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

Optimal. Leaf size=109 \[ \frac{2 a b \sin (x)}{\left (a^2+b^2\right )^2}-\frac{\left (a^2-b^2\right ) \cos (x)}{\left (a^2+b^2\right )^2}+\frac{a b^2}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}-\frac{b \left (b^2-2 a^2\right ) \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.259293, antiderivative size = 151, normalized size of antiderivative = 1.39, 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, 3100, 2637, 3074, 206, 3109, 2638, 3155} \[ \frac{2 a b \sin (x)}{\left (a^2+b^2\right )^2}+\frac{b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac{a^2 \cos (x)}{\left (a^2+b^2\right )^2}+\frac{a b^2}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}-\frac{b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{2 a^2 b \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[x]^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 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 3100

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 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 2638

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

Rule 3155

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*} \int \frac{\cos ^2(x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx &=\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 ) \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{b^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}\\ &=\frac{2 a^2 b \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{b^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{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]  time = 0.682217, size = 110, normalized size = 1.01 \[ \frac{2 b \left (b^2-2 a^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{-b \left (a^2+b^2\right ) \sin (2 x)+a \left (a^2+b^2\right ) \cos (2 x)+a^3-5 a b^2}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[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]))

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Maple [A]  time = 0.121, size = 144, normalized size = 1.3 \begin{align*} -4\,{\frac{-ab\tan \left ( x/2 \right ) +1/2\,{a}^{2}-1/2\,{b}^{2}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+4\,{\frac{b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ({\frac{-1/2\,\tan \left ( x/2 \right ){b}^{2}-1/2\,ab}{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a}}-1/2\,{\frac{2\,{a}^{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)^2*sin(x)/(a*cos(x)+b*sin(x))^2,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.542812, size = 586, normalized size = 5.38 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

[Out]

Timed out

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Giac [A]  time = 1.24049, size = 275, normalized size = 2.52 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))