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)*ln(a*cos(x)+b*sin(x))/(a^2+b^2)^2-b*sin(x)/(a^2+b^2)/(a*cos(x)+b*sin(x))

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Rubi [A]  time = 0.17, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 0.26, 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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fricas [A]  time = 0.51, size = 138, normalized size = 1.97 \[ \frac {2 \, {\left (2 \, a^{2} b x + a b^{2}\right )} \cos \relax (x) - {\left ({\left (a^{3} - a b^{2}\right )} \cos \relax (x) + {\left (a^{2} b - b^{3}\right )} \sin \relax (x)\right )} \log \left (2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}\right ) + 2 \, {\left (2 \, a b^{2} x - a^{2} b\right )} \sin \relax (x)}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \relax (x)\right )}} \]

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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giac [B]  time = 0.19, size = 144, normalized size = 2.06 \[ \frac {2 \, a b x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \tan \relax (x) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {a^{2} b \tan \relax (x) - b^{3} \tan \relax (x) + 2 \, a^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \relax (x) + a\right )}} \]

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))

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maple [A]  time = 0.10, size = 120, normalized size = 1.71 \[ \frac {\ln \left (\tan ^{2}\relax (x )+1\right ) a^{2}}{2 \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (\tan ^{2}\relax (x )+1\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{2}}+\frac {2 a b \arctan \left (\tan \relax (x )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {a}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \relax (x )\right )}-\frac {\ln \left (a +b \tan \relax (x )\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \relax (x )\right ) b^{2}}{\left (a^{2}+b^{2}\right )^{2}} \]

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]

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))+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

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maxima [A]  time = 0.45, size = 118, normalized size = 1.69 \[ \frac {2 \, a b x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (b \tan \relax (x) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {a}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \relax (x)} \]

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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mupad [B]  time = 5.16, size = 1017, normalized size = 14.53 \[ \text {result too large to display} \]

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]

-(b^3*sin(x) + a^3*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*cos(x) - b^3*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*si
n(x) + a^2*b*sin(x) - a^3*log(-(65536*a^4*b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^1
2*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 14*a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*
a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x))/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 3
5*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) + 14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*cos(x) + b^3*log(-(65536*a^4*
b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^12*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 14*
a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x))
/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 35*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) + 14
*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*sin(x) - 4*a^2*b*atan(sin(x/2)/cos(x/2))*cos(x) - 4*a*b^2*atan(sin(x/2)
/cos(x/2))*sin(x) + a*b^2*log(-(65536*a^4*b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^1
2*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 14*a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*
a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x))/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 3
5*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) + 14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*cos(x) - a^2*b*log(-(65536*a^
4*b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^12*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 1
4*a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x
))/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 35*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) +
14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*sin(x) - a*b^2*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*cos(x) + a^2*b*l
og((a*cos(x) + b*sin(x))/cos(x/2)^2)*sin(x))/(b^5*sin(x) + a^5*cos(x) + a*b^4*cos(x) + a^4*b*sin(x) + 2*a^3*b^
2*cos(x) + 2*a^2*b^3*sin(x))

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sympy [A]  time = 2.10, size = 991, normalized size = 14.16 \[ \text {result too large to display} \]

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]

Piecewise((zoo*log(sin(x)), Eq(a, 0) & Eq(b, 0)), (-log(cos(x))/a**2, Eq(b, 0)), (2*I*x*sin(x)**2/(8*b**2*sin(
x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) + 4*x*sin(x)*cos(x)/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*c
os(x) - 8*b**2*cos(x)**2) - 2*I*x*cos(x)**2/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) +
sin(x)**2/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) - cos(x)**2/(8*b**2*sin(x)**2 - 16*I
*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2), Eq(a, -I*b)), (-2*I*x*sin(x)**2/(8*b**2*sin(x)**2 + 16*I*b**2*sin(x)*
cos(x) - 8*b**2*cos(x)**2) + 4*x*sin(x)*cos(x)/(8*b**2*sin(x)**2 + 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2)
+ 2*I*x*cos(x)**2/(8*b**2*sin(x)**2 + 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) + sin(x)**2/(8*b**2*sin(x)*
*2 + 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) - cos(x)**2/(8*b**2*sin(x)**2 + 16*I*b**2*sin(x)*cos(x) - 8*b
**2*cos(x)**2), Eq(a, I*b)), (-a**3*log(a*cos(x)/b + sin(x))*cos(x)/(a**5*cos(x) + a**4*b*sin(x) + 2*a**3*b**2
*cos(x) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x)) + a**3*cos(x)/(a**5*cos(x) + a**4*b*sin(x) + 2*a**
3*b**2*cos(x) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x)) + 2*a**2*b*x*cos(x)/(a**5*cos(x) + a**4*b*si
n(x) + 2*a**3*b**2*cos(x) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x)) - a**2*b*log(a*cos(x)/b + sin(x)
)*sin(x)/(a**5*cos(x) + a**4*b*sin(x) + 2*a**3*b**2*cos(x) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x))
+ 2*a*b**2*x*sin(x)/(a**5*cos(x) + a**4*b*sin(x) + 2*a**3*b**2*cos(x) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) +
b**5*sin(x)) + a*b**2*log(a*cos(x)/b + sin(x))*cos(x)/(a**5*cos(x) + a**4*b*sin(x) + 2*a**3*b**2*cos(x) + 2*a*
*2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x)) + a*b**2*cos(x)/(a**5*cos(x) + a**4*b*sin(x) + 2*a**3*b**2*cos(x
) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x)) + b**3*log(a*cos(x)/b + sin(x))*sin(x)/(a**5*cos(x) + a*
*4*b*sin(x) + 2*a**3*b**2*cos(x) + 2*a**2*b**3*sin(x) + a*b**4*cos(x) + b**5*sin(x)), True))

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