3.1.14 \(\int \frac {\cos (x)}{a+b \cot (x)} \, dx\) [14]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {3599, 3188, 2717, 2718, 3153, 212} \begin {gather*} \frac {a \sin (x)}{a^2+b^2}-\frac {b \cos (x)}{a^2+b^2}+\frac {a b \tanh ^{-1}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(a + b*Cot[x]),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3153

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 3188

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]

Rule 3599

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Sin[e + f*x]^
m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/Cos[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 61, normalized size = 0.94 \begin {gather*} -\frac {2 a b \tanh ^{-1}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {-b \cos (x)+a \sin (x)}{a^2+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]/(a + b*Cot[x]),x]

[Out]

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

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Maple [A]
time = 0.29, size = 81, normalized size = 1.25

method	result	size
default	\(-\frac {2 \left (-a \tan \left (\frac {x}{2}\right )+b \right )}{\left (a^{2}+b^{2}\right ) \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}+\frac {4 a b \arctanh \left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}\)	\(81\)
risch	\(-\frac {i {\mathrm e}^{i x}}{2 \left (i b +a \right )}+\frac {i {\mathrm e}^{-i x}}{-2 i b +2 a}+\frac {b a \ln \left ({\mathrm e}^{i x}-\frac {i a^{2} b +i b^{3}-a^{3}-a \,b^{2}}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {b a \ln \left ({\mathrm e}^{i x}+\frac {i a^{2} b +i b^{3}-a^{3}-a \,b^{2}}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(a+b*cot(x)),x,method=_RETURNVERBOSE)

[Out]

-2/(a^2+b^2)*(-a*tan(1/2*x)+b)/(1+tan(1/2*x)^2)+4*a*b/(2*a^2+2*b^2)/(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 [A]
time = 0.51, size = 105, normalized size = 1.62 \begin {gather*} \frac {a b \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}}{a^{2} + b^{2} + \frac {{\left (a^{2} + b^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (61) = 122\).
time = 4.09, size = 144, normalized size = 2.22 \begin {gather*} \frac {\sqrt {a^{2} + b^{2}} a b \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 ) - 2 \, {\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(x)/(a + b*cot(x)), x)

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Giac [A]
time = 0.44, size = 94, normalized size = 1.45 \begin {gather*} \frac {a b \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 )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, x\right ) - b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.38, size = 93, normalized size = 1.43 \begin {gather*} \frac {2\,a\,b\,\mathrm {atanh}\left (\frac {2\,a\,b^2+2\,a^3-2\,b\,\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\,b}{a^2+b^2}-\frac {2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(a + b*cot(x)),x)

[Out]

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

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