Integrand size = 19, antiderivative size = 94 \[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\log (\cos (c+d x))}{a d}+\frac {\log (1-\sec (c+d x))}{2 (a+b) d}+\frac {\log (1+\sec (c+d x))}{2 (a-b) d}-\frac {b^2 \log (a+b \sec (c+d x))}{a \left (a^2-b^2\right ) d} \]
ln(cos(d*x+c))/a/d+1/2*ln(1-sec(d*x+c))/(a+b)/d+1/2*ln(1+sec(d*x+c))/(a-b) /d-b^2*ln(a+b*sec(d*x+c))/a/(a^2-b^2)/d
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 \log (\cos (c+d x))}{a}+\frac {\log (1-\sec (c+d x))}{a+b}+\frac {\log (1+\sec (c+d x))}{a-b}-\frac {2 b^2 \log (a+b \sec (c+d x))}{a^3-a b^2}}{2 d} \]
((2*Log[Cos[c + d*x]])/a + Log[1 - Sec[c + d*x]]/(a + b) + Log[1 + Sec[c + d*x]]/(a - b) - (2*b^2*Log[a + b*Sec[c + d*x]])/(a^3 - a*b^2))/(2*d)
Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 25, 4373, 615, 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.
\(\displaystyle \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -\frac {b^2 \int \frac {\cos (c+d x)}{b (a+b \sec (c+d x)) \left (b^2-b^2 \sec ^2(c+d x)\right )}d(b \sec (c+d x))}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle -\frac {b^2 \int \left (\frac {\cos (c+d x)}{a b^3}+\frac {1}{2 b^2 (a+b) (b-b \sec (c+d x))}+\frac {1}{a (a-b) (a+b) (a+b \sec (c+d x))}-\frac {1}{2 (a-b) b^2 (\sec (c+d x) b+b)}\right )d(b \sec (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^2 \left (\frac {\log (a+b \sec (c+d x))}{a \left (a^2-b^2\right )}+\frac {\log (b \sec (c+d x))}{a b^2}-\frac {\log (b-b \sec (c+d x))}{2 b^2 (a+b)}-\frac {\log (b \sec (c+d x)+b)}{2 b^2 (a-b)}\right )}{d}\) |
-((b^2*(Log[b*Sec[c + d*x]]/(a*b^2) - Log[b - b*Sec[c + d*x]]/(2*b^2*(a + b)) + Log[a + b*Sec[c + d*x]]/(a*(a^2 - b^2)) - Log[b + b*Sec[c + d*x]]/(2 *(a - b)*b^2)))/d)
3.3.91.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {-\frac {b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(75\) |
default | \(\frac {-\frac {b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(75\) |
risch | \(\frac {i x}{a}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}+\frac {2 i b^{2} x}{a \left (a^{2}-b^{2}\right )}+\frac {2 i b^{2} c}{d a \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d a \left (a^{2}-b^{2}\right )}\) | \(193\) |
1/d*(-b^2/(a+b)/(a-b)/a*ln(b+a*cos(d*x+c))+1/(2*a+2*b)*ln(cos(d*x+c)-1)+1/ (2*a-2*b)*ln(cos(d*x+c)+1))
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d} \]
-1/2*(2*b^2*log(a*cos(d*x + c) + b) - (a^2 + a*b)*log(1/2*cos(d*x + c) + 1 /2) - (a^2 - a*b)*log(-1/2*cos(d*x + c) + 1/2))/((a^3 - a*b^2)*d)
\[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\cot {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{3} - a b^{2}} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]
-1/2*(2*b^2*log(a*cos(d*x + c) + b)/(a^3 - a*b^2) - log(cos(d*x + c) + 1)/ (a - b) - log(cos(d*x + c) - 1)/(a + b))/d
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (90) = 180\).
Time = 0.31 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.73 \[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {a \log \left ({\left | -a - b + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{2} - b^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\frac {{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | a \right |} \right |}}{{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | a \right |} \right |}}\right )}{{\left (a^{2} - b^{2}\right )} {\left | a \right |}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]
-1/2*(a*log(abs(-a - b + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(co s(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/(a^2 - b^2) - (a^2 - 2*b^2)*log(abs(-2*b - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*ab s(a))/abs(-2*b - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*abs(a)))/((a^2 - b^2)*abs(a)) - log(abs(- cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d
Time = 14.75 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {\cot (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {b^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a\,b^2-a^3\right )} \]