3.2.95 \(\int \frac {\sin ^3(x)}{(a+b \sin (x))^3} \, dx\) [195]

3.2.95.1 Optimal result
3.2.95.2 Mathematica [A] (verified)
3.2.95.3 Rubi [A] (verified)
3.2.95.4 Maple [A] (verified)
3.2.95.5 Fricas [B] (verification not implemented)
3.2.95.6 Sympy [F(-1)]
3.2.95.7 Maxima [F(-2)]
3.2.95.8 Giac [A] (verification not implemented)
3.2.95.9 Mupad [B] (verification not implemented)

3.2.95.1 Optimal result

Integrand size = 13, antiderivative size = 144 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^3} \, dx=\frac {x}{b^3}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{5/2}}+\frac {a^2 \cos (x) \sin (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \]

output
x/b^3-a*(2*a^4-5*a^2*b^2+6*b^4)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/b 
^3/(a^2-b^2)^(5/2)+1/2*a^2*cos(x)*sin(x)/b/(a^2-b^2)/(a+b*sin(x))^2+1/2*a^ 
2*(2*a^2-5*b^2)*cos(x)/b^2/(a^2-b^2)^2/(a+b*sin(x))
 
3.2.95.2 Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^3} \, dx=\frac {2 x-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {a^3 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}+\frac {3 a^2 b \left (a^2-2 b^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}}{2 b^3} \]

input
Integrate[Sin[x]^3/(a + b*Sin[x])^3,x]
 
output
(2*x - (2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - 
 b^2]])/(a^2 - b^2)^(5/2) - (a^3*b*Cos[x])/((a - b)*(a + b)*(a + b*Sin[x]) 
^2) + (3*a^2*b*(a^2 - 2*b^2)*Cos[x])/((a - b)^2*(a + b)^2*(a + b*Sin[x]))) 
/(2*b^3)
 
3.2.95.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.36, 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, 3500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\int \frac {2 \sin (x) \left (a^2-b^2\right )^2+a b \left (a^2-4 b^2\right )}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\int \frac {2 \sin (x) \left (a^2-b^2\right )^2+a b \left (a^2-4 b^2\right )}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{a+b \sin (x)}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{a+b \sin (x)}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\frac {4 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{b}+\frac {2 x \left (a^2-b^2\right )^2}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a^2 \sin (x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {a^2 \left (2 a^2-5 b^2\right ) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}\)

input
Int[Sin[x]^3/(a + b*Sin[x])^3,x]
 
output
(a^2*Cos[x]*Sin[x])/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2) - (-(((2*(a^2 - b^2 
)^2*x)/b - (2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[(2*b + 2*a*Tan[x/2])/(2 
*Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]))/(b*(a^2 - b^2))) - (a^2*(2*a^2 - 
5*b^2)*Cos[x])/(b*(a^2 - b^2)*(a + b*Sin[x])))/(2*b*(a^2 - b^2))
 

3.2.95.3.1 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 3500
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^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 
3.2.95.4 Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.87

method result size
default \(\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{3}}-\frac {2 a \left (\frac {-\frac {a \,b^{2} \left (a^{2}-4 b^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (2 a^{4}-a^{2} b^{2}-10 b^{4}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {a \,b^{2} \left (7 a^{2}-16 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {a^{2} b \left (2 a^{2}-5 b^{2}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}}{{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{4}-5 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\right )}{b^{3}}\) \(269\)
risch \(\frac {x}{b^{3}}-\frac {i a^{2} \left (-4 i a^{3} b \,{\mathrm e}^{3 i x}+7 i a \,b^{3} {\mathrm e}^{3 i x}+8 i a^{3} b \,{\mathrm e}^{i x}-17 i a \,b^{3} {\mathrm e}^{i x}+6 a^{4} {\mathrm e}^{2 i x}-9 a^{2} b^{2} {\mathrm e}^{2 i x}-6 b^{4} {\mathrm e}^{2 i x}-3 a^{2} b^{2}+6 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2} b^{3}}+\frac {i a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{3}}-\frac {5 i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b}+\frac {3 i a b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {i a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{3}}+\frac {5 i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b}-\frac {3 i a b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\) \(606\)

input
int(sin(x)^3/(a+b*sin(x))^3,x,method=_RETURNVERBOSE)
 
output
2/b^3*arctan(tan(1/2*x))-2/b^3*a*((-1/2*a*b^2*(a^2-4*b^2)/(a^4-2*a^2*b^2+b 
^4)*tan(1/2*x)^3-1/2*b*(2*a^4-a^2*b^2-10*b^4)/(a^4-2*a^2*b^2+b^4)*tan(1/2* 
x)^2-1/2*a*b^2*(7*a^2-16*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-1/2*a^2*b*(2* 
a^2-5*b^2)/(a^4-2*a^2*b^2+b^4))/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^2+1/2*(2 
*a^4-5*a^2*b^2+6*b^4)/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a* 
tan(1/2*x)+2*b)/(a^2-b^2)^(1/2)))
 
3.2.95.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (134) = 268\).

Time = 0.36 (sec) , antiderivative size = 819, normalized size of antiderivative = 5.69 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^3} \, dx=\left [-\frac {4 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} x \cos \left (x\right )^{2} + {\left (2 \, a^{7} - 3 \, a^{5} b^{2} + a^{3} b^{4} + 6 \, a b^{6} - {\left (2 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (x\right )^{2} + 2 \, {\left (2 \, a^{6} b - 5 \, a^{4} b^{3} + 6 \, a^{2} b^{5}\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^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} x - 2 \, {\left (2 \, a^{7} b - 7 \, a^{5} b^{3} + 5 \, a^{3} b^{5}\right )} \cos \left (x\right ) - 2 \, {\left (4 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} x + 3 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 2 \, a^{2} b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{4 \, {\left (a^{8} b^{3} - 2 \, a^{6} b^{5} + 2 \, a^{2} b^{9} - b^{11} - {\left (a^{6} b^{5} - 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} - b^{11}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b^{4} - 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} - a b^{10}\right )} \sin \left (x\right )\right )}}, -\frac {2 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} x \cos \left (x\right )^{2} - {\left (2 \, a^{7} - 3 \, a^{5} b^{2} + a^{3} b^{4} + 6 \, a b^{6} - {\left (2 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (x\right )^{2} + 2 \, {\left (2 \, a^{6} b - 5 \, a^{4} b^{3} + 6 \, a^{2} b^{5}\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^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} x - {\left (2 \, a^{7} b - 7 \, a^{5} b^{3} + 5 \, a^{3} b^{5}\right )} \cos \left (x\right ) - {\left (4 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} x + 3 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 2 \, a^{2} b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \, {\left (a^{8} b^{3} - 2 \, a^{6} b^{5} + 2 \, a^{2} b^{9} - b^{11} - {\left (a^{6} b^{5} - 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} - b^{11}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b^{4} - 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} - a b^{10}\right )} \sin \left (x\right )\right )}}\right ] \]

input
integrate(sin(x)^3/(a+b*sin(x))^3,x, algorithm="fricas")
 
output
[-1/4*(4*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*x*cos(x)^2 + (2*a^7 - 3*a 
^5*b^2 + a^3*b^4 + 6*a*b^6 - (2*a^5*b^2 - 5*a^3*b^4 + 6*a*b^6)*cos(x)^2 + 
2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*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^8 - 
2*a^6*b^2 + 2*a^2*b^6 - b^8)*x - 2*(2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*cos(x 
) - 2*(4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*x + 3*(a^6*b^2 - 3*a^4*b^ 
4 + 2*a^2*b^6)*cos(x))*sin(x))/(a^8*b^3 - 2*a^6*b^5 + 2*a^2*b^9 - b^11 - ( 
a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*cos(x)^2 + 2*(a^7*b^4 - 3*a^5*b^6 
+ 3*a^3*b^8 - a*b^10)*sin(x)), -1/2*(2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - 
b^8)*x*cos(x)^2 - (2*a^7 - 3*a^5*b^2 + a^3*b^4 + 6*a*b^6 - (2*a^5*b^2 - 5* 
a^3*b^4 + 6*a*b^6)*cos(x)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*sin(x))* 
sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - 2*(a^8 
- 2*a^6*b^2 + 2*a^2*b^6 - b^8)*x - (2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*cos(x 
) - (4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*x + 3*(a^6*b^2 - 3*a^4*b^4 
+ 2*a^2*b^6)*cos(x))*sin(x))/(a^8*b^3 - 2*a^6*b^5 + 2*a^2*b^9 - b^11 - (a^ 
6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*cos(x)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 
3*a^3*b^8 - a*b^10)*sin(x))]
 
3.2.95.6 Sympy [F(-1)]

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

input
integrate(sin(x)**3/(a+b*sin(x))**3,x)
 
output
Timed out
 
3.2.95.7 Maxima [F(-2)]

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

input
integrate(sin(x)^3/(a+b*sin(x))^3,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
 
3.2.95.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.62 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^3} \, dx=-\frac {{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{4} b \tan \left (\frac {1}{2} \, x\right )^{3} - 4 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a^{5} \tan \left (\frac {1}{2} \, x\right )^{2} - a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 10 \, a b^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 7 \, a^{4} b \tan \left (\frac {1}{2} \, x\right ) - 16 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, a^{5} - 5 \, a^{3} b^{2}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} + \frac {x}{b^{3}} \]

input
integrate(sin(x)^3/(a+b*sin(x))^3,x, algorithm="giac")
 
output
-(2*a^5 - 5*a^3*b^2 + 6*a*b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan(( 
a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(a^2 
- b^2)) + (a^4*b*tan(1/2*x)^3 - 4*a^2*b^3*tan(1/2*x)^3 + 2*a^5*tan(1/2*x)^ 
2 - a^3*b^2*tan(1/2*x)^2 - 10*a*b^4*tan(1/2*x)^2 + 7*a^4*b*tan(1/2*x) - 16 
*a^2*b^3*tan(1/2*x) + 2*a^5 - 5*a^3*b^2)/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*t 
an(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2) + x/b^3
 
3.2.95.9 Mupad [B] (verification not implemented)

Time = 14.84 (sec) , antiderivative size = 5756, normalized size of antiderivative = 39.97 \[ \int \frac {\sin ^3(x)}{(a+b \sin (x))^3} \, dx=\text {Too large to display} \]

input
int(sin(x)^3/(a + b*sin(x))^3,x)
 
output
(2*atan((((8*(4*a^2*b^10 - 16*a^4*b^8 + 24*a^6*b^6 - 16*a^8*b^4 + 4*a^10*b 
^2))/(b^13 - 4*a^2*b^11 + 6*a^4*b^9 - 4*a^6*b^7 + a^8*b^5) - (((8*(4*a*b^1 
4 - 8*a^3*b^12 + 6*a^5*b^10 - 4*a^7*b^8 + 2*a^9*b^6))/(b^13 - 4*a^2*b^11 + 
 6*a^4*b^9 - 4*a^6*b^7 + a^8*b^5) - (((8*(4*a^2*b^16 - 16*a^4*b^14 + 24*a^ 
6*b^12 - 16*a^8*b^10 + 4*a^10*b^8))/(b^13 - 4*a^2*b^11 + 6*a^4*b^9 - 4*a^6 
*b^7 + a^8*b^5) + (8*tan(x/2)*(12*a*b^18 - 56*a^3*b^16 + 104*a^5*b^14 - 96 
*a^7*b^12 + 44*a^9*b^10 - 8*a^11*b^8))/(b^14 - 4*a^2*b^12 + 6*a^4*b^10 - 4 
*a^6*b^8 + a^8*b^6))*1i)/b^3 + (8*tan(x/2)*(24*a^2*b^14 - 68*a^4*b^12 + 72 
*a^6*b^10 - 36*a^8*b^8 + 8*a^10*b^6))/(b^14 - 4*a^2*b^12 + 6*a^4*b^10 - 4* 
a^6*b^8 + a^8*b^6))*1i)/b^3 + (8*tan(x/2)*(8*a*b^12 - 72*a^3*b^10 + 124*a^ 
5*b^8 - 105*a^7*b^6 + 44*a^9*b^4 - 8*a^11*b^2))/(b^14 - 4*a^2*b^12 + 6*a^4 
*b^10 - 4*a^6*b^8 + a^8*b^6))/b^3 + ((8*(4*a^2*b^10 - 16*a^4*b^8 + 24*a^6* 
b^6 - 16*a^8*b^4 + 4*a^10*b^2))/(b^13 - 4*a^2*b^11 + 6*a^4*b^9 - 4*a^6*b^7 
 + a^8*b^5) + (((((8*(4*a^2*b^16 - 16*a^4*b^14 + 24*a^6*b^12 - 16*a^8*b^10 
 + 4*a^10*b^8))/(b^13 - 4*a^2*b^11 + 6*a^4*b^9 - 4*a^6*b^7 + a^8*b^5) + (8 
*tan(x/2)*(12*a*b^18 - 56*a^3*b^16 + 104*a^5*b^14 - 96*a^7*b^12 + 44*a^9*b 
^10 - 8*a^11*b^8))/(b^14 - 4*a^2*b^12 + 6*a^4*b^10 - 4*a^6*b^8 + a^8*b^6)) 
*1i)/b^3 + (8*(4*a*b^14 - 8*a^3*b^12 + 6*a^5*b^10 - 4*a^7*b^8 + 2*a^9*b^6) 
)/(b^13 - 4*a^2*b^11 + 6*a^4*b^9 - 4*a^6*b^7 + a^8*b^5) + (8*tan(x/2)*(24* 
a^2*b^14 - 68*a^4*b^12 + 72*a^6*b^10 - 36*a^8*b^8 + 8*a^10*b^6))/(b^14 ...