\(\int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx\) [185]

Optimal result

Integrand size = 13, antiderivative size = 169 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\frac {\left (6 a^2+b^2\right ) x}{2 b^4}-\frac {2 a^3 \left (3 a^2-4 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac {a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2-b^2\right ) \cos (x) \sin (x)}{2 b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (x) \sin ^2(x)}{b \left (a^2-b^2\right ) (a+b \sin (x))} \] Output:

1/2*(6*a^2+b^2)*x/b^4-2*a^3*(3*a^2-4*b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2 
)^(1/2))/b^4/(a^2-b^2)^(3/2)+a*(3*a^2-2*b^2)*cos(x)/b^3/(a^2-b^2)-1/2*(3*a 
^2-b^2)*cos(x)*sin(x)/b^2/(a^2-b^2)+a^2*cos(x)*sin(x)^2/b/(a^2-b^2)/(a+b*s 
in(x))
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.68 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\frac {12 a^2 x+2 b^2 x-\frac {8 a^3 \left (3 a^2-4 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+4 a b \cos (x) \left (2+\frac {a^3}{(a-b) (a+b) (a+b \sin (x))}\right )-b^2 \sin (2 x)}{4 b^4} \] Input:

Integrate[Sin[x]^4/(a + b*Sin[x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.11, 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, 3528, 25, 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.

\(\displaystyle \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^4}{(a+b \sin (x))^2}dx\)

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a^2 \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\frac {\left (3 a^2-b^2\right ) \sin (x) \cos (x)}{2 b}-\frac {\frac {8 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )+\frac {x \left (a^2-b^2\right ) \left (6 a^2+b^2\right )}{b}}{b}+\frac {2 a \left (3 a^2-2 b^2\right ) \cos (x)}{b}}{2 b}}{b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 217

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

Input:

Int[Sin[x]^4/(a + b*Sin[x])^2,x]
 

Output:

(a^2*Cos[x]*Sin[x]^2)/(b*(a^2 - b^2)*(a + b*Sin[x])) - (-1/2*((((a^2 - b^2 
)*(6*a^2 + b^2)*x)/b - (4*a^3*((3*a^2)/b - 4*b)*ArcTan[(2*b + 2*a*Tan[x/2] 
)/(2*Sqrt[a^2 - b^2])])/Sqrt[a^2 - b^2])/b + (2*a*(3*a^2 - 2*b^2)*Cos[x])/ 
b)/b + ((3*a^2 - b^2)*Cos[x]*Sin[x])/(2*b))/(b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 = 0.80 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {2 a^{3} \left (\frac {-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{a^{2}-b^{2}}-\frac {a b}{a^{2}-b^{2}}}{a \tan \left (\frac {x}{2}\right )^{2}+2 b \tan \left (\frac {x}{2}\right )+a}+\frac {\left (3 a^{2}-4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}+\frac {\frac {2 \left (\frac {b^{2} \tan \left (\frac {x}{2}\right )^{3}}{2}+2 a b \tan \left (\frac {x}{2}\right )^{2}-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2}+2 a b \right )}{\left (\tan \left (\frac {x}{2}\right )^{2}+1\right )^{2}}+\left (6 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{4}}\)	\(183\)
risch	\(\frac {3 x \,a^{2}}{b^{4}}+\frac {x}{2 b^{2}}+\frac {i {\mathrm e}^{2 i x}}{8 b^{2}}+\frac {a \,{\mathrm e}^{i x}}{b^{3}}+\frac {a \,{\mathrm e}^{-i x}}{b^{3}}-\frac {i {\mathrm e}^{-2 i x}}{8 b^{2}}-\frac {2 i a^{4} \left (i b +a \,{\mathrm e}^{i x}\right )}{b^{4} \left (a^{2}-b^{2}\right ) \left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )}-\frac {3 a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{2}}+\frac {3 a^{5} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b^{2}}\)	\(423\)

Input:

int(sin(x)^4/(a+b*sin(x))^2,x,method=_RETURNVERBOSE)
 

Output:

-2*a^3/b^4*((-b^2/(a^2-b^2)*tan(1/2*x)-a*b/(a^2-b^2))/(a*tan(1/2*x)^2+2*b* 
tan(1/2*x)+a)+(3*a^2-4*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)))+2/b^4*((1/2*b^2*tan(1/2*x)^3+2*a*b*tan(1/2*x)^2-1/2*b^ 
2*tan(1/2*x)+2*a*b)/(tan(1/2*x)^2+1)^2+1/2*(6*a^2+b^2)*arctan(tan(1/2*x)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 580, normalized size of antiderivative = 3.43 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\left [\frac {{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} - {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + {\left (3 \, a^{5} b - 4 \, a^{3} 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 ) + {\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} x + {\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \cos \left (x\right ) + {\left ({\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} x + 3 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8} + {\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}}, \frac {{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} + 2 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + {\left (3 \, a^{5} b - 4 \, a^{3} 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 ) + {\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} x + {\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \cos \left (x\right ) + {\left ({\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} x + 3 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \, {\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8} + {\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}}\right ] \] Input:

integrate(sin(x)^4/(a+b*sin(x))^2,x, algorithm="fricas")
 

Output:

[1/2*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(x)^3 - (3*a^6 - 4*a^4*b^2 + (3*a^5*b 
 - 4*a^3*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)) + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 
 + a*b^6)*x + (6*a^6*b - 11*a^4*b^3 + 6*a^2*b^5 - b^7)*cos(x) + ((6*a^6*b 
- 11*a^4*b^3 + 4*a^2*b^5 + b^7)*x + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(x) 
)*sin(x))/(a^5*b^4 - 2*a^3*b^6 + a*b^8 + (a^4*b^5 - 2*a^2*b^7 + b^9)*sin(x 
)), 1/2*((a^4*b^3 - 2*a^2*b^5 + b^7)*cos(x)^3 + 2*(3*a^6 - 4*a^4*b^2 + (3* 
a^5*b - 4*a^3*b^3)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^ 
2 - b^2)*cos(x))) + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*x + (6*a^6*b 
- 11*a^4*b^3 + 6*a^2*b^5 - b^7)*cos(x) + ((6*a^6*b - 11*a^4*b^3 + 4*a^2*b^ 
5 + b^7)*x + 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(x))*sin(x))/(a^5*b^4 - 2* 
a^3*b^6 + a*b^8 + (a^4*b^5 - 2*a^2*b^7 + b^9)*sin(x))]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\text {Timed out} \] Input:

integrate(sin(x)**4/(a+b*sin(x))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sin(x)^4/(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 = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=-\frac {2 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} b \tan \left (\frac {1}{2} \, x\right ) + a^{4}\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}} + \frac {{\left (6 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} + \frac {b \tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right ) + 4 \, a}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} b^{3}} \] Input:

integrate(sin(x)^4/(a+b*sin(x))^2,x, algorithm="giac")
 

Output:

-2*(3*a^5 - 4*a^3*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^4 - b^6)*sqrt(a^2 - b^2)) + 2*(a^3*b*t 
an(1/2*x) + a^4)/((a^2*b^3 - b^5)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)) + 
 1/2*(6*a^2 + b^2)*x/b^4 + (b*tan(1/2*x)^3 + 4*a*tan(1/2*x)^2 - b*tan(1/2* 
x) + 4*a)/((tan(1/2*x)^2 + 1)^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 20.71 (sec) , antiderivative size = 4368, normalized size of antiderivative = 25.85 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\text {Too large to display} \] Input:

int(sin(x)^4/(a + b*sin(x))^2,x)
 

Output:

(atan((((a^2*6i + b^2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8 
*b^5 + 36*a^10*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(2*a*b^13 
 + 19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b^3))/(b 
^13 - 2*a^2*b^11 + a^4*b^9) - ((a^2*6i + b^2*1i)*((8*(2*a*b^14 + 6*a^3*b^1 
2 - 14*a^5*b^10 + 6*a^7*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) - (((8*(4*a^2* 
b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/ 
2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^1 
1 + a^4*b^9))*(a^2*6i + b^2*1i))/(2*b^4) + (8*tan(x/2)*(32*a^4*b^12 - 56*a 
^6*b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)))/(2*b^4))*1i)/(2*b^4 
) + ((a^2*6i + b^2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 - 60*a^8*b^ 
5 + 36*a^10*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)*(2*a*b^13 + 
19*a^3*b^11 + 16*a^5*b^9 - 197*a^7*b^7 + 228*a^9*b^5 - 72*a^11*b^3))/(b^13 
 - 2*a^2*b^11 + a^4*b^9) + ((a^2*6i + b^2*1i)*((8*(2*a*b^14 + 6*a^3*b^12 - 
 14*a^5*b^10 + 6*a^7*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (((8*(4*a^2*b^1 
5 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(x/2)* 
(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + 
 a^4*b^9))*(a^2*6i + b^2*1i))/(2*b^4) + (8*tan(x/2)*(32*a^4*b^12 - 56*a^6* 
b^10 + 24*a^8*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9)))/(2*b^4))*1i)/(2*b^4))/ 
((16*(54*a^11 + 4*a^5*b^6 + 9*a^7*b^4 - 81*a^9*b^2))/(b^12 - 2*a^2*b^10 + 
a^4*b^8) - ((a^2*6i + b^2*1i)*((8*(a^2*b^11 + 10*a^4*b^9 + 13*a^6*b^7 -...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.52 \[ \int \frac {\sin ^4(x)}{(a+b \sin (x))^2} \, dx=\frac {6 a^{7} x +3 a^{6} b -6 a^{4} b^{3}+3 a^{2} b^{5}-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a^{5} b +16 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a^{3} b^{3}-\cos \left (x \right ) \sin \left (x \right )^{2} b^{7}-\cos \left (x \right ) \sin \left (x \right )^{2} a^{4} b^{3}+\sin \left (x \right ) b^{7} x +a \,b^{6} x +16 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4} b^{2}+2 \cos \left (x \right ) \sin \left (x \right )^{2} a^{2} b^{5}+3 \cos \left (x \right ) \sin \left (x \right ) a^{5} b^{2}-6 \cos \left (x \right ) \sin \left (x \right ) a^{3} b^{4}+3 \cos \left (x \right ) \sin \left (x \right ) a \,b^{6}+6 \sin \left (x \right ) a^{6} b x -11 \sin \left (x \right ) a^{4} b^{3} x +4 \sin \left (x \right ) a^{2} b^{5} x -12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{6}+6 \cos \left (x \right ) a^{6} b -10 \cos \left (x \right ) a^{4} b^{3}+4 \cos \left (x \right ) a^{2} b^{5}+3 \sin \left (x \right ) a^{5} b^{2}-6 \sin \left (x \right ) a^{3} b^{4}+3 \sin \left (x \right ) a \,b^{6}-11 a^{5} b^{2} x +4 a^{3} b^{4} x}{2 b^{4} \left (\sin \left (x \right ) a^{4} b -2 \sin \left (x \right ) a^{2} b^{3}+\sin \left (x \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int(sin(x)^4/(a+b*sin(x))^2,x)
 

Output:

( - 12*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)*a 
**5*b + 16*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin( 
x)*a**3*b**3 - 12*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2 
))*a**6 + 16*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a* 
*4*b**2 - cos(x)*sin(x)**2*a**4*b**3 + 2*cos(x)*sin(x)**2*a**2*b**5 - cos( 
x)*sin(x)**2*b**7 + 3*cos(x)*sin(x)*a**5*b**2 - 6*cos(x)*sin(x)*a**3*b**4 
+ 3*cos(x)*sin(x)*a*b**6 + 6*cos(x)*a**6*b - 10*cos(x)*a**4*b**3 + 4*cos(x 
)*a**2*b**5 + 6*sin(x)*a**6*b*x + 3*sin(x)*a**5*b**2 - 11*sin(x)*a**4*b**3 
*x - 6*sin(x)*a**3*b**4 + 4*sin(x)*a**2*b**5*x + 3*sin(x)*a*b**6 + sin(x)* 
b**7*x + 6*a**7*x + 3*a**6*b - 11*a**5*b**2*x - 6*a**4*b**3 + 4*a**3*b**4* 
x + 3*a**2*b**5 + a*b**6*x)/(2*b**4*(sin(x)*a**4*b - 2*sin(x)*a**2*b**3 + 
sin(x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))