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

Optimal. Leaf size=60 \[ \frac {a}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac {b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3154, 3074, 206} \[ \frac {a}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac {b \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

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

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

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Mathematica [A]  time = 0.16, size = 62, normalized size = 1.03 \[ \frac {a}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac {2 b \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 164, normalized size = 2.73 \[ \frac {2 \, a^{3} + 2 \, a b^{2} + {\left (a b \cos \relax (x) + b^{2} \sin \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^3 + 2*a*b^2 + (a*b*cos(x) + b^2*sin(x))*sqrt(a^2 + b^2)*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^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 [A]  time = 0.26, size = 103, normalized size = 1.72 \[ -\frac {b \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )} {\left (a^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.55, size = 97, normalized size = 1.62 \[ \frac {8 \tan \left (\frac {x}{2}\right ) b +8 a}{\left (-4 a^{2}-4 b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {8 b \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (-4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(2*tan(1/2*x)*b+2*a)/(-4*a^2-4*b^2)/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)-8*b/(-4*a^2-4*b^2)/(a^2+b^2)^(1/2)*arc
tanh(1/2*(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2))

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maxima [B]  time = 0.42, size = 128, normalized size = 2.13 \[ -\frac {b \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a + \frac {b \sin \relax (x)}{\cos \relax (x) + 1}\right )}}{a^{3} + a b^{2} + \frac {2 \, {\left (a^{2} b + b^{3}\right )} \sin \relax (x)}{\cos \relax (x) + 1} - \frac {{\left (a^{3} + a b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.61, size = 86, normalized size = 1.43 \[ \frac {\frac {2\,a}{a^2+b^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}}{-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}-\frac {2\,b\,\mathrm {atanh}\left (\frac {2\,b-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )}{2\,\sqrt {a^2+b^2}}\right )}{{\left (a^2+b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a*cos(x) + b*sin(x))^2,x)

[Out]

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

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

[Out]

Timed out

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