\(\int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx\) [287]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=-\frac {b \left (-2 a^2+b^2\right ) \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {\left (a^2-b^2\right ) \cos (x)}{\left (a^2+b^2\right )^2}+\frac {2 a b \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a b^2}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.39, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3190, 3179, 2717, 3153, 212, 3188, 2718, 3234} \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 a^2 b \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b^3 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {2 a b \sin (x)}{\left (a^2+b^2\right )^2}+\frac {b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 \cos (x)}{\left (a^2+b^2\right )^2}+\frac {a b^2}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

[In]

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

[Out]

(2*a^2*b*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (b^3*ArcTanh[(b*Cos[x] - a*Sin[x]
)/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a^2*Cos[x])/(a^2 + b^2)^2 + (b^2*Cos[x])/(a^2 + b^2)^2 + (2*a*b*Sin[x
])/(a^2 + b^2)^2 + (a*b^2)/((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 3179

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[b*(Cos[c + d*x]^(m - 1)/(d*(a^2 + b^2)*(m - 1))), x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1), x]
, x] + Dist[b^2/(a^2 + b^2), Int[Cos[c + d*x]^(m - 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), 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 3234

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

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

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 b \left (-2 a^2+b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^3-5 a b^2+a \left (a^2+b^2\right ) \cos (2 x)-b \left (a^2+b^2\right ) \sin (2 x)}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

[In]

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

[Out]

(2*b*(-2*a^2 + b^2)*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a^3 - 5*a*b^2 + a*(a^2 +
b^2)*Cos[2*x] - b*(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 = 143, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {4 b \left (\frac {-\frac {b^{2} \tan \left (\frac {x}{2}\right )}{2}-\frac {a b}{2}}{\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a}-\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {4 \tan \left (\frac {x}{2}\right ) a b -2 a^{2}+2 b^{2}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}\)	\(143\)
risch	\(-\frac {{\mathrm e}^{i x}}{2 \left (-2 i b a +a^{2}-b^{2}\right )}-\frac {{\mathrm e}^{-i x}}{2 \left (2 i b a +a^{2}-b^{2}\right )}+\frac {2 i a \,b^{2} {\mathrm e}^{i x}}{\left (-i a +b \right )^{2} \left (i a +b \right )^{2} \left (b \,{\mathrm e}^{2 i x}+i a \,{\mathrm e}^{2 i x}-b +i a \right )}+\frac {2 i b \ln \left ({\mathrm e}^{i x}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}-\frac {i b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}-\frac {2 i b \ln \left ({\mathrm e}^{i x}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}+\frac {i b^{3} \ln \left ({\mathrm e}^{i x}-\frac {i b +a}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}\)	\(326\)

[In]

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

[Out]

4*b/(a^2+b^2)^2*((-1/2*b^2*tan(1/2*x)-1/2*a*b)/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-a)-1/2*(2*a^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)))+4/(a^4+2*a^2*b^2+b^4)*(tan(1/2*x)*a*b-1/2*a^2+1/2*b^2)/(
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 = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.31 \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {6 \, a^{3} b^{2} + 6 \, a b^{4} - 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right ) \sin \left (x\right ) - \sqrt {a^{2} + b^{2}} {\left ({\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (x\right ) + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )\right )} \log \left (-\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right ) + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )\right )}} \]

[In]

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

[Out]

1/2*(6*a^3*b^2 + 6*a*b^4 - 2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(x)^2 + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(x)*sin(x) -
sqrt(a^2 + b^2)*((2*a^3*b - a*b^3)*cos(x) + (2*a^2*b^2 - b^4)*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)*cos(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. \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(x)**2*sin(x)/(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. 264 vs. \(2 (106) = 212\).

Time = 0.32 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.42 \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {{\left (2 \, a^{2} b - b^{3}\right )} \log \left (\frac {b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (a^{3} - 2 \, a b^{2} - \frac {3 \, b^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {{\left (a^{3} + 4 \, a b^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {{\left (2 \, a^{2} b - b^{3}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} + \frac {2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac {{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}} \]

[In]

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

[Out]

(2*a^2*b - b^3)*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*(a^3 - 2*a*b^2 - 3*b^3*sin(x)/(cos(x) + 1) - (a^3 + 4*a*b^2)*
sin(x)^2/(cos(x) + 1)^2 + (2*a^2*b - b^3)*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 = 0.31 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.87 \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {{\left (2 \, a^{2} b - b^{3}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (2 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right ) + a^{3} - 2 \, a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{4} - 2 \, b \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )} {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \]

[In]

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

[Out]

(2*a^2*b - b^3)*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*(2*a^2*b*tan(1/2*x)^3 - b^3*tan(1/2*x)^3 - a^3*tan(1/2*x)^2 -
4*a*b^2*tan(1/2*x)^2 - 3*b^3*tan(1/2*x) + a^3 - 2*a*b^2)/((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 = 24.45 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.32 \[ \int \frac {\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {\frac {2\,\left (2\,a\,b^2-a^3\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {6\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^3+4\,a\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (2\,a^2-b^2\right )}{a^4+2\,a^2\,b^2+b^4}}{-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}+\frac {b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^5-a^4\,b\,1{}\mathrm {i}+2{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^2-a^2\,b^3\,2{}\mathrm {i}+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^4-b^5\,1{}\mathrm {i}}{{\left (a^2+b^2\right )}^{5/2}}\right )\,\left (2\,a^2-b^2\right )\,2{}\mathrm {i}}{{\left (a^2+b^2\right )}^{5/2}} \]

[In]

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

[Out]

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