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

Optimal. Leaf size=48 \[ \frac{a \tanh ^{-1}\left (\frac{a \cos (x)-b \sin (x)}{\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2}}-\frac{\tanh ^{-1}(\cos (x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0783539, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3110, 3770, 3074, 206} \[ \frac{a \tanh ^{-1}\left (\frac{a \cos (x)-b \sin (x)}{\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2}}-\frac{\tanh ^{-1}(\cos (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]/(b*Cos[x] + a*Sin[x]),x]

[Out]

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

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[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 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{\cot (x)}{b \cos (x)+a \sin (x)} \, dx &=\int \left (\frac{\csc (x)}{b}-\frac{a}{b (b \cos (x)+a \sin (x))}\right ) \, dx\\ &=\frac{\int \csc (x) \, dx}{b}-\frac{a \int \frac{1}{b \cos (x)+a \sin (x)} \, dx}{b}\\ &=-\frac{\tanh ^{-1}(\cos (x))}{b}+\frac{a \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,a \cos (x)-b \sin (x)\right )}{b}\\ &=-\frac{\tanh ^{-1}(\cos (x))}{b}+\frac{a \tanh ^{-1}\left (\frac{a \cos (x)-b \sin (x)}{\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2}}\\ \end{align*}

Mathematica [A]  time = 0.0829771, size = 60, normalized size = 1.25 \[ \frac{-\frac{2 a \tanh ^{-1}\left (\frac{b \tan \left (\frac{x}{2}\right )-a}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}+\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.106, size = 49, normalized size = 1. \begin{align*}{\frac{1}{b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{a}{b\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.573723, size = 363, normalized size = 7.56 \begin{align*} \frac{\sqrt{a^{2} + b^{2}} a \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} - a^{2} - 2 \, b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cos \left (x\right ) - b \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} + a^{2}}\right ) -{\left (a^{2} + b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (a^{2} + b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} b + b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (x \right )}}{a \sin{\left (x \right )} + b \cos{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(x)/(a*sin(x) + b*cos(x)), x)

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Giac [A]  time = 1.1932, size = 101, normalized size = 2.1 \begin{align*} \frac{a \log \left (\frac{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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