \cos \relax (x) + 3 \, \cos \relax (x)^{2} + 12 \, x + 1\right )} \sin \relax (x) - 12 \, x + 1}{3 \, {\left (a^{2} \cos \relax (x)^{2} - a^{2} \cos \relax (x) - 2 \, a^{2} - {\left (a^{2} \cos \relax (x) + 2 \, a^{2}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((6*x - 11)*cos(x)^2 + 3*cos(x)^3 - (6*x + 13)*cos(x) - (2*(3*x + 7)*cos(x) + 3*cos(x)^2 + 12*x + 1)*sin(
x) - 12*x + 1)/(a^2*cos(x)^2 - a^2*cos(x) - 2*a^2 - (a^2*cos(x) + 2*a^2)*sin(x))

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giac [A]  time = 0.38, size = 51, normalized size = 1.09 \[ -\frac {2 \, x}{a^{2}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} a^{2}} - \frac {2 \, {\left (6 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, x\right ) + 7\right )}}{3 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x/a^2 - 2/((tan(1/2*x)^2 + 1)*a^2) - 2/3*(6*tan(1/2*x)^2 + 15*tan(1/2*x) + 7)/(a^2*(tan(1/2*x) + 1)^3)

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maple [A]  time = 0.10, size = 66, normalized size = 1.40 \[ -\frac {2}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}-\frac {4 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}+\frac {4}{3 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {4}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.87, size = 144, normalized size = 3.06 \[ -\frac {4 \, {\left (\frac {12 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {11 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {9 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + 5\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {4 \, a^{2} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {4 \, a^{2} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {a^{2} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}}\right )}} - \frac {4 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(12*sin(x)/(cos(x) + 1) + 11*sin(x)^2/(cos(x) + 1)^2 + 9*sin(x)^3/(cos(x) + 1)^3 + 3*sin(x)^4/(cos(x) + 1
)^4 + 5)/(a^2 + 3*a^2*sin(x)/(cos(x) + 1) + 4*a^2*sin(x)^2/(cos(x) + 1)^2 + 4*a^2*sin(x)^3/(cos(x) + 1)^3 + 3*
a^2*sin(x)^4/(cos(x) + 1)^4 + a^2*sin(x)^5/(cos(x) + 1)^5) - 4*arctan(sin(x)/(cos(x) + 1))/a^2

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mupad [B]  time = 6.46, size = 62, normalized size = 1.32 \[ -\frac {2\,x}{a^2}-\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+12\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {44\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{3}+16\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {20}{3}}{a^2\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a + a*sin(x))^2,x)

[Out]

- (2*x)/a^2 - (16*tan(x/2) + (44*tan(x/2)^2)/3 + 12*tan(x/2)^3 + 4*tan(x/2)^4 + 20/3)/(a^2*(tan(x/2)^2 + 1)*(t
an(x/2) + 1)^3)

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sympy [B]  time = 6.79, size = 779, normalized size = 16.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x*tan(x/2)**5/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2
*tan(x/2) + 3*a**2) - 18*x*tan(x/2)**4/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**
2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 24*x*tan(x/2)**3/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**
2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 24*x*tan(x/2)**2/(3*a**2*tan(x/2)**5 + 9*a**
2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 18*x*tan(x/2)/(3*a**2*
tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 6*x
/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a*
*2) - 12*tan(x/2)**4/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*
a**2*tan(x/2) + 3*a**2) - 36*tan(x/2)**3/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a
**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 44*tan(x/2)**2/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**
2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 48*tan(x/2)/(3*a**2*tan(x/2)**5 + 9*a**2*tan
(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 20/(3*a**2*tan(x/2)**5 + 9*
a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2)

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