Integrand size = 19, antiderivative size = 81 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {1}{4 a^2 d (1+\cos (c+d x))^2}+\frac {5}{4 a^2 d (1+\cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{8 a^2 d}+\frac {7 \log (1+\cos (c+d x))}{8 a^2 d} \] Output:
-1/4/a^2/d/(1+cos(d*x+c))^2+5/4/a^2/d/(1+cos(d*x+c))+1/8*ln(1-cos(d*x+c))/ a^2/d+7/8*ln(1+cos(d*x+c))/a^2/d
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\left (-1+10 \cos ^2\left (\frac {1}{2} (c+d x)\right )+4 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (7 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right ) \sec ^2(c+d x)}{4 a^2 d (1+\sec (c+d x))^2} \] Input:
Integrate[Cot[c + d*x]/(a + a*Sec[c + d*x])^2,x]
Output:
((-1 + 10*Cos[(c + d*x)/2]^2 + 4*Cos[(c + d*x)/2]^4*(7*Log[Cos[(c + d*x)/2 ]] + Log[Sin[(c + d*x)/2]]))*Sec[c + d*x]^2)/(4*a^2*d*(1 + Sec[c + d*x])^2 )
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 25, 4367, 27, 99, 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.
\(\displaystyle \int \frac {\cot (c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {a^2 \int \frac {\cos ^3(c+d x)}{a^4 (1-\cos (c+d x)) (\cos (c+d x)+1)^3}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cos ^3(c+d x)}{(1-\cos (c+d x)) (\cos (c+d x)+1)^3}d\cos (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {7}{8 (\cos (c+d x)+1)}+\frac {5}{4 (\cos (c+d x)+1)^2}-\frac {1}{2 (\cos (c+d x)+1)^3}-\frac {1}{8 (\cos (c+d x)-1)}\right )d\cos (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {5}{4 (\cos (c+d x)+1)}+\frac {1}{4 (\cos (c+d x)+1)^2}-\frac {1}{8} \log (1-\cos (c+d x))-\frac {7}{8} \log (\cos (c+d x)+1)}{a^2 d}\) |
Input:
Int[Cot[c + d*x]/(a + a*Sec[c + d*x])^2,x]
Output:
-((1/(4*(1 + Cos[c + d*x])^2) - 5/(4*(1 + Cos[c + d*x])) - Log[1 - Cos[c + d*x]]/8 - (7*Log[1 + Cos[c + d*x]])/8)/(a^2*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[1/(a^(m - n - 1)*b^n*d) Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {5}{4 \left (1+\cos \left (d x +c \right )\right )}+\frac {7 \ln \left (1+\cos \left (d x +c \right )\right )}{8}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{8}}{d \,a^{2}}\) | \(55\) |
default | \(\frac {-\frac {1}{4 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {5}{4 \left (1+\cos \left (d x +c \right )\right )}+\frac {7 \ln \left (1+\cos \left (d x +c \right )\right )}{8}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{8}}{d \,a^{2}}\) | \(55\) |
risch | \(-\frac {i x}{a^{2}}-\frac {2 i c}{d \,a^{2}}+\frac {5 \,{\mathrm e}^{3 i \left (d x +c \right )}+8 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4}}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d \,a^{2}}\) | \(114\) |
Input:
int(cot(d*x+c)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(-1/4/(1+cos(d*x+c))^2+5/4/(1+cos(d*x+c))+7/8*ln(1+cos(d*x+c))+1/8 *ln(-1+cos(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.31 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 10 \, \cos \left (d x + c\right ) + 8}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/8*(7*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + (cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 10*c os(d*x + c) + 8)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (5 \, \cos \left (d x + c\right ) + 4\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}} + \frac {7 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
1/8*(2*(5*cos(d*x + c) + 4)/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^2 ) + 7*log(cos(d*x + c) + 1)/a^2 + log(cos(d*x + c) - 1)/a^2)/d
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{8 \, a^{2} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{8 \, a^{2} d} + \frac {5 \, \cos \left (d x + c\right ) + 4}{4 \, a^{2} d {\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
7/8*log(abs(cos(d*x + c) + 1))/(a^2*d) + 1/8*log(abs(cos(d*x + c) - 1))/(a ^2*d) + 1/4*(5*cos(d*x + c) + 4)/(a^2*d*(cos(d*x + c) + 1)^2)
Time = 11.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16}}{a^2\,d} \] Input:
int(cot(c + d*x)/(a + a/cos(c + d*x))^2,x)
Output:
(log(tan(c/2 + (d*x)/2))/4 - log(tan(c/2 + (d*x)/2)^2 + 1) + tan(c/2 + (d* x)/2)^2/2 - tan(c/2 + (d*x)/2)^4/16)/(a^2*d)
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {-16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16 a^{2} d} \] Input:
int(cot(d*x+c)/(a+a*sec(d*x+c))^2,x)
Output:
( - 16*log(tan((c + d*x)/2)**2 + 1) + 4*log(tan((c + d*x)/2)) - tan((c + d *x)/2)**4 + 8*tan((c + d*x)/2)**2)/(16*a**2*d)