\(\int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx\) [193]

Optimal result

Integrand size = 13, antiderivative size = 243 \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx=\frac {\left (12 a^2+b^2\right ) x}{2 b^5}-\frac {a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^5 \left (a^2-b^2\right )^{5/2}}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{2 b^4 \left (a^2-b^2\right )^2}-\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \cos (x) \sin (x)}{2 b^3 \left (a^2-b^2\right )^2}+\frac {a^2 \cos (x) \sin ^3(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {a^2 \left (4 a^2-7 b^2\right ) \cos (x) \sin ^2(x)}{2 b^2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \] Output:

1/2*(12*a^2+b^2)*x/b^5-a^3*(12*a^4-29*a^2*b^2+20*b^4)*arctan((b+a*tan(1/2* 
x))/(a^2-b^2)^(1/2))/b^5/(a^2-b^2)^(5/2)+3/2*a*(4*a^4-7*a^2*b^2+2*b^4)*cos 
(x)/b^4/(a^2-b^2)^2-1/2*(6*a^4-10*a^2*b^2+b^4)*cos(x)*sin(x)/b^3/(a^2-b^2) 
^2+1/2*a^2*cos(x)*sin(x)^3/b/(a^2-b^2)/(a+b*sin(x))^2+1/2*a^2*(4*a^2-7*b^2 
)*cos(x)*sin(x)^2/b^2/(a^2-b^2)^2/(a+b*sin(x))
 

Mathematica [A] (verified)

Time = 5.94 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.67 \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx=\frac {2 \left (12 a^2+b^2\right ) x-\frac {4 a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+12 a b \cos (x)-\frac {2 a^5 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}+\frac {2 a^4 b \left (7 a^2-10 b^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}-b^2 \sin (2 x)}{4 b^5} \] Input:

Integrate[Sin[x]^5/(a + b*Sin[x])^3,x]
 

Output:

(2*(12*a^2 + b^2)*x - (4*a^3*(12*a^4 - 29*a^2*b^2 + 20*b^4)*ArcTan[(b + a* 
Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + 12*a*b*Cos[x] - (2*a^5*b*C 
os[x])/((a - b)*(a + b)*(a + b*Sin[x])^2) + (2*a^4*b*(7*a^2 - 10*b^2)*Cos[ 
x])/((a - b)^2*(a + b)^2*(a + b*Sin[x])) - b^2*Sin[2*x])/(4*b^5)
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.13, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.231, Rules used = {3042, 3271, 3042, 3526, 25, 3042, 3528, 27, 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.

\(\displaystyle \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^5}{(a+b \sin (x))^3}dx\)

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {-\frac {\int -\frac {\sin (x) \left (2 \left (4 a^2-7 b^2\right ) a^2-b \left (a^2-4 b^2\right ) \sin (x) a-2 \left (6 a^4-10 b^2 a^2+b^4\right ) \sin ^2(x)\right )}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\int \frac {\sin (x) \left (2 \left (4 a^2-7 b^2\right ) a^2-b \left (a^2-4 b^2\right ) \sin (x) a-2 \left (6 a^4-10 b^2 a^2+b^4\right ) \sin ^2(x)\right )}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\int \frac {\sin (x) \left (2 \left (4 a^2-7 b^2\right ) a^2-b \left (a^2-4 b^2\right ) \sin (x) a-2 \left (6 a^4-10 b^2 a^2+b^4\right ) \sin (x)^2\right )}{a+b \sin (x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\int -\frac {2 \left (-3 a \left (4 a^4-7 b^2 a^2+2 b^4\right ) \sin ^2(x)-b \left (2 a^4-4 b^2 a^2-b^4\right ) \sin (x)+a \left (6 a^4-10 b^2 a^2+b^4\right )\right )}{a+b \sin (x)}dx}{2 b}+\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\int \frac {-3 a \left (4 a^4-7 b^2 a^2+2 b^4\right ) \sin ^2(x)-b \left (2 a^4-4 b^2 a^2-b^4\right ) \sin (x)+a \left (6 a^4-10 b^2 a^2+b^4\right )}{a+b \sin (x)}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\int \frac {-3 a \left (4 a^4-7 b^2 a^2+2 b^4\right ) \sin (x)^2-b \left (2 a^4-4 b^2 a^2-b^4\right ) \sin (x)+a \left (6 a^4-10 b^2 a^2+b^4\right )}{a+b \sin (x)}dx}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {\int \frac {\left (12 a^2+b^2\right ) \sin (x) \left (a^2-b^2\right )^2+a b \left (6 a^4-10 b^2 a^2+b^4\right )}{a+b \sin (x)}dx}{b}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {\int \frac {\left (12 a^2+b^2\right ) \sin (x) \left (a^2-b^2\right )^2+a b \left (6 a^4-10 b^2 a^2+b^4\right )}{a+b \sin (x)}dx}{b}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {\frac {x \left (a^2-b^2\right )^2 \left (12 a^2+b^2\right )}{b}-\frac {a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \int \frac {1}{a+b \sin (x)}dx}{b}}{b}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {\frac {x \left (a^2-b^2\right )^2 \left (12 a^2+b^2\right )}{b}-\frac {a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \int \frac {1}{a+b \sin (x)}dx}{b}}{b}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {\frac {x \left (a^2-b^2\right )^2 \left (12 a^2+b^2\right )}{b}-\frac {2 a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{b}}{b}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {\frac {4 a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{b}+\frac {x \left (a^2-b^2\right )^2 \left (12 a^2+b^2\right )}{b}}{b}+\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a^2 \sin ^3(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \sin (x) \cos (x)}{b}-\frac {\frac {3 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (x)}{b}+\frac {\frac {x \left (a^2-b^2\right )^2 \left (12 a^2+b^2\right )}{b}-\frac {2 a^3 \left (12 a^4-29 a^2 b^2+20 b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2}}}{b}}{b}}{b \left (a^2-b^2\right )}-\frac {a^2 \left (4 a^2-7 b^2\right ) \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

Input:

Int[Sin[x]^5/(a + b*Sin[x])^3,x]
 

Output:

(a^2*Cos[x]*Sin[x]^3)/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2) - (-((a^2*(4*a^2 
- 7*b^2)*Cos[x]*Sin[x]^2)/(b*(a^2 - b^2)*(a + b*Sin[x]))) + (-(((((a^2 - b 
^2)^2*(12*a^2 + b^2)*x)/b - (2*a^3*(12*a^4 - 29*a^2*b^2 + 20*b^4)*ArcTan[( 
2*b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]))/b + (3*a*(4 
*a^4 - 7*a^2*b^2 + 2*b^4)*Cos[x])/b)/b) + ((6*a^4 - 10*a^2*b^2 + b^4)*Cos[ 
x]*Sin[x])/b)/(b*(a^2 - b^2)))/(2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 3526

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((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 - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d 
, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.36

method	result	size
default	\(-\frac {4 a^{3} \left (\frac {-\frac {a \,b^{2} \left (5 a^{2}-8 b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}}{4 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {3 b \left (2 a^{4}+b^{2} a^{2}-6 b^{4}\right ) \tan \left (\frac {x}{2}\right )^{2}}{4 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {b^{2} a \left (19 a^{2}-28 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{4 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {3 a^{2} b \left (2 a^{2}-3 b^{2}\right )}{4 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}}{\left (a \tan \left (\frac {x}{2}\right )^{2}+2 b \tan \left (\frac {x}{2}\right )+a \right )^{2}}+\frac {\left (12 a^{4}-29 b^{2} a^{2}+20 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\right )}{b^{5}}+\frac {\frac {4 \left (\frac {b^{2} \tan \left (\frac {x}{2}\right )^{3}}{4}+\frac {3 a b \tan \left (\frac {x}{2}\right )^{2}}{2}-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{4}+\frac {3 a b}{2}\right )}{\left (\tan \left (\frac {x}{2}\right )^{2}+1\right )^{2}}+\left (12 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{5}}\)	\(330\)
risch	\(\frac {6 x \,a^{2}}{b^{5}}+\frac {x}{2 b^{3}}+\frac {i {\mathrm e}^{2 i x}}{8 b^{3}}+\frac {3 a \,{\mathrm e}^{i x}}{2 b^{4}}+\frac {3 a \,{\mathrm e}^{-i x}}{2 b^{4}}-\frac {i {\mathrm e}^{-2 i x}}{8 b^{3}}-\frac {i a^{4} \left (-8 i a^{3} b \,{\mathrm e}^{3 i x}+11 i a \,b^{3} {\mathrm e}^{3 i x}+20 i b \,a^{3} {\mathrm e}^{i x}-29 i a \,b^{3} {\mathrm e}^{i x}+14 a^{4} {\mathrm e}^{2 i x}-13 a^{2} b^{2} {\mathrm e}^{2 i x}-10 b^{4} {\mathrm e}^{2 i x}-7 b^{2} a^{2}+10 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2} b^{5}}-\frac {6 i a^{7} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{5}}+\frac {29 i a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{3}}-\frac {10 i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b}+\frac {6 i a^{7} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{5}}-\frac {29 i a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{3}}+\frac {10 i a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b}\)	\(668\)

Input:

int(sin(x)^5/(a+b*sin(x))^3,x,method=_RETURNVERBOSE)
 

Output:

-4*a^3/b^5*((-1/4*a*b^2*(5*a^2-8*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-3/4 
*b*(2*a^4+a^2*b^2-6*b^4)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-1/4*b^2*a*(19*a^ 
2-28*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-3/4*a^2*b*(2*a^2-3*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+1/4*(12*a^4-29*a^2*b^2+20* 
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)))+4/b^5*((1/4*b^2*tan(1/2*x)^3+3/2*a*b*tan(1/2*x)^2-1/4*b^2 
*tan(1/2*x)+3/2*a*b)/(tan(1/2*x)^2+1)^2+1/4*(12*a^2+b^2)*arctan(tan(1/2*x) 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (227) = 454\).

Time = 0.16 (sec) , antiderivative size = 1090, normalized size of antiderivative = 4.49 \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx=\text {Too large to display} \] Input:

integrate(sin(x)^5/(a+b*sin(x))^3,x, algorithm="fricas")
 

Output:

[-1/4*(2*(12*a^8*b^2 - 35*a^6*b^4 + 33*a^4*b^6 - 9*a^2*b^8 - b^10)*x*cos(x 
)^2 + 8*(a^7*b^3 - 3*a^5*b^5 + 3*a^3*b^7 - a*b^9)*cos(x)^3 + (12*a^9 - 17* 
a^7*b^2 - 9*a^5*b^4 + 20*a^3*b^6 - (12*a^7*b^2 - 29*a^5*b^4 + 20*a^3*b^6)* 
cos(x)^2 + 2*(12*a^8*b - 29*a^6*b^3 + 20*a^4*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 
)) - 2*(12*a^10 - 23*a^8*b^2 - 2*a^6*b^4 + 24*a^4*b^6 - 10*a^2*b^8 - b^10) 
*x - 2*(12*a^9*b - 29*a^7*b^3 + 15*a^5*b^5 + 6*a^3*b^7 - 4*a*b^9)*cos(x) - 
 2*((a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*cos(x)^3 + 2*(12*a^9*b - 35*a 
^7*b^3 + 33*a^5*b^5 - 9*a^3*b^7 - a*b^9)*x + (18*a^8*b^2 - 51*a^6*b^4 + 46 
*a^4*b^6 - 14*a^2*b^8 + b^10)*cos(x))*sin(x))/(a^8*b^5 - 2*a^6*b^7 + 2*a^2 
*b^11 - b^13 - (a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*cos(x)^2 + 2*(a^7 
*b^6 - 3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*sin(x)), -1/2*((12*a^8*b^2 - 35*a^ 
6*b^4 + 33*a^4*b^6 - 9*a^2*b^8 - b^10)*x*cos(x)^2 + 4*(a^7*b^3 - 3*a^5*b^5 
 + 3*a^3*b^7 - a*b^9)*cos(x)^3 - (12*a^9 - 17*a^7*b^2 - 9*a^5*b^4 + 20*a^3 
*b^6 - (12*a^7*b^2 - 29*a^5*b^4 + 20*a^3*b^6)*cos(x)^2 + 2*(12*a^8*b - 29* 
a^6*b^3 + 20*a^4*b^5)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt 
(a^2 - b^2)*cos(x))) - (12*a^10 - 23*a^8*b^2 - 2*a^6*b^4 + 24*a^4*b^6 - 10 
*a^2*b^8 - b^10)*x - (12*a^9*b - 29*a^7*b^3 + 15*a^5*b^5 + 6*a^3*b^7 - 4*a 
*b^9)*cos(x) - ((a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*cos(x)^3 + 2*(...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx=\text {Timed out} \] Input:

integrate(sin(x)**5/(a+b*sin(x))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sin(x)^5/(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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (227) = 454\).

Time = 0.14 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.12 \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sin(x)^5/(a+b*sin(x))^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-(12*a^7 - 29*a^5*b^2 + 20*a^3*b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arc 
tan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^4*b^5 - 2*a^2*b^7 + b^9)*sqrt 
(a^2 - b^2)) + (6*a^6*b*tan(1/2*x)^7 - 10*a^4*b^3*tan(1/2*x)^7 + a^2*b^5*t 
an(1/2*x)^7 + 12*a^7*tan(1/2*x)^6 - 5*a^5*b^2*tan(1/2*x)^6 - 20*a^3*b^4*ta 
n(1/2*x)^6 + 4*a*b^6*tan(1/2*x)^6 + 54*a^6*b*tan(1/2*x)^5 - 90*a^4*b^3*tan 
(1/2*x)^5 + 17*a^2*b^5*tan(1/2*x)^5 + 4*b^7*tan(1/2*x)^5 + 36*a^7*tan(1/2* 
x)^4 - 15*a^5*b^2*tan(1/2*x)^4 - 66*a^3*b^4*tan(1/2*x)^4 + 24*a*b^6*tan(1/ 
2*x)^4 + 90*a^6*b*tan(1/2*x)^3 - 162*a^4*b^3*tan(1/2*x)^3 + 55*a^2*b^5*tan 
(1/2*x)^3 - 4*b^7*tan(1/2*x)^3 + 36*a^7*tan(1/2*x)^2 - 31*a^5*b^2*tan(1/2* 
x)^2 - 40*a^3*b^4*tan(1/2*x)^2 + 20*a*b^6*tan(1/2*x)^2 + 42*a^6*b*tan(1/2* 
x) - 74*a^4*b^3*tan(1/2*x) + 23*a^2*b^5*tan(1/2*x) + 12*a^7 - 21*a^5*b^2 + 
 6*a^3*b^4)/((a^4*b^4 - 2*a^2*b^6 + b^8)*(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)^2) + 1/2*(12*a^2 + b^2)*x/b^5
 

Mupad [B] (verification not implemented)

Time = 22.66 (sec) , antiderivative size = 6640, normalized size of antiderivative = 27.33 \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx=\text {Too large to display} \] Input:

int(sin(x)^5/(a + b*sin(x))^3,x)
 

Output:

((3*(4*a^7 + 2*a^3*b^4 - 7*a^5*b^2))/(b^4*(a^4 + b^4 - 2*a^2*b^2)) + (tan( 
x/2)^7*(6*a^6 + a^2*b^4 - 10*a^4*b^2))/(b^3*(a^4 + b^4 - 2*a^2*b^2)) + (ta 
n(x/2)^5*(54*a^6 + 4*b^6 + 17*a^2*b^4 - 90*a^4*b^2))/(b^3*(a^4 + b^4 - 2*a 
^2*b^2)) + (tan(x/2)^3*(90*a^6 - 4*b^6 + 55*a^2*b^4 - 162*a^4*b^2))/(b^3*( 
a^4 + b^4 - 2*a^2*b^2)) + (tan(x/2)^6*(4*a*b^6 + 12*a^7 - 20*a^3*b^4 - 5*a 
^5*b^2))/(b^4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(x/2)^2*(20*a*b^6 + 36*a^7 - 
40*a^3*b^4 - 31*a^5*b^2))/(b^4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(x/2)*(42*a^ 
6 + 23*a^2*b^4 - 74*a^4*b^2))/(b^3*(a^4 + b^4 - 2*a^2*b^2)) + (3*tan(x/2)^ 
4*(3*a^2 + 4*b^2)*(2*a*b^4 + 4*a^5 - 7*a^3*b^2))/(b^4*(a^4 + b^4 - 2*a^2*b 
^2)))/(tan(x/2)^2*(4*a^2 + 4*b^2) + tan(x/2)^6*(4*a^2 + 4*b^2) + tan(x/2)^ 
4*(6*a^2 + 8*b^2) + a^2 + a^2*tan(x/2)^8 + 4*a*b*tan(x/2) + 12*a*b*tan(x/2 
)^3 + 12*a*b*tan(x/2)^5 + 4*a*b*tan(x/2)^7) + (atan((((a^2*12i + b^2*1i)*( 
(4*(2*a^2*b^16 + 40*a^4*b^14 + 108*a^6*b^12 - 872*a^8*b^10 + 1538*a^10*b^8 
 - 1104*a^12*b^6 + 288*a^14*b^4))/(b^19 - 4*a^2*b^17 + 6*a^4*b^15 - 4*a^6* 
b^13 + a^8*b^11) - ((a^2*12i + b^2*1i)*((4*(4*a*b^20 + 28*a^3*b^18 - 120*a 
^5*b^16 + 164*a^7*b^14 - 100*a^9*b^12 + 24*a^11*b^10))/(b^19 - 4*a^2*b^17 
+ 6*a^4*b^15 - 4*a^6*b^13 + a^8*b^11) - (((4*(8*a^2*b^22 - 32*a^4*b^20 + 4 
8*a^6*b^18 - 32*a^8*b^16 + 8*a^10*b^14))/(b^19 - 4*a^2*b^17 + 6*a^4*b^15 - 
 4*a^6*b^13 + a^8*b^11) + (8*tan(x/2)*(12*a*b^24 - 56*a^3*b^22 + 104*a^5*b 
^20 - 96*a^7*b^18 + 44*a^9*b^16 - 8*a^11*b^14))/(b^20 - 4*a^2*b^18 + 6*...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 953, normalized size of antiderivative = 3.92 \[ \int \frac {\sin ^5(x)}{(a+b \sin (x))^3} \, dx =\text {Too large to display} \] Input:

int(sin(x)^5/(a+b*sin(x))^3,x)
 

Output:

( - 48*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)** 
2*a**7*b**2 + 116*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2 
))*sin(x)**2*a**5*b**4 - 80*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a 
**2 - b**2))*sin(x)**2*a**3*b**6 - 96*sqrt(a**2 - b**2)*atan((tan(x/2)*a + 
 b)/sqrt(a**2 - b**2))*sin(x)*a**8*b + 232*sqrt(a**2 - b**2)*atan((tan(x/2 
)*a + b)/sqrt(a**2 - b**2))*sin(x)*a**6*b**3 - 160*sqrt(a**2 - b**2)*atan( 
(tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)*a**4*b**5 - 48*sqrt(a**2 - b**2 
)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**9 + 116*sqrt(a**2 - b**2)*at 
an((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**7*b**2 - 80*sqrt(a**2 - b**2)*at 
an((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**5*b**4 - 2*cos(x)*sin(x)**3*a**6 
*b**4 + 6*cos(x)*sin(x)**3*a**4*b**6 - 6*cos(x)*sin(x)**3*a**2*b**8 + 2*co 
s(x)*sin(x)**3*b**10 + 8*cos(x)*sin(x)**2*a**7*b**3 - 24*cos(x)*sin(x)**2* 
a**5*b**5 + 24*cos(x)*sin(x)**2*a**3*b**7 - 8*cos(x)*sin(x)**2*a*b**9 + 36 
*cos(x)*sin(x)*a**8*b**2 - 100*cos(x)*sin(x)*a**6*b**4 + 86*cos(x)*sin(x)* 
a**4*b**6 - 22*cos(x)*sin(x)*a**2*b**8 + 24*cos(x)*a**9*b - 66*cos(x)*a**7 
*b**3 + 54*cos(x)*a**5*b**5 - 12*cos(x)*a**3*b**7 + 24*sin(x)**2*a**8*b**2 
*x + 18*sin(x)**2*a**7*b**3 - 70*sin(x)**2*a**6*b**4*x - 50*sin(x)**2*a**5 
*b**5 + 66*sin(x)**2*a**4*b**6*x + 43*sin(x)**2*a**3*b**7 - 18*sin(x)**2*a 
**2*b**8*x - 11*sin(x)**2*a*b**9 - 2*sin(x)**2*b**10*x + 48*sin(x)*a**9*b* 
x + 36*sin(x)*a**8*b**2 - 140*sin(x)*a**7*b**3*x - 100*sin(x)*a**6*b**4...