\(\int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx\) [186]

Optimal result

Integrand size = 13, antiderivative size = 124 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=-\frac {2 a x}{b^3}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}-\frac {\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))} \] Output:

-2*a*x/b^3+2*a^2*(2*a^2-3*b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/b^ 
3/(a^2-b^2)^(3/2)-(2*a^2-b^2)*cos(x)/b^2/(a^2-b^2)+a^2*cos(x)*sin(x)/b/(a^ 
2-b^2)/(a+b*sin(x))
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.76 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\frac {-2 a x+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+b \cos (x) \left (-1-\frac {a^3}{(a-b) (a+b) (a+b \sin (x))}\right )}{b^3} \] Input:

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

Output:

(-2*a*x + (2*a^2*(2*a^2 - 3*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]]) 
/(a^2 - b^2)^(3/2) + b*Cos[x]*(-1 - a^3/((a - b)*(a + b)*(a + b*Sin[x])))) 
/b^3
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 3271, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^3}{(a+b \sin (x))^2}dx\)

\(\Big \downarrow \) 3271

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {a^2-b \sin (x) a-\left (2 a^2-b^2\right ) \sin ^2(x)}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int \frac {a^2-b \sin (x) a-\left (2 a^2-b^2\right ) \sin (x)^2}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\int \frac {b a^2+2 \left (a^2-b^2\right ) \sin (x) a}{a+b \sin (x)}dx}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\int \frac {b a^2+2 \left (a^2-b^2\right ) \sin (x) a}{a+b \sin (x)}dx}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\frac {2 a x \left (a^2-b^2\right )}{b}-\frac {a^2 \left (2 a^2-3 b^2\right ) \int \frac {1}{a+b \sin (x)}dx}{b}}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\frac {2 a x \left (a^2-b^2\right )}{b}-\frac {a^2 \left (2 a^2-3 b^2\right ) \int \frac {1}{a+b \sin (x)}dx}{b}}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\frac {2 a x \left (a^2-b^2\right )}{b}-\frac {2 a^2 \left (2 a^2-3 b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{b}}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\frac {4 a^2 \left (2 a^2-3 b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{b}+\frac {2 a x \left (a^2-b^2\right )}{b}}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\frac {2 a x \left (a^2-b^2\right )}{b}-\frac {2 a^2 \left (2 a^2-3 b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}}}{b}+\frac {\left (2 a^2-b^2\right ) \cos (x)}{b}}{b \left (a^2-b^2\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[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] + Simp[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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15

method	result	size
default	\(-\frac {4 \left (\frac {b}{2 \tan \left (\frac {x}{2}\right )^{2}+2}+a \arctan \left (\tan \left (\frac {x}{2}\right )\right )\right )}{b^{3}}+\frac {4 a^{2} \left (\frac {-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a b}{2 \left (a^{2}-b^{2}\right )}}{a \tan \left (\frac {x}{2}\right )^{2}+2 b \tan \left (\frac {x}{2}\right )+a}+\frac {\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{3}}\)	\(142\)
risch	\(-\frac {2 a x}{b^{3}}-\frac {{\mathrm e}^{i x}}{2 b^{2}}-\frac {{\mathrm e}^{-i x}}{2 b^{2}}+\frac {2 i a^{3} \left (-i a \,{\mathrm e}^{i x}+b \right )}{b^{3} \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}+\frac {2 i a^{4} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{3}}-\frac {3 i a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}-\frac {2 i a^{4} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{3}}+\frac {3 i a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}\)	\(394\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 483, normalized size of antiderivative = 3.90 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\left [-\frac {{\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (x\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} x + 2 \, {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (x\right ) + 2 \, {\left (2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} x + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \, {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sin \left (x\right )\right )}}, -\frac {{\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (x\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} x + {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (x\right ) + {\left (2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} x + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sin \left (x\right )}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.65 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\frac {2 \, {\left (2 \, a^{4} - 3 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {2 \, a x}{b^{3}} \] Input:

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

Output:

2*(2*a^4 - 3*a^2*b^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2 
*x) + b)/sqrt(a^2 - b^2)))/((a^2*b^3 - b^5)*sqrt(a^2 - b^2)) - 2*(a^2*b*ta 
n(1/2*x)^3 + 2*a^3*tan(1/2*x)^2 - a*b^2*tan(1/2*x)^2 + 3*a^2*b*tan(1/2*x) 
- 2*b^3*tan(1/2*x) + 2*a^3 - a*b^2)/((a*tan(1/2*x)^4 + 2*b*tan(1/2*x)^3 + 
2*a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)*(a^2*b^2 - b^4)) - 2*a*x/b^3
 

Mupad [B] (verification not implemented)

Time = 19.11 (sec) , antiderivative size = 2578, normalized size of antiderivative = 20.79 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(a*b^2 - 2*a^3))/(b^2*(a^2 - b^2)) - (2*a^2*tan(x/2)^3)/(b*(a^2 - b^2) 
) + (2*tan(x/2)^2*(a*b^2 - 2*a^3))/(b^2*(a^2 - b^2)) - (2*tan(x/2)*(3*a^2 
- 2*b^2))/(b*(a^2 - b^2)))/(a + 2*b*tan(x/2) + 2*a*tan(x/2)^2 + a*tan(x/2) 
^4 + 2*b*tan(x/2)^3) - (4*a*atan((512*a^4*b^5*tan(x/2))/((512*a^4*b^14)/(b 
^9 - 2*a^2*b^7 + a^4*b^5) - (1408*a^6*b^12)/(b^9 - 2*a^2*b^7 + a^4*b^5) + 
(1280*a^8*b^10)/(b^9 - 2*a^2*b^7 + a^4*b^5) - (384*a^10*b^8)/(b^9 - 2*a^2* 
b^7 + a^4*b^5)) - (384*a^6*b^3*tan(x/2))/((512*a^4*b^14)/(b^9 - 2*a^2*b^7 
+ a^4*b^5) - (1408*a^6*b^12)/(b^9 - 2*a^2*b^7 + a^4*b^5) + (1280*a^8*b^10) 
/(b^9 - 2*a^2*b^7 + a^4*b^5) - (384*a^10*b^8)/(b^9 - 2*a^2*b^7 + a^4*b^5)) 
))/b^3 - (a^2*atan(((a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32 
*(4*a^4*b^6 - 8*a^6*b^4 + 4*a^8*b^2))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*ta 
n(x/2)*(8*a^3*b^8 - 29*a^5*b^6 + 28*a^7*b^4 - 8*a^9*b^2))/(b^10 - 2*a^2*b^ 
8 + a^4*b^6) + (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(2*a 
^2*b^10 - 3*a^4*b^8 + a^6*b^6))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2) 
*(6*a^3*b^10 - 10*a^5*b^8 + 4*a^7*b^6))/(b^10 - 2*a^2*b^8 + a^4*b^6) + (a^ 
2*((32*(a^2*b^12 - 2*a^4*b^10 + a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (3 
2*tan(x/2)*(3*a*b^14 - 8*a^3*b^12 + 7*a^5*b^10 - 2*a^7*b^8))/(b^10 - 2*a^2 
*b^8 + a^4*b^6))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(b^9 - 3*a^ 
2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*1i 
)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + (a^2*(2*a^2 - 3*b^2)*(-(a +...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 365, normalized size of antiderivative = 2.94 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx=\frac {4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a^{4} b -6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a^{2} b^{3}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{5}-6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} b^{2}-\cos \left (x \right ) \sin \left (x \right ) a^{4} b^{2}+2 \cos \left (x \right ) \sin \left (x \right ) a^{2} b^{4}-\cos \left (x \right ) \sin \left (x \right ) b^{6}-2 \cos \left (x \right ) a^{5} b +3 \cos \left (x \right ) a^{3} b^{3}-\cos \left (x \right ) a \,b^{5}-2 \sin \left (x \right ) a^{5} b x -\sin \left (x \right ) a^{4} b^{2}+4 \sin \left (x \right ) a^{3} b^{3} x +2 \sin \left (x \right ) a^{2} b^{4}-2 \sin \left (x \right ) a \,b^{5} x -\sin \left (x \right ) b^{6}-2 a^{6} x -a^{5} b +4 a^{4} b^{2} x +2 a^{3} b^{3}-2 a^{2} b^{4} x -a \,b^{5}}{b^{3} \left (\sin \left (x \right ) a^{4} b -2 \sin \left (x \right ) a^{2} b^{3}+\sin \left (x \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

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

Output:

(4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)*a**4* 
b - 6*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)*a* 
*2*b**3 + 4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a** 
5 - 6*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**3*b**2 
 - cos(x)*sin(x)*a**4*b**2 + 2*cos(x)*sin(x)*a**2*b**4 - cos(x)*sin(x)*b** 
6 - 2*cos(x)*a**5*b + 3*cos(x)*a**3*b**3 - cos(x)*a*b**5 - 2*sin(x)*a**5*b 
*x - sin(x)*a**4*b**2 + 4*sin(x)*a**3*b**3*x + 2*sin(x)*a**2*b**4 - 2*sin( 
x)*a*b**5*x - sin(x)*b**6 - 2*a**6*x - a**5*b + 4*a**4*b**2*x + 2*a**3*b** 
3 - 2*a**2*b**4*x - a*b**5)/(b**3*(sin(x)*a**4*b - 2*sin(x)*a**2*b**3 + si 
n(x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))