\(\int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx\) [194]

Optimal result

Integrand size = 13, antiderivative size = 179 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx=-\frac {3 a x}{b^4}+\frac {3 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{5/2}}-\frac {\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))} \] Output:

-3*a*x/b^4+3*a^2*(2*a^4-5*a^2*b^2+4*b^4)*arctan((b+a*tan(1/2*x))/(a^2-b^2) 
^(1/2))/b^4/(a^2-b^2)^(5/2)-1/2*(3*a^2-2*b^2)*cos(x)/b^3/(a^2-b^2)+1/2*a^2 
*cos(x)*sin(x)^2/b/(a^2-b^2)/(a+b*sin(x))^2-3/2*a^3*(a^2-2*b^2)*cos(x)/b^3 
/(a^2-b^2)^2/(a+b*sin(x))
 

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx=\frac {-6 a x+\frac {6 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-2 b \cos (x)+\frac {a^4 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}+\frac {a^3 b \left (-5 a^2+8 b^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}}{2 b^4} \] Input:

Integrate[Sin[x]^4/(a + b*Sin[x])^3,x]
 

Output:

(-6*a*x + (6*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[ 
a^2 - b^2]])/(a^2 - b^2)^(5/2) - 2*b*Cos[x] + (a^4*b*Cos[x])/((a - b)*(a + 
 b)*(a + b*Sin[x])^2) + (a^3*b*(-5*a^2 + 8*b^2)*Cos[x])/((a - b)^2*(a + b) 
^2*(a + b*Sin[x])))/(2*b^4)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.23, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3271, 3042, 3510, 3042, 3502, 27, 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.

\(\displaystyle \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^4}{(a+b \sin (x))^3}dx\)

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3510

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

\(\displaystyle \frac {a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {3 \left (2 a x \left (a^2-b^2\right )^2-a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \int \frac {1}{a+b \sin (x)}dx\right )}{b}+\left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {3 \left (2 a x \left (a^2-b^2\right )^2-a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \int \frac {1}{a+b \sin (x)}dx\right )}{b}+\left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {3 \left (2 a x \left (a^2-b^2\right )^2-2 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )\right )}{b}+\left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\frac {3 \left (4 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )+2 a x \left (a^2-b^2\right )^2\right )}{b}+\left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\frac {\left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \cos (x)+\frac {3 \left (2 a x \left (a^2-b^2\right )^2-\frac {2 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}\right )}{b}}{b^2 \left (a^2-b^2\right )}+\frac {3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right ) (a+b \sin (x))}}{2 b \left (a^2-b^2\right )}\)

Input:

Int[Sin[x]^4/(a + b*Sin[x])^3,x]
 

Output:

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

Defintions of rubi rules used

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 3510

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {2 a^{2} \left (\frac {-\frac {3 a \,b^{2} \left (a^{2}-2 b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}+b^{2} a^{2}-14 b^{4}\right ) \tan \left (\frac {x}{2}\right )^{2}}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {b^{2} a \left (13 a^{2}-22 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {a^{2} b \left (4 a^{2}-7 b^{2}\right )}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}}{\left (a \tan \left (\frac {x}{2}\right )^{2}+2 b \tan \left (\frac {x}{2}\right )+a \right )^{2}}+\frac {3 \left (2 a^{4}-5 b^{2} a^{2}+4 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\right )}{b^{4}}-\frac {2 \left (\frac {b}{\tan \left (\frac {x}{2}\right )^{2}+1}+3 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )\right )}{b^{4}}\)	\(286\)
risch	\(-\frac {3 a x}{b^{4}}-\frac {{\mathrm e}^{i x}}{2 b^{3}}-\frac {{\mathrm e}^{-i x}}{2 b^{3}}+\frac {i a^{3} \left (-6 i a^{3} b \,{\mathrm e}^{3 i x}+9 i a \,b^{3} {\mathrm e}^{3 i x}+14 i b \,a^{3} {\mathrm e}^{i x}-23 i a \,b^{3} {\mathrm e}^{i x}+10 a^{4} {\mathrm e}^{2 i x}-11 a^{2} b^{2} {\mathrm e}^{2 i x}-8 b^{4} {\mathrm e}^{2 i x}-5 b^{2} a^{2}+8 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2} b^{4}}-\frac {3 a^{6} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{4}}+\frac {15 a^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{2}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {3 a^{6} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{4}}-\frac {15 a^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} b^{2}}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\)	\(624\)

Input:

int(sin(x)^4/(a+b*sin(x))^3,x,method=_RETURNVERBOSE)
 

Output:

2*a^2/b^4*((-3/2*a*b^2*(a^2-2*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-1/2*b* 
(4*a^4+a^2*b^2-14*b^4)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-1/2*b^2*a*(13*a^2- 
22*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-1/2*a^2*b*(4*a^2-7*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+3/2*(2*a^4-5*a^2*b^2+4*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)))-2/b^4*(b/(tan(1/2*x)^2+1)+3*a*arctan(tan(1/2*x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (167) = 334\).

Time = 0.16 (sec) , antiderivative size = 945, normalized size of antiderivative = 5.28 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sin(x)^4/(a+b*sin(x))^3,x, algorithm="fricas")
 

Output:

[1/4*(12*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*x*cos(x)^2 + 4*(a^6*b^3 
 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(x)^3 - 3*(2*a^8 - 3*a^6*b^2 - a^4*b^4 
+ 4*a^2*b^6 - (2*a^6*b^2 - 5*a^4*b^4 + 4*a^2*b^6)*cos(x)^2 + 2*(2*a^7*b - 
5*a^5*b^3 + 4*a^3*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)) - 12*(a^9 - 2*a^7*b^2 + 2 
*a^3*b^6 - a*b^8)*x - 2*(6*a^8*b - 15*a^6*b^3 + 7*a^4*b^5 + 4*a^2*b^7 - 2* 
b^9)*cos(x) - 2*(12*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*x + (9*a^7*b 
^2 - 25*a^5*b^4 + 20*a^3*b^6 - 4*a*b^8)*cos(x))*sin(x))/(a^8*b^4 - 2*a^6*b 
^6 + 2*a^2*b^10 - b^12 - (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*cos(x)^ 
2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*sin(x)), 1/2*(6*(a^7*b^2 
- 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*x*cos(x)^2 + 2*(a^6*b^3 - 3*a^4*b^5 + 3*a 
^2*b^7 - b^9)*cos(x)^3 - 3*(2*a^8 - 3*a^6*b^2 - a^4*b^4 + 4*a^2*b^6 - (2*a 
^6*b^2 - 5*a^4*b^4 + 4*a^2*b^6)*cos(x)^2 + 2*(2*a^7*b - 5*a^5*b^3 + 4*a^3* 
b^5)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x 
))) - 6*(a^9 - 2*a^7*b^2 + 2*a^3*b^6 - a*b^8)*x - (6*a^8*b - 15*a^6*b^3 + 
7*a^4*b^5 + 4*a^2*b^7 - 2*b^9)*cos(x) - (12*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 
 - a^2*b^7)*x + (9*a^7*b^2 - 25*a^5*b^4 + 20*a^3*b^6 - 4*a*b^8)*cos(x))*si 
n(x))/(a^8*b^4 - 2*a^6*b^6 + 2*a^2*b^10 - b^12 - (a^6*b^6 - 3*a^4*b^8 + 3* 
a^2*b^10 - b^12)*cos(x)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx=\text {Timed out} \] Input:

integrate(sin(x)**4/(a+b*sin(x))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sin(x)^4/(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 [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.43 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx=\frac {3 \, {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{5} b \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, a^{6} \tan \left (\frac {1}{2} \, x\right )^{2} + a^{4} b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 14 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 13 \, a^{5} b \tan \left (\frac {1}{2} \, x\right ) - 22 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, x\right ) + 4 \, a^{6} - 7 \, a^{4} b^{2}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} - \frac {3 \, a x}{b^{4}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} b^{3}} \] Input:

integrate(sin(x)^4/(a+b*sin(x))^3,x, algorithm="giac")
 

Output:

3*(2*a^6 - 5*a^4*b^2 + 4*a^2*b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arcta 
n((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)*sqrt(a 
^2 - b^2)) - (3*a^5*b*tan(1/2*x)^3 - 6*a^3*b^3*tan(1/2*x)^3 + 4*a^6*tan(1/ 
2*x)^2 + a^4*b^2*tan(1/2*x)^2 - 14*a^2*b^4*tan(1/2*x)^2 + 13*a^5*b*tan(1/2 
*x) - 22*a^3*b^3*tan(1/2*x) + 4*a^6 - 7*a^4*b^2)/((a^4*b^3 - 2*a^2*b^5 + b 
^7)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2) - 3*a*x/b^4 - 2/((tan(1/2*x)^ 
2 + 1)*b^3)
 

Mupad [B] (verification not implemented)

Time = 22.14 (sec) , antiderivative size = 5945, normalized size of antiderivative = 33.21 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx=\text {Too large to display} \] Input:

int(sin(x)^4/(a + b*sin(x))^3,x)
 

Output:

- ((a^2*(6*a^4 + 2*b^4 - 11*a^2*b^2))/(b^3*(a^2 - b^2)^2) + (3*tan(x/2)^5* 
(a^5 - 2*a^3*b^2))/(b^2*(a^2 - b^2)^2) - (3*tan(x/2)^4*(4*a^2*b^4 - 2*a^6 
+ a^4*b^2))/(b^3*(a^2 - b^2)^2) + (2*tan(x/2)^2*(6*a^6 + 4*b^6 - 13*a^2*b^ 
4 - 3*a^4*b^2))/(b^3*(a^2 - b^2)^2) + (4*a*tan(x/2)^3*(6*a^4 + 2*b^4 - 11* 
a^2*b^2))/(b^2*(a^2 - b^2)^2) + (a*tan(x/2)*(21*a^4 + 8*b^4 - 38*a^2*b^2)) 
/(b^2*(a^2 - b^2)^2))/(tan(x/2)^2*(3*a^2 + 4*b^2) + tan(x/2)^4*(3*a^2 + 4* 
b^2) + a^2 + a^2*tan(x/2)^6 + 4*a*b*tan(x/2) + 8*a*b*tan(x/2)^3 + 4*a*b*ta 
n(x/2)^5) - (6*a*atan(((3*a*((8*(36*a^4*b^11 - 144*a^6*b^9 + 216*a^8*b^7 - 
 144*a^10*b^5 + 36*a^12*b^3))/(b^16 - 4*a^2*b^14 + 6*a^4*b^12 - 4*a^6*b^10 
 + a^8*b^8) + (8*tan(x/2)*(72*a^3*b^13 - 468*a^5*b^11 + 936*a^7*b^9 - 873* 
a^9*b^7 + 396*a^11*b^5 - 72*a^13*b^3))/(b^17 - 4*a^2*b^15 + 6*a^4*b^13 - 4 
*a^6*b^11 + a^8*b^9) - (a*((8*(12*a^2*b^16 - 36*a^4*b^14 + 42*a^6*b^12 - 2 
4*a^8*b^10 + 6*a^10*b^8))/(b^16 - 4*a^2*b^14 + 6*a^4*b^12 - 4*a^6*b^10 + a 
^8*b^8) - (a*((8*(4*a^2*b^19 - 16*a^4*b^17 + 24*a^6*b^15 - 16*a^8*b^13 + 4 
*a^10*b^11))/(b^16 - 4*a^2*b^14 + 6*a^4*b^12 - 4*a^6*b^10 + a^8*b^8) + (8* 
tan(x/2)*(12*a*b^21 - 56*a^3*b^19 + 104*a^5*b^17 - 96*a^7*b^15 + 44*a^9*b^ 
13 - 8*a^11*b^11))/(b^17 - 4*a^2*b^15 + 6*a^4*b^13 - 4*a^6*b^11 + a^8*b^9) 
)*3i)/b^4 + (8*tan(x/2)*(48*a^3*b^16 - 156*a^5*b^14 + 192*a^7*b^12 - 108*a 
^9*b^10 + 24*a^11*b^8))/(b^17 - 4*a^2*b^15 + 6*a^4*b^13 - 4*a^6*b^11 + a^8 
*b^9))*3i)/b^4))/b^4 + (3*a*((8*(36*a^4*b^11 - 144*a^6*b^9 + 216*a^8*b^...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 864, normalized size of antiderivative = 4.83 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^3} \, dx =\text {Too large to display} \] Input:

int(sin(x)^4/(a+b*sin(x))^3,x)
 

Output:

(24*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)**2*a 
**6*b**2 - 60*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*s 
in(x)**2*a**4*b**4 + 48*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 
- b**2))*sin(x)**2*a**2*b**6 + 48*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/ 
sqrt(a**2 - b**2))*sin(x)*a**7*b - 120*sqrt(a**2 - b**2)*atan((tan(x/2)*a 
+ b)/sqrt(a**2 - b**2))*sin(x)*a**5*b**3 + 96*sqrt(a**2 - b**2)*atan((tan( 
x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)*a**3*b**5 + 24*sqrt(a**2 - b**2)*ata 
n((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**8 - 60*sqrt(a**2 - b**2)*atan((ta 
n(x/2)*a + b)/sqrt(a**2 - b**2))*a**6*b**2 + 48*sqrt(a**2 - b**2)*atan((ta 
n(x/2)*a + b)/sqrt(a**2 - b**2))*a**4*b**4 - 4*cos(x)*sin(x)**2*a**6*b**3 
+ 12*cos(x)*sin(x)**2*a**4*b**5 - 12*cos(x)*sin(x)**2*a**2*b**7 + 4*cos(x) 
*sin(x)**2*b**9 - 18*cos(x)*sin(x)*a**7*b**2 + 50*cos(x)*sin(x)*a**5*b**4 
- 40*cos(x)*sin(x)*a**3*b**6 + 8*cos(x)*sin(x)*a*b**8 - 12*cos(x)*a**8*b + 
 34*cos(x)*a**6*b**3 - 26*cos(x)*a**4*b**5 + 4*cos(x)*a**2*b**7 - 12*sin(x 
)**2*a**7*b**2*x - 9*sin(x)**2*a**6*b**3 + 36*sin(x)**2*a**5*b**4*x + 25*s 
in(x)**2*a**4*b**5 - 36*sin(x)**2*a**3*b**6*x - 20*sin(x)**2*a**2*b**7 + 1 
2*sin(x)**2*a*b**8*x + 4*sin(x)**2*b**9 - 24*sin(x)*a**8*b*x - 18*sin(x)*a 
**7*b**2 + 72*sin(x)*a**6*b**3*x + 50*sin(x)*a**5*b**4 - 72*sin(x)*a**4*b* 
*5*x - 40*sin(x)*a**3*b**6 + 24*sin(x)*a**2*b**7*x + 8*sin(x)*a*b**8 - 12* 
a**9*x - 9*a**8*b + 36*a**7*b**2*x + 25*a**6*b**3 - 36*a**5*b**4*x - 20...