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))/(a^2+b^2)^(1/2))/b/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 48, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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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fricas [B]  time = 0.47, size = 142, normalized size = 2.96 \[ \frac {\sqrt {a^{2} + b^{2}} a \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \relax (x) - b \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}}\right ) - {\left (a^{2} + b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (a^{2} + b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b + b^{3}\right )}} \]

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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giac [A]  time = 0.21, size = 75, normalized size = 1.56 \[ \frac {a \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{b} \]

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

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maple [A]  time = 0.10, size = 49, normalized size = 1.02 \[ -\frac {2 a \arctanh \left (\frac {2 b \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{b \sqrt {a^{2}+b^{2}}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))+1/b*ln(tan(1/2*x))

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maxima [A]  time = 0.46, size = 79, normalized size = 1.65 \[ \frac {a \log \left (\frac {a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b} + \frac {\log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.98, size = 123, normalized size = 2.56 \[ \frac {\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{b}-\frac {2\,a\,\mathrm {atanh}\left (\frac {\sqrt {a^2+b^2}\,\left (4{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,a^2+2{}\mathrm {i}\,\cos \left (\frac {x}{2}\right )\,a\,b+1{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,b^2\right )}{a^3\,\sin \left (\frac {x}{2}\right )\,4{}\mathrm {i}+a^2\,b\,\cos \left (\frac {x}{2}\right )\,1{}\mathrm {i}+a\,b^2\,\sin \left (\frac {x}{2}\right )\,3{}\mathrm {i}+b\,\cos \left (\frac {x}{2}\right )\,\left (a^2+b^2\right )\,1{}\mathrm {i}}\right )}{b\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(b*cos(x) + a*sin(x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\relax (x )}}{a \sin {\relax (x )} + b \cos {\relax (x )}}\, dx \]

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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