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

Optimal. Leaf size=92 \[ \frac {a x}{2 \left (a^2+b^2\right )}-\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {b \sin ^2(x)}{2 \left (a^2+b^2\right )}-\frac {a \sin (x) \cos (x)}{2 \left (a^2+b^2\right )}+\frac {a^2 b \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

[Out]

-a*b^2*x/(a^2+b^2)^2+1/2*a*x/(a^2+b^2)+a^2*b*ln(a*cos(x)+b*sin(x))/(a^2+b^2)^2-1/2*a*cos(x)*sin(x)/(a^2+b^2)+1
/2*b*sin(x)^2/(a^2+b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3109, 2564, 30, 2635, 8, 3097, 3133} \[ \frac {a x}{2 \left (a^2+b^2\right )}-\frac {a b^2 x}{\left (a^2+b^2\right )^2}+\frac {b \sin ^2(x)}{2 \left (a^2+b^2\right )}-\frac {a \sin (x) \cos (x)}{2 \left (a^2+b^2\right )}+\frac {a^2 b \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b^2*x)/(a^2 + b^2)^2) + (a*x)/(2*(a^2 + b^2)) + (a^2*b*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^2 - (a*Cos[x
]*Sin[x])/(2*(a^2 + b^2)) + (b*Sin[x]^2)/(2*(a^2 + b^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3097

Int[sin[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(b*x)/(a^2 + b^2), x] - Dist[a/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.34, size = 153, normalized size = 1.66 \[ -\frac {-2 a^3 x+2 a^3 \sin (2 x)+2 b \left (a^2+b^2\right ) \cos (2 x)-2 i b \left (b^2-3 a^2\right ) \tan ^{-1}(\tan (x))-2 \left (a^2+b^2\right ) (b \log (a \cos (x)+b \sin (x))+a x)-6 i a^2 b x-3 a^2 b \log \left ((a \cos (x)+b \sin (x))^2\right )+b^3 \log \left ((a \cos (x)+b \sin (x))^2\right )+6 a b^2 x+2 a b^2 \sin (2 x)+2 i b^3 x}{8 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(-2*a^3*x - (6*I)*a^2*b*x + 6*a*b^2*x + (2*I)*b^3*x - (2*I)*b*(-3*a^2 + b^2)*ArcTan[Tan[x]] + 2*b*(a^2 +
b^2)*Cos[2*x] - 2*(a^2 + b^2)*(a*x + b*Log[a*Cos[x] + b*Sin[x]]) - 3*a^2*b*Log[(a*Cos[x] + b*Sin[x])^2] + b^3*
Log[(a*Cos[x] + b*Sin[x])^2] + 2*a^3*Sin[2*x] + 2*a*b^2*Sin[2*x])/(a^2 + b^2)^2

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 94, normalized size = 1.02 \[ \frac {a^{2} b \log \left (2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}\right ) - {\left (a^{2} b + b^{3}\right )} \cos \relax (x)^{2} - {\left (a^{3} + a b^{2}\right )} \cos \relax (x) \sin \relax (x) + {\left (a^{3} - a b^{2}\right )} x}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 2.01, size = 152, normalized size = 1.65 \[ \frac {a^{2} b^{2} \log \left ({\left | b \tan \relax (x) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {a^{2} b \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (a^{3} - a b^{2}\right )} x}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {a^{2} b \tan \relax (x)^{2} - a^{3} \tan \relax (x) - a b^{2} \tan \relax (x) - b^{3}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.08, size = 174, normalized size = 1.89 \[ -\frac {\tan \relax (x ) a^{3}}{2 \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\relax (x )+1\right )}-\frac {\tan \relax (x ) a \,b^{2}}{2 \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\relax (x )+1\right )}-\frac {a^{2} b}{2 \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\relax (x )+1\right )}-\frac {b^{3}}{2 \left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\relax (x )+1\right )}-\frac {\ln \left (\tan ^{2}\relax (x )+1\right ) a^{2} b}{2 \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \relax (x )\right ) a^{3}}{2 \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \relax (x )\right ) a \,b^{2}}{2 \left (a^{2}+b^{2}\right )^{2}}+\frac {b \,a^{2} \ln \left (a +b \tan \relax (x )\right )}{\left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.46, size = 211, normalized size = 2.29 \[ \frac {a^{2} b \log \left (-a - \frac {2 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a^{2} b \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} - a b^{2}\right )} \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {\frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \frac {2 \, b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {a \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}}{a^{2} + b^{2} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {{\left (a^{2} + b^{2}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.23, size = 3401, normalized size = 36.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*tan(x/2)^3)/(a^2 + b^2) - (a*tan(x/2))/(a^2 + b^2) + (2*b*tan(x/2)^2)/(a^2 + b^2))/(2*tan(x/2)^2 + tan(x/2
)^4 + 1) + (a^2*b*log(a + 2*b*tan(x/2) - a*tan(x/2)^2))/(a^4 + b^4 + 2*a^2*b^2) - (4*a^2*b*log(1/(cos(x) + 1))
)/(4*a^4 + 4*b^4 + 8*a^2*b^2) - (a*atan((tan(x/2)*((((4*a^2*b*((a*(a + b)*((8*(12*a^9*b + 12*a^5*b^5 + 24*a^7*
b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*a^2*b*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a
^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a - b))/(2*(a^4 + b^4 + 2*a^2*b^2
)) - (16*a^3*b*(a + b)*(a - b)*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^
4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2) - (a
*((8*(a^9 + 2*a^3*b^6 - 7*a^5*b^4 - 8*a^7*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*a^2*b*((8*(12*a^9*b +
12*a^5*b^5 + 24*a^7*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*a^2*b*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^
6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b
^4 + 8*a^2*b^2))*(a + b)*(a - b))/(2*(a^4 + b^4 + 2*a^2*b^2)) + (a^3*(a + b)^3*(a - b)^3*(12*a*b^10 + 48*a^3*b
^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((a^4 + b^4 + 2*a^2*b^2)^3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(
a^6 - b^6 + 35*a^2*b^4 - 35*a^4*b^2))/(a^6 + b^6 + 15*a^2*b^4 + 15*a^4*b^2)^2 - (2*a*b*(5*a^4 + 5*b^4 - 26*a^2
*b^2)*((8*(a^7*b + 2*a^5*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (4*a^2*b*((8*(a^9 + 2*a^3*b^6 - 7*a^5*b^4
- 8*a^7*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*a^2*b*((8*(12*a^9*b + 12*a^5*b^5 + 24*a^7*b^3))/(a^6 +
b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (32*a^2*b*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4
*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2)))/(4*a^4 + 4*b^4
+ 8*a^2*b^2) + (a*(a + b)*(a - b)*((a*(a + b)*((8*(12*a^9*b + 12*a^5*b^5 + 24*a^7*b^3))/(a^6 + b^6 + 3*a^2*b^4
+ 3*a^4*b^2) - (32*a^2*b*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8
*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a - b))/(2*(a^4 + b^4 + 2*a^2*b^2)) - (16*a^3*b*(a + b)*(a -
b)*(12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^4 + b^4 +
2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(2*(a^4 + b^4 + 2*a^2*b^2)) - (8*a^4*b*(a + b)^2*(a - b)^2*(
12*a*b^10 + 48*a^3*b^8 + 72*a^5*b^6 + 48*a^7*b^4 + 12*a^9*b^2))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^
2*b^2)^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(a^6 + b^6 + 15*a^2*b^4 + 15*a^4*b^2)^2)*(a^10 + b^10 + 5*a^2*
b^8 + 10*a^4*b^6 + 10*a^6*b^4 + 5*a^8*b^2))/(4*a^6 - 4*a^4*b^2) + (((a*(a + b)*((8*(3*a^8*b + 3*a^4*b^5 + 6*a^
6*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*a^2*b*((8*(4*a^4*b^6 - 2*a^2*b^8 - 2*a^10 + 12*a^6*b^4 + 4*a^
8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (32*a^2*b*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48
*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2))*(a
- b))/(2*(a^4 + b^4 + 2*a^2*b^2)) - (4*a^2*b*((a*(a + b)*(a - b)*((8*(4*a^4*b^6 - 2*a^2*b^8 - 2*a^10 + 12*a^6
*b^4 + 4*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (32*a^2*b*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a
^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(2*(a^4 + b^4 + 2*a^
2*b^2)) + (16*a^3*b*(a + b)*(a - b)*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^4 +
4*b^4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*a^2*b^2)
+ (a^3*(a + b)^3*(a - b)^3*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((a^4 + b^4 + 2*a
^2*b^2)^3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a^6 - b^6 + 35*a^2*b^4 - 35*a^4*b^2)*(a^10 + b^10 + 5*a^2*b^8
+ 10*a^4*b^6 + 10*a^6*b^4 + 5*a^8*b^2))/((4*a^6 - 4*a^4*b^2)*(a^6 + b^6 + 15*a^2*b^4 + 15*a^4*b^2)^2) + (2*a*
b*(5*a^4 + 5*b^4 - 26*a^2*b^2)*((8*a^6*b^2)/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (4*a^2*b*((8*(3*a^8*b + 3*a^
4*b^5 + 6*a^6*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*a^2*b*((8*(4*a^4*b^6 - 2*a^2*b^8 - 2*a^10 + 12*a^
6*b^4 + 4*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (32*a^2*b*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*
a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(4*a^4 + 4*b^4 + 8*
a^2*b^2)))/(4*a^4 + 4*b^4 + 8*a^2*b^2) + (a*((a*(a + b)*(a - b)*((8*(4*a^4*b^6 - 2*a^2*b^8 - 2*a^10 + 12*a^6*b
^4 + 4*a^8*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (32*a^2*b*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6
*b^5 + 48*a^8*b^3))/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))))/(2*(a^4 + b^4 + 2*a^2*
b^2)) + (16*a^3*b*(a + b)*(a - b)*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))/((4*a^4 + 4
*b^4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a + b)*(a - b))/(2*(a^4 + b^4
+ 2*a^2*b^2)) + (8*a^4*b*(a + b)^2*(a - b)^2*(12*a^10*b + 12*a^2*b^9 + 48*a^4*b^7 + 72*a^6*b^5 + 48*a^8*b^3))
/((4*a^4 + 4*b^4 + 8*a^2*b^2)*(a^4 + b^4 + 2*a^2*b^2)^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))*(a^10 + b^10 + 5
*a^2*b^8 + 10*a^4*b^6 + 10*a^6*b^4 + 5*a^8*b^2))/((4*a^6 - 4*a^4*b^2)*(a^6 + b^6 + 15*a^2*b^4 + 15*a^4*b^2)^2)
)*(a + b)*(a - b))/(a^4 + b^4 + 2*a^2*b^2)

________________________________________________________________________________________

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(cos(x)*sin(x)**2/(a*cos(x)+b*sin(x)),x)

[Out]

Timed out

________________________________________________________________________________________