\(\int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx\) [13]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}-\frac {2 \cos (x)}{a^2 (1+\sin (x))}+\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2} \]

[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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2844, 3047, 3102, 12, 2814, 2727} \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}-\frac {2 \cos (x)}{a^2 (\sin (x)+1)}+\frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2} \]

[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 2727

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 2814

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 2844

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 3047

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 3102

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.79 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right ) \left (6 (-5+6 x) \cos \left (\frac {x}{2}\right )+(41-12 x) \cos \left (\frac {3 x}{2}\right )-3 \cos \left (\frac {5 x}{2}\right )+6 (-9+8 x+4 (1+x) \cos (x)+\cos (2 x)) \sin \left (\frac {x}{2}\right )\right )}{12 a^2 (1+\sin (x))^2} \]

[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)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79

method	result	size
parallelrisch	\(\frac {-2 \tan \left (x \right ) \left (\sec ^{2}\left (x \right )\right )+2 \left (\sec ^{3}\left (x \right )\right )-3 \cos \left (x \right )+8 \tan \left (x \right )-9 \sec \left (x \right )-6 x -10}{3 a^{2}}\)	\(37\)
default	\(\frac {\frac {4}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {4}{\tan \left (\frac {x}{2}\right )+1}-\frac {2}{1+\tan ^{2}\left (\frac {x}{2}\right )}-4 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\)	\(56\)
risch	\(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i x}}{2 a^{2}}-\frac {{\mathrm e}^{-i x}}{2 a^{2}}-\frac {2 \left (15 i {\mathrm e}^{i x}+9 \,{\mathrm e}^{2 i x}-8\right )}{3 a^{2} \left ({\mathrm e}^{i x}+i\right )^{3}}\)	\(60\)
norman	\(\frac {-\frac {16 \tan \left (\frac {x}{2}\right )}{a}-\frac {4 \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{a}-\frac {2 x}{a}-\frac {20}{3 a}-\frac {6 x \tan \left (\frac {x}{2}\right )}{a}-\frac {12 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a}-\frac {20 x \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{a}-\frac {24 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a}-\frac {24 x \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{a}-\frac {20 x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{a}-\frac {12 x \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{a}-\frac {6 x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{a}-\frac {12 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{a}-\frac {28 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a}-\frac {40 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a}-\frac {40 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{a}-\frac {44 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{a}-\frac {68 \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{3 a}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{3} a \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}\)	\(227\)

[In]

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

[Out]

1/3*(-2*tan(x)*sec(x)^2+2*sec(x)^3-3*cos(x)+8*tan(x)-9*sec(x)-6*x-10)/a^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (43) = 86\).

Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.02 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {{\left (6 \, x - 11\right )} \cos \left (x\right )^{2} + 3 \, \cos \left (x\right )^{3} - {\left (6 \, x + 13\right )} \cos \left (x\right ) - {\left (2 \, {\left (3 \, x + 7\right )} \cos \left (x\right ) + 3 \, \cos \left (x\right )^{2} + 12 \, x + 1\right )} \sin \left (x\right ) - 12 \, x + 1}{3 \, {\left (a^{2} \cos \left (x\right )^{2} - a^{2} \cos \left (x\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (x\right ) + 2 \, a^{2}\right )} \sin \left (x\right )\right )}} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 779 vs. \(2 (48) = 96\).

Time = 1.94 (sec) , antiderivative size = 779, normalized size of antiderivative = 16.57 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=\text {Too large to display} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (43) = 86\).

Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 3.06 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {4 \, {\left (\frac {12 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {11 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {9 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + 5\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {4 \, a^{2} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {4 \, a^{2} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}\right )}} - \frac {4 \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2 \, x}{a^{2}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} a^{2}} - \frac {2 \, {\left (6 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, x\right ) + 7\right )}}{3 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2\,x}{a^2}-\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+12\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {44\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{3}+16\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {20}{3}}{a^2\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \]

[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)