\(\int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx\) [24]

Optimal result

Integrand size = 13, antiderivative size = 104 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac {\sin ^3(x)}{3 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx=\frac {-12 a^3 x+18 a b^2 x-24 \left (-a^2+b^2\right )^{3/2} \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+3 b \left (4 a^2-5 b^2\right ) \sin (x)-3 a b^2 \sin (2 x)+b^3 \sin (3 x)}{12 b^4} \] Input:

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

Output:

(-12*a^3*x + 18*a*b^2*x - 24*(-a^2 + b^2)^(3/2)*ArcTanh[((a - b)*Tan[x/2]) 
/Sqrt[-a^2 + b^2]] + 3*b*(4*a^2 - 5*b^2)*Sin[x] - 3*a*b^2*Sin[2*x] + b^3*S 
in[3*x])/(12*b^4)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.20, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 3174, 25, 3042, 3344, 25, 3042, 3214, 3042, 3138, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3174

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3344

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 218

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

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*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 3174

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(b*(m + p)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F 
reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 
 0] && IntegersQ[2*m, 2*p]
 

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 3344

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* 
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* 
p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( 
(p - 1)/(b^2*(m + p)*(m + p + 1)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si 
n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 
2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 
, 0] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {2 \left (\frac {\left (-a^{2} b -\frac {1}{2} b^{2} a +b^{3}\right ) \tan \left (\frac {x}{2}\right )^{5}+\left (-2 a^{2} b +\frac {10}{3} b^{3}\right ) \tan \left (\frac {x}{2}\right )^{3}+\left (-a^{2} b +b^{3}+\frac {1}{2} b^{2} a \right ) \tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2}\right )}{b^{4}}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}\)	\(152\)
risch	\(-\frac {a^{3} x}{b^{4}}+\frac {3 a x}{2 b^{2}}-\frac {i {\mathrm e}^{i x} a^{2}}{2 b^{3}}+\frac {5 i {\mathrm e}^{i x}}{8 b}+\frac {i {\mathrm e}^{-i x} a^{2}}{2 b^{3}}-\frac {5 i {\mathrm e}^{-i x}}{8 b}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right ) a^{2}}{b^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right ) a^{2}}{b^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{b^{2}}+\frac {\sin \left (3 x \right )}{12 b}-\frac {a \sin \left (2 x \right )}{4 b^{2}}\)	\(269\)

Input:

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

Output:

-2/b^4*(((-a^2*b-1/2*b^2*a+b^3)*tan(1/2*x)^5+(-2*a^2*b+10/3*b^3)*tan(1/2*x 
)^3+(-a^2*b+b^3+1/2*b^2*a)*tan(1/2*x))/(1+tan(1/2*x)^2)^3+1/2*a*(2*a^2-3*b 
^2)*arctan(tan(1/2*x)))+2*(a+b)^2*(a-b)^2/b^4/((a-b)*(a+b))^(1/2)*arctan(( 
a-b)*tan(1/2*x)/((a-b)*(a+b))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.34 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx=\left [-\frac {3 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} x - {\left (2 \, b^{3} \cos \left (x\right )^{2} - 3 \, a b^{2} \cos \left (x\right ) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \left (x\right )}{6 \, b^{4}}, \frac {6 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} x + {\left (2 \, b^{3} \cos \left (x\right )^{2} - 3 \, a b^{2} \cos \left (x\right ) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \left (x\right )}{6 \, b^{4}}\right ] \] Input:

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

Output:

[-1/6*(3*(a^2 - b^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*co 
s(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos( 
x)^2 + 2*a*b*cos(x) + a^2)) + 3*(2*a^3 - 3*a*b^2)*x - (2*b^3*cos(x)^2 - 3* 
a*b^2*cos(x) + 6*a^2*b - 8*b^3)*sin(x))/b^4, 1/6*(6*(a^2 - b^2)^(3/2)*arct 
an(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x))) - 3*(2*a^3 - 3*a*b^2)*x + (2* 
b^3*cos(x)^2 - 3*a*b^2*cos(x) + 6*a^2*b - 8*b^3)*sin(x))/b^4]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(sin(x)^4/(a+b*cos(x)),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. 194 vs. \(2 (86) = 172\).

Time = 0.11 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.87 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx=-\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x}{2 \, b^{4}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 20 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) - 3 \, a b \tan \left (\frac {1}{2} \, x\right ) - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3} b^{3}} \] Input:

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

Output:

-1/2*(2*a^3 - 3*a*b^2)*x/b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*(pi*floor(1/2*x/p 
i + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 
- b^2)))/(sqrt(a^2 - b^2)*b^4) + 1/3*(6*a^2*tan(1/2*x)^5 + 3*a*b*tan(1/2*x 
)^5 - 6*b^2*tan(1/2*x)^5 + 12*a^2*tan(1/2*x)^3 - 20*b^2*tan(1/2*x)^3 + 6*a 
^2*tan(1/2*x) - 3*a*b*tan(1/2*x) - 6*b^2*tan(1/2*x))/((tan(1/2*x)^2 + 1)^3 
*b^3)
 

Mupad [B] (verification not implemented)

Time = 44.97 (sec) , antiderivative size = 1677, normalized size of antiderivative = 16.12 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((4*tan(x/2)^3*(3*a^2 - 5*b^2))/(3*b^3) - (tan(x/2)*(a*b - 2*a^2 + 2*b^2)) 
/b^3 + (tan(x/2)^5*(a*b + 2*a^2 - 2*b^2))/b^3)/(3*tan(x/2)^2 + 3*tan(x/2)^ 
4 + tan(x/2)^6 + 1) - (2*atanh((64*tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4 
*b^2)^(1/2))/(128*a*b^2 + 112*a^2*b - 352*a^3 - 64*b^3 + (16*a^4)/b + (320 
*a^5)/b^2 - (112*a^6)/b^3 - (96*a^7)/b^4 + (48*a^8)/b^5) + (144*a^2*tan(x/ 
2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(128*a*b^4 + 16*a^4*b + 320* 
a^5 - 64*b^5 + 112*a^2*b^3 - 352*a^3*b^2 - (112*a^6)/b - (96*a^7)/b^2 + (4 
8*a^8)/b^3) + (80*a^3*tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/ 
(128*a*b^5 + 320*a^5*b - 112*a^6 - 64*b^6 + 112*a^2*b^4 - 352*a^3*b^3 + 16 
*a^4*b^2 - (96*a^7)/b + (48*a^8)/b^2) - (144*a^4*tan(x/2)*(b^6 - a^6 - 3*a 
^2*b^4 + 3*a^4*b^2)^(1/2))/(128*a*b^6 - 112*a^6*b - 96*a^7 - 64*b^7 + 112* 
a^2*b^5 - 352*a^3*b^4 + 16*a^4*b^3 + 320*a^5*b^2 + (48*a^8)/b) + (48*a^5*t 
an(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(128*a*b^7 - 96*a^7*b + 
 48*a^8 - 64*b^8 + 112*a^2*b^6 - 352*a^3*b^5 + 16*a^4*b^4 + 320*a^5*b^3 - 
112*a^6*b^2) - (192*a*tan(x/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/ 
(128*a*b^3 - 352*a^3*b + 16*a^4 - 64*b^4 + 112*a^2*b^2 + (320*a^5)/b - (11 
2*a^6)/b^2 - (96*a^7)/b^3 + (48*a^8)/b^4))*(-(a + b)^3*(a - b)^3)^(1/2))/b 
^4 + (a*atan(((a*(2*a^2 - 3*b^2)*((8*tan(x/2)*(4*a*b^8 - 16*a^8*b + 8*a^9 
- 4*b^9 + 7*a^2*b^7 + 11*a^3*b^6 - 39*a^4*b^5 - 3*a^5*b^4 + 48*a^6*b^3 - 1 
6*a^7*b^2))/b^6 - (a*((8*(2*a*b^12 - 4*b^13 + 10*a^2*b^11 - 6*a^3*b^10 ...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.34 \[ \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) b^{2}-3 \cos \left (x \right ) \sin \left (x \right ) a \,b^{2}-2 \sin \left (x \right )^{3} b^{3}+6 \sin \left (x \right ) a^{2} b -6 \sin \left (x \right ) b^{3}-6 a^{3} x +9 a \,b^{2} x}{6 b^{4}} \] Input:

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

Output:

(12*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a**2 - b**2))*a* 
*2 - 12*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a**2 - b**2) 
)*b**2 - 3*cos(x)*sin(x)*a*b**2 - 2*sin(x)**3*b**3 + 6*sin(x)*a**2*b - 6*s 
in(x)*b**3 - 6*a**3*x + 9*a*b**2*x)/(6*b**4)