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

Optimal. Leaf size=104 \[ -\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 b^3}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}-\frac {\sin ^3(x)}{3 b} \]

[Out]

-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)*tan(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

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Rubi [A]  time = 0.26, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2695, 2865, 2735, 2659, 205} \[ -\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 b^3}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}-\frac {\sin ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Cos[x]),x]

[Out]

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

Rule 205

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(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 2695

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] + Dist[(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] /; FreeQ[{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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(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 2865

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] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[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]

Rubi steps

\begin {align*} \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx &=-\frac {\sin ^3(x)}{3 b}-\frac {\int \frac {(-b-a \cos (x)) \sin ^2(x)}{a+b \cos (x)} \, dx}{b}\\ &=\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}-\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^3}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 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}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \cos (x)} \, dx}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 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}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\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}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 96, normalized size = 0.92 \[ \frac {-12 a^3 x+3 b \left (4 a^2-5 b^2\right ) \sin (x)-24 \left (b^2-a^2\right )^{3/2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )+18 a b^2 x-3 a b^2 \sin (2 x)+b^3 \sin (3 x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Cos[x]),x]

[Out]

(-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*Sin[3*x])/(12*b^4)

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fricas [A]  time = 1.06, size = 243, normalized size = 2.34 \[ \left [-\frac {3 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) + {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \relax (x) + b\right )} \sin \relax (x) - a^{2} + 2 \, b^{2}}{b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}}\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} x - {\left (2 \, b^{3} \cos \relax (x)^{2} - 3 \, a b^{2} \cos \relax (x) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \relax (x)}{6 \, b^{4}}, \frac {6 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \cos \relax (x) + b}{\sqrt {a^{2} - b^{2}} \sin \relax (x)}\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} x + {\left (2 \, b^{3} \cos \relax (x)^{2} - 3 \, a b^{2} \cos \relax (x) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \relax (x)}{6 \, b^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*(a^2 - b^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(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)*arctan(-(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]

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giac [B]  time = 0.52, size = 194, normalized size = 1.87 \[ -\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x}{2 \, b^{4}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 20 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) - 3 \, a b \tan \left (\frac {1}{2} \, x\right ) - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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/pi + 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)

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maple [B]  time = 0.04, size = 315, normalized size = 3.03 \[ \frac {2 \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a^{4}}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a^{2}}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \left (\tan ^{5}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) a}{b^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{b \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {20 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3 b \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2 \tan \left (\frac {x}{2}\right ) a^{2}}{b^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \tan \left (\frac {x}{2}\right )}{b \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {\tan \left (\frac {x}{2}\right ) a}{b^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) a^{3}}{b^{4}}+\frac {3 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) a}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^4/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))*a^4-4/b^2/((a-b)*(a+b))^(1/2)*arctan(ta
n(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))*a^2+2/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))+2/b
^3/(tan(1/2*x)^2+1)^3*tan(1/2*x)^5*a^2+1/b^2/(tan(1/2*x)^2+1)^3*tan(1/2*x)^5*a-2/b/(tan(1/2*x)^2+1)^3*tan(1/2*
x)^5+4/b^3/(tan(1/2*x)^2+1)^3*tan(1/2*x)^3*a^2-20/3/b/(tan(1/2*x)^2+1)^3*tan(1/2*x)^3+2/b^3/(tan(1/2*x)^2+1)^3
*tan(1/2*x)*a^2-2/b/(tan(1/2*x)^2+1)^3*tan(1/2*x)-1/b^2/(tan(1/2*x)^2+1)^3*tan(1/2*x)*a-2/b^4*arctan(tan(1/2*x
))*a^3+3/b^2*arctan(tan(1/2*x))*a

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.11, size = 1677, normalized size = 16.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a + b*cos(x)),x)

[Out]

((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 + (48*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))/(12
8*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*tan(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 - (112*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 - 16*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 - 6*a^4*b^9 + 4*a^5*b^8))/b^9 - (a*tan(x/2)*(2*a^2 - 3*b^2)*(8*
a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2 - 3*b^2)*1i)/(2*b^4)))/(2*b^4) + (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 - 16*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 - 6*a^4*b^9 + 4*a^5*b^8))/b^9 + (a*tan(x/2
)*(2*a^2 - 3*b^2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2 - 3*b^2)*1i)/(2*b^4)))/(2*b^4))/((16*(6
*a^10*b - 6*a*b^10 - 4*a^11 + 15*a^2*b^9 + 10*a^3*b^8 - 49*a^4*b^7 + 8*a^5*b^6 + 59*a^6*b^5 - 26*a^7*b^4 - 31*
a^8*b^3 + 18*a^9*b^2))/b^9 + (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 - 16*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 - 6*a^4*b^9 + 4*a^5*b^8))/b^9 - (a*tan(x/2)*(2*a^2 - 3*b^2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*
4i)/b^10)*(2*a^2 - 3*b^2)*1i)/(2*b^4))*1i)/(2*b^4) - (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 - 16*a^7*b^2))/b^6 + (a*((8*(2*a*b^1
2 - 4*b^13 + 10*a^2*b^11 - 6*a^3*b^10 - 6*a^4*b^9 + 4*a^5*b^8))/b^9 + (a*tan(x/2)*(2*a^2 - 3*b^2)*(8*a*b^10 -
16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2 - 3*b^2)*1i)/(2*b^4))*1i)/(2*b^4)))*(2*a^2 - 3*b^2))/b^4

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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