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

Optimal. Leaf size=70 \[ \frac{2 a b x}{\left (a^2+b^2\right )^2}-\frac{b \sin (x)}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac{\left (a^2-b^2\right ) \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.166461, antiderivative size = 87, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3111, 3098, 3133, 3097, 3075} \[ \frac{2 a b x}{\left (a^2+b^2\right )^2}-\frac{b \sin (x)}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac{a^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2}+\frac{b^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*x)/(a^2 + b^2)^2 - (a^2*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^2 + (b^2*Log[a*Cos[x] + b*Sin[x]])/(a^2 +
b^2)^2 - (b*Sin[x])/((a^2 + b^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 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(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 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]

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 3075

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

Rubi steps

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

Mathematica [C]  time = 0.239182, size = 144, normalized size = 2.06 \[ \frac{a \cos (x) \left (\left (b^2-a^2\right ) \log \left ((a \cos (x)+b \sin (x))^2\right )-2 i x (a+i b)^2\right )+b \sin (x) \left (\left (b^2-a^2\right ) \log \left ((a \cos (x)+b \sin (x))^2\right )+2 (a+i b) (a (-1-i x)+b (x+i))\right )+2 i \left (a^2-b^2\right ) \tan ^{-1}(\tan (x)) (a \cos (x)+b \sin (x))}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.089, size = 120, normalized size = 1.7 \begin{align*}{\frac{a}{ \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( x \right ) \right ) }}-{\frac{\ln \left ( a+b\tan \left ( x \right ) \right ){a}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( a+b\tan \left ( x \right ) \right ){b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ){a}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ){b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{ab\arctan \left ( \tan \left ( x \right ) \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.71112, size = 159, normalized size = 2.27 \begin{align*} \frac{2 \, a b x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{2} - b^{2}\right )} \log \left (b \tan \left (x\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac{a}{a^{3} + a b^{2} +{\left (a^{2} b + b^{3}\right )} \tan \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.506245, size = 323, normalized size = 4.61 \begin{align*} \frac{2 \,{\left (2 \, a^{2} b x + a b^{2}\right )} \cos \left (x\right ) -{\left ({\left (a^{3} - a b^{2}\right )} \cos \left (x\right ) +{\left (a^{2} b - b^{3}\right )} \sin \left (x\right )\right )} \log \left (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 (2 \, a b^{2} x - a^{2} b\right )} \sin \left (x\right )}{2 \,{\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right ) +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.09075, size = 194, normalized size = 2.77 \begin{align*} \frac{2 \, a b x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac{a^{2} b \tan \left (x\right ) - b^{3} \tan \left (x\right ) + 2 \, a^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \tan \left (x\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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