Integrand size = 11, antiderivative size = 65 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\frac {a b \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {b \cos (x)}{a^2+b^2}+\frac {a \sin (x)}{a^2+b^2} \] Output:
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)
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=-\frac {2 a b \text {arctanh}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {-b \cos (x)+a \sin (x)}{a^2+b^2} \] Input:
Integrate[Cos[x]/(a + b*Cot[x]),x]
Output:
(-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)
Time = 0.45 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4001, 25, 25, 3042, 3588, 3042, 3117, 3118, 3553, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos (x)}{a+b \cot (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (x+\frac {\pi }{2}\right )}{a-b \tan \left (x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4001 |
\(\displaystyle \int -\frac {\sin (x) \cos (x)}{-a \sin (x)-b \cos (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos (x) \sin (x)}{b \cos (x)+a \sin (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sin (x) \cos (x)}{a \sin (x)+b \cos (x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x) \cos (x)}{a \sin (x)+b \cos (x)}dx\) |
\(\Big \downarrow \) 3588 |
\(\displaystyle \frac {b \int \sin (x)dx}{a^2+b^2}+\frac {a \int \cos (x)dx}{a^2+b^2}-\frac {a b \int \frac {1}{b \cos (x)+a \sin (x)}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \int \sin (x)dx}{a^2+b^2}+\frac {a \int \sin \left (x+\frac {\pi }{2}\right )dx}{a^2+b^2}-\frac {a b \int \frac {1}{b \cos (x)+a \sin (x)}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {b \int \sin (x)dx}{a^2+b^2}-\frac {a b \int \frac {1}{b \cos (x)+a \sin (x)}dx}{a^2+b^2}+\frac {a \sin (x)}{a^2+b^2}\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle -\frac {a b \int \frac {1}{b \cos (x)+a \sin (x)}dx}{a^2+b^2}+\frac {a \sin (x)}{a^2+b^2}-\frac {b \cos (x)}{a^2+b^2}\) |
\(\Big \downarrow \) 3553 |
\(\displaystyle \frac {a b \int \frac {1}{a^2+b^2-(a \cos (x)-b \sin (x))^2}d(a \cos (x)-b \sin (x))}{a^2+b^2}+\frac {a \sin (x)}{a^2+b^2}-\frac {b \cos (x)}{a^2+b^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a b \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {a \sin (x)}{a^2+b^2}-\frac {b \cos (x)}{a^2+b^2}\) |
Input:
Int[Cos[x]/(a + b*Cot[x]),x]
Output:
(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)
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x _Symbol] :> Simp[-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]
Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. ) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[b /(a^2 + b^2) Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Simp[a/(a ^2 + b^2) Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Simp[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]
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/C os[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]))
Time = 0.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25
method | result | size |
default | \(\frac {4 a b \,\operatorname {arctanh}\left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (-a \tan \left (\frac {x}{2}\right )+b \right )}{\left (a^{2}+b^{2}\right ) \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}\) | \(81\) |
risch | \(-\frac {i {\mathrm e}^{i x}}{2 \left (i b +a \right )}+\frac {i {\mathrm e}^{-i x}}{-2 i b +2 a}+\frac {b a \ln \left ({\mathrm e}^{i x}-\frac {i a^{2} b +i b^{3}-a^{3}-a \,b^{2}}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {b a \ln \left ({\mathrm e}^{i x}+\frac {i a^{2} b +i b^{3}-a^{3}-a \,b^{2}}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\) | \(144\) |
Input:
int(cos(x)/(a+b*cot(x)),x,method=_RETURNVERBOSE)
Output:
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))-2/(a^2+b^2)*(-a*tan(1/2*x)+b)/(1+tan(1/2*x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (61) = 122\).
Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.22 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} a b \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \] Input:
integrate(cos(x)/(a+b*cot(x)),x, algorithm="fricas")
Output:
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)
\[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\int \frac {\cos {\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \] Input:
integrate(cos(x)/(a+b*cot(x)),x)
Output:
Integral(cos(x)/(a + b*cot(x)), x)
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.62 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\frac {a b \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}}{a^{2} + b^{2} + \frac {{\left (a^{2} + b^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}} \] Input:
integrate(cos(x)/(a+b*cot(x)),x, algorithm="maxima")
Output:
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)
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.45 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\frac {a b \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, x\right ) - b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}} \] Input:
integrate(cos(x)/(a+b*cot(x)),x, algorithm="giac")
Output:
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))
Time = 9.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\frac {2\,a\,b\,\mathrm {atanh}\left (\frac {2\,a\,b^2+2\,a^3-2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+b^2\right )}{2\,{\left (a^2+b^2\right )}^{3/2}}\right )}{{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,b}{a^2+b^2}-\frac {2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1} \] Input:
int(cos(x)/(a + b*cot(x)),x)
Output:
(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)
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.42 \[ \int \frac {\cos (x)}{a+b \cot (x)} \, dx=\frac {2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i -\cos \left (x \right ) a^{2} b -\cos \left (x \right ) b^{3}+\sin \left (x \right ) a^{3}+\sin \left (x \right ) a \,b^{2}+a^{2} b +b^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}} \] Input:
int(cos(x)/(a+b*cot(x)),x)
Output:
(2*sqrt(a**2 + b**2)*atan((tan(x/2)*b*i - a*i)/sqrt(a**2 + b**2))*a*b*i - cos(x)*a**2*b - cos(x)*b**3 + sin(x)*a**3 + sin(x)*a*b**2 + a**2*b + b**3) /(a**4 + 2*a**2*b**2 + b**4)