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

Optimal. Leaf size=110 \[ -\frac{\left (a^2-b^2\right ) \sin (x)}{\left (a^2+b^2\right )^2}-\frac{2 a b \cos (x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}-\frac{a \left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.239339, antiderivative size = 152, normalized size of antiderivative = 1.38, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3111, 3109, 2637, 2638, 3074, 206, 3099, 3154} \[ -\frac{a^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac{b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac{2 a b \cos (x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}-\frac{a^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{2 a b^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[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 3111

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 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 2637

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

Rule 2638

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

Rule 3074

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 206

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

Rule 3099

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 3154

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*} \int \frac{\cos (x) \sin ^2(x)}{(a \cos (x)+b \sin (x))^2} \, dx &=\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 \operatorname{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 ) \operatorname{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 \tanh ^{-1}\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 \tanh ^{-1}\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]  time = 0.589082, size = 111, normalized size = 1.01 \[ \frac{2 a \left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{a \left (a^2+b^2\right ) \sin (2 x)+b \left (a^2+b^2\right ) \cos (2 x)+5 a^2 b-b^3}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[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]))

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Maple [A]  time = 0.114, size = 142, normalized size = 1.3 \begin{align*} 2\,{\frac{ \left ( -{a}^{2}+{b}^{2} \right ) \tan \left ( x/2 \right ) -2\,ab}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{a}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ({\frac{-\tan \left ( x/2 \right ){b}^{2}-ab}{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a}}-{\frac{{a}^{2}-2\,{b}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.544414, size = 597, normalized size = 5.43 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.20592, size = 282, normalized size = 2.56 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))