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])/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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Rubi [A]  time = 0.0464099, antiderivative size = 60, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.154449, 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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Maple [A]  time = 0.092, size = 97, normalized size = 1.6 \begin{align*} 4\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{ \left ( -4\,{a}^{2}-4\,{b}^{2} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a \right ) }}-8\,{\frac{b}{ \left ( -4\,{a}^{2}-4\,{b}^{2} \right ) \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 ) } \end{align*}

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*b*tan(1/2*x)+2*a)/(-4*a^2-4*b^2)/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-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 [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(sin(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.498819, size = 398, normalized size = 6.63 \begin{align*} \frac{2 \, a^{3} + 2 \, a b^{2} +{\left (a b \cos \left (x\right ) + b^{2} \sin \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \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^{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(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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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(sin(x)/(a*cos(x)+b*sin(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19466, size = 139, normalized size = 2.32 \begin{align*} -\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 )}} \end{align*}

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))