\(\int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx\) [285]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {a \left (a^2-2 b^2\right ) \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {2 a b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {\left (a^2-b^2\right ) \sin (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

[Out]

-a*(a^2-2*b^2)*arctanh((b*cos(x)-a*sin(x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)-2*a*b*cos(x)/(a^2+b^2)^2-(a^2-b^2)
*sin(x)/(a^2+b^2)^2-a^2*b/(a^2+b^2)^2/(a*cos(x)+b*sin(x))

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.38, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3190, 3188, 2717, 2718, 3153, 212, 3178, 3233} \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 a b^2 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {2 a b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}-\frac {a^3 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[In]

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

[Out]

-((a^3*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2)) + (2*a*b^2*ArcTanh[(b*Cos[x] - a*Sin
[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (2*a*b*Cos[x])/(a^2 + b^2)^2 - (a^2*Sin[x])/(a^2 + b^2)^2 + (b^2*Si
n[x])/(a^2 + b^2)^2 - (a^2*b)/((a^2 + b^2)^2*(a*Cos[x] + b*Sin[x]))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3153

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

Rule 3178

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

Rule 3188

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 3190

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

Rule 3233

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> Simp[-(b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Co
s[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - c*C)/(a^2 - b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d +
e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - c*C, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {a \int \frac {\sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}+\frac {b \int \frac {\cos (x) \sin (x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}-\frac {(a b) \int \frac {\sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx}{a^2+b^2} \\ & = -\frac {a^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}+\frac {a^3 \int \frac {1}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^2}+2 \frac {(a b) \int \sin (x) \, dx}{\left (a^2+b^2\right )^2}+\frac {b^2 \int \cos (x) \, dx}{\left (a^2+b^2\right )^2}-2 \frac {\left (a b^2\right ) \int \frac {1}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^2} \\ & = -\frac {2 a b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}-\frac {a^3 \text {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{\left (a^2+b^2\right )^2}+2 \frac {\left (a b^2\right ) \text {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{\left (a^2+b^2\right )^2} \\ & = -\frac {a^3 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {2 a b^2 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {2 a b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01 \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 a \left (a^2-2 b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {5 a^2 b-b^3+b \left (a^2+b^2\right ) \cos (2 x)+a \left (a^2+b^2\right ) \sin (2 x)}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.29

method	result	size
default	\(-\frac {2 a \left (\frac {-b^{2} \tan \left (\frac {x}{2}\right )-a b}{\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a}-\frac {\left (a^{2}-2 b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {2 \left (-a^{2}+b^{2}\right ) \tan \left (\frac {x}{2}\right )-4 a b}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}\)	\(142\)
risch	\(\frac {i {\mathrm e}^{i x}}{-4 i b a +2 a^{2}-2 b^{2}}-\frac {i {\mathrm e}^{-i x}}{2 \left (2 i b a +a^{2}-b^{2}\right )}-\frac {2 b \,a^{2} {\mathrm e}^{i x}}{\left (i b +a \right )^{2} \left (-i b +a \right )^{2} \left (-i b \,{\mathrm e}^{2 i x}+a \,{\mathrm e}^{2 i x}+i b +a \right )}-\frac {a^{3} \ln \left ({\mathrm e}^{i x}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {2 a \ln \left ({\mathrm e}^{i x}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {a^{3} \ln \left ({\mathrm e}^{i x}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {2 a \ln \left ({\mathrm e}^{i x}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\)	\(396\)

[In]

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

[Out]

-2*a/(a^2+b^2)^2*((-b^2*tan(1/2*x)-a*b)/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-a)-(a^2-2*b^2)/(a^2+b^2)^(1/2)*arctanh(
1/2*(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2)))+2/(a^4+2*a^2*b^2+b^4)*((-a^2+b^2)*tan(1/2*x)-2*a*b)/(1+tan(1/2*x)^2
)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (106) = 212\).

Time = 0.26 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.29 \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {4 \, a^{4} b + 2 \, a^{2} b^{3} - 2 \, b^{5} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + \sqrt {a^{2} + b^{2}} {\left ({\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (x\right ) + {\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (x\right )\right )} \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right ) + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )\right )}} \]

[In]

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

[Out]

-1/2*(4*a^4*b + 2*a^2*b^3 - 2*b^5 + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(x)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(x)*
sin(x) + sqrt(a^2 + b^2)*((a^4 - 2*a^2*b^2)*cos(x) + (a^3*b - 2*a*b^3)*sin(x))*log((2*a*b*cos(x)*sin(x) + (a^2
- b^2)*cos(x)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(x) - a*sin(x)))/(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*c
os(x)^2 + b^2)))/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(x) + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sin(x))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (106) = 212\).

Time = 0.30 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.41 \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {{\left (a^{2} - 2 \, b^{2}\right )} a \log \left (\frac {b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a^{2} b + \frac {{\left (a^{3} + 4 \, a b^{2}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {{\left (a^{2} b - 2 \, b^{3}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {{\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} + \frac {2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac {{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.90 \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {{\left (a^{3} - 2 \, a b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, x\right ) - 4 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - 3 \, a^{2} b\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{4} - 2 \, b \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )} {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 23.44 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.26 \[ \int \frac {\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^3+4\,a\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {6\,a^2\,b}{a^4+2\,a^2\,b^2+b^4}-\frac {2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a^2-2\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^2-2\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}}{-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}-\frac {a\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^5-a^4\,b\,1{}\mathrm {i}+2{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^2-a^2\,b^3\,2{}\mathrm {i}+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^4-b^5\,1{}\mathrm {i}}{{\left (a^2+b^2\right )}^{5/2}}\right )\,\left (a^2-2\,b^2\right )\,2{}\mathrm {i}}{{\left (a^2+b^2\right )}^{5/2}} \]

[In]

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

[Out]

- ((2*tan(x/2)*(4*a*b^2 + a^3))/(a^4 + b^4 + 2*a^2*b^2) + (6*a^2*b)/(a^4 + b^4 + 2*a^2*b^2) - (2*a*tan(x/2)^3*
(a^2 - 2*b^2))/(a^4 + b^4 + 2*a^2*b^2) + (2*b*tan(x/2)^2*(a^2 - 2*b^2))/(a^4 + b^4 + 2*a^2*b^2))/(a + 2*b*tan(
x/2) - a*tan(x/2)^4 + 2*b*tan(x/2)^3) - (a*atan((a^5*tan(x/2)*1i - a^4*b*1i - b^5*1i - a^2*b^3*2i + a^3*b^2*ta
n(x/2)*2i + a*b^4*tan(x/2)*1i)/(a^2 + b^2)^(5/2))*(a^2 - 2*b^2)*2i)/(a^2 + b^2)^(5/2)