3.275 \(\int \frac {\cos (x) \sin (x)}{a \cos (x)+b \sin (x)} \, dx\)

Optimal. Leaf size=65 \[ \frac {b \sin (x)}{a^2+b^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {a b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

[Out]

a*b*arctanh((b*cos(x)-a*sin(x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)-a*cos(x)/(a^2+b^2)+b*sin(x)/(a^2+b^2)

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3109, 2637, 2638, 3074, 206} \[ \frac {b \sin (x)}{a^2+b^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {a b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

\begin {align*} \int \frac {\cos (x) \sin (x)}{a \cos (x)+b \sin (x)} \, dx &=\frac {a \int \sin (x) \, dx}{a^2+b^2}+\frac {b \int \cos (x) \, dx}{a^2+b^2}-\frac {(a b) \int \frac {1}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}\\ &=-\frac {a \cos (x)}{a^2+b^2}+\frac {b \sin (x)}{a^2+b^2}+\frac {(a b) \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{a^2+b^2}\\ &=\frac {a b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {a \cos (x)}{a^2+b^2}+\frac {b \sin (x)}{a^2+b^2}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 61, normalized size = 0.94 \[ \frac {b \sin (x)-a \cos (x)}{a^2+b^2}-\frac {2 a b \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.65, size = 142, normalized size = 2.18 \[ \frac {\sqrt {a^{2} + b^{2}} a b \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 2 \, {\left (a^{3} + a b^{2}\right )} \cos \relax (x) + 2 \, {\left (a^{2} b + b^{3}\right )} \sin \relax (x)}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 5.98, size = 94, normalized size = 1.45 \[ \frac {a b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) - a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 81, normalized size = 1.25 \[ -\frac {4 a b \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (-b \tan \left (\frac {x}{2}\right )+a \right )}{\left (a^{2}+b^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 105, normalized size = 1.62 \[ \frac {a b \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1}\right )}}{a^{2} + b^{2} + \frac {{\left (a^{2} + b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.38, size = 93, normalized size = 1.43 \[ \frac {2\,a\,b\,\mathrm {atanh}\left (\frac {2\,a^2\,b+2\,b^3-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+b^2\right )}{2\,{\left (a^2+b^2\right )}^{3/2}}\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,a}{a^2+b^2}-\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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