Integrand size = 11, antiderivative size = 47 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {\text {arctanh}(\sin (x))}{a}+\frac {b \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}} \] Output:
arctanh(sin(x))/a+b*arctanh((a*cos(x)-b*sin(x))/(a^2+b^2)^(1/2))/a/(a^2+b^ 2)^(1/2)
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.62 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {-\frac {2 b \text {arctanh}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )}{a} \] Input:
Integrate[Sec[x]/(a + b*Cot[x]),x]
Output:
((-2*b*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] - Log[C os[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]])/a
Time = 0.36 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 4001, 25, 25, 3042, 3589, 2009}
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.
\(\displaystyle \int \frac {\sec (x)}{a+b \cot (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right ) \left (a-b \tan \left (x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4001 |
\(\displaystyle \int -\frac {\tan (x)}{-a \sin (x)-b \cos (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\tan (x)}{b \cos (x)+a \sin (x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\tan (x)}{a \sin (x)+b \cos (x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{\cos (x) (a \sin (x)+b \cos (x))}dx\) |
\(\Big \downarrow \) 3589 |
\(\displaystyle \int \left (\frac {\sec (x)}{a}-\frac {b}{a (a \sin (x)+b \cos (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2}}+\frac {\text {arctanh}(\sin (x))}{a}\) |
Input:
Int[Sec[x]/(a + b*Cot[x]),x]
Output:
ArcTanh[Sin[x]]/a + (b*ArcTanh[(a*Cos[x] - b*Sin[x])/Sqrt[a^2 + b^2]])/(a* Sqrt[a^2 + b^2])
Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. ) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[Ex pandTrig[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]
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 = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.34
method | result | size |
default | \(\frac {2 b \,\operatorname {arctanh}\left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \sqrt {a^{2}+b^{2}}}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a}\) | \(63\) |
risch | \(\frac {b \ln \left ({\mathrm e}^{i x}-\frac {i b -a}{\sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}\, a}-\frac {b \ln \left ({\mathrm e}^{i x}+\frac {i b -a}{\sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}\, a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{a}\) | \(109\) |
Input:
int(sec(x)/(a+b*cot(x)),x,method=_RETURNVERBOSE)
Output:
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 /a*ln(tan(1/2*x)-1)+1/a*ln(tan(1/2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (43) = 86\).
Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.98 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} b \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) + {\left (a^{2} + b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a^{2} + b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{3} + a b^{2}\right )}} \] Input:
integrate(sec(x)/(a+b*cot(x)),x, algorithm="fricas")
Output:
1/2*(sqrt(a^2 + b^2)*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)) + (a^2 + b^2)*log(sin(x) + 1) - (a^2 + b^2 )*log(-sin(x) + 1))/(a^3 + a*b^2)
\[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\int \frac {\sec {\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \] Input:
integrate(sec(x)/(a+b*cot(x)),x)
Output:
Integral(sec(x)/(a + b*cot(x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (43) = 86\).
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.09 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {b \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a} \] Input:
integrate(sec(x)/(a+b*cot(x)),x, algorithm="maxima")
Output:
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)))/(sqrt(a^2 + b^2)*a) + log(sin(x)/(cos(x) + 1) + 1 )/a - log(sin(x)/(cos(x) + 1) - 1)/a
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (43) = 86\).
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.91 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {b \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a} \] Input:
integrate(sec(x)/(a+b*cot(x)),x, algorithm="giac")
Output:
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)))/(sqrt(a^2 + b^2)*a) + log(abs(tan(1/2*x) + 1))/a - log(abs(tan(1/2*x) - 1))/a
Time = 9.74 (sec) , antiderivative size = 408, normalized size of antiderivative = 8.68 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^3}{\sqrt {a^2+b^2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}-\frac {64\,b^5}{{\left (a^2+b^2\right )}^{3/2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}+\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}\,\left (\frac {64\,a^2\,b^3}{a^2+b^2}+128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}\right )}-\frac {128\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a^2+b^2\right )}^{3/2}\,\left (\frac {64\,a^2\,b^3}{a^2+b^2}+128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}\right )}+\frac {128\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}-\frac {192\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a^2+b^2\right )}^{3/2}\,\left (128\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {128\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}+\frac {64\,a\,b^3}{a^2+b^2}\right )}\right )}{a\,\sqrt {a^2+b^2}} \] Input:
int(1/(cos(x)*(a + b*cot(x))),x)
Output:
(2*atanh(tan(x/2)))/a - (2*b*atanh((64*b^3)/((a^2 + b^2)^(1/2)*(128*b^2*ta n(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2))) - (64*b ^5)/((a^2 + b^2)^(3/2)*(128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2))) + (128*b^4*tan(x/2))/((a^2 + b^2)^(1/2)*((64*a^ 2*b^3)/(a^2 + b^2) + 128*a*b^2*tan(x/2) - (128*a*b^4*tan(x/2))/(a^2 + b^2) )) - (128*b^6*tan(x/2))/((a^2 + b^2)^(3/2)*((64*a^2*b^3)/(a^2 + b^2) + 128 *a*b^2*tan(x/2) - (128*a*b^4*tan(x/2))/(a^2 + b^2))) + (128*a*b^2*tan(x/2) )/((a^2 + b^2)^(1/2)*(128*b^2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2))) - (192*a*b^4*tan(x/2))/((a^2 + b^2)^(3/2)*(128*b^ 2*tan(x/2) - (128*b^4*tan(x/2))/(a^2 + b^2) + (64*a*b^3)/(a^2 + b^2)))))/( a*(a^2 + b^2)^(1/2))
Time = 0.17 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.04 \[ \int \frac {\sec (x)}{a+b \cot (x)} \, dx=\frac {2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -a i}{\sqrt {a^{2}+b^{2}}}\right ) b i -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b^{2}}{a \left (a^{2}+b^{2}\right )} \] Input:
int(sec(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))*b*i - lo g(tan(x/2) - 1)*a**2 - log(tan(x/2) - 1)*b**2 + log(tan(x/2) + 1)*a**2 + l og(tan(x/2) + 1)*b**2)/(a*(a**2 + b**2))