Integrand size = 17, antiderivative size = 43 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\frac {(a+b) \log (1-\cos (c+d x))}{2 d}+\frac {(a-b) \log (1+\cos (c+d x))}{2 d} \] Output:
1/2*(a+b)*ln(1-cos(d*x+c))/d+1/2*(a-b)*ln(1+cos(d*x+c))/d
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.60 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=-\frac {b \text {arctanh}(\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d} \] Input:
Integrate[Cot[c + d*x]*(a + b*Sec[c + d*x]),x]
Output:
-((b*ArcTanh[Cos[c + d*x]])/d) + (a*Log[Sin[c + d*x]])/d
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 25, 4371, 3042, 25, 3147, 452, 219, 240}
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.
\(\displaystyle \int \cot (c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}{\cot \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 4371 |
\(\displaystyle -\int \sec \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int -\frac {b-a \sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -\frac {a \int \frac {b+a \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle -\frac {a \left (b \int \frac {1}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))+\int \frac {a \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {a \left (\int \frac {a \cos (c+d x)}{a^2-a^2 \cos ^2(c+d x)}d(a \cos (c+d x))+\frac {b \text {arctanh}(\cos (c+d x))}{a}\right )}{d}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -\frac {a \left (\frac {b \text {arctanh}(\cos (c+d x))}{a}-\frac {1}{2} \log \left (a^2-a^2 \cos ^2(c+d x)\right )\right )}{d}\) |
Input:
Int[Cot[c + d*x]*(a + b*Sec[c + d*x]),x]
Output:
-((a*((b*ArcTanh[Cos[c + d*x]])/a - Log[a^2 - a^2*Cos[c + d*x]^2]/2))/d)
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[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))/cot[(c_.) + (d_.)*(x_)], x_Symbo l] :> Int[(b + a*Sin[c + d*x])/Cos[c + d*x], x] /; FreeQ[{a, b, c, d}, x]
Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {a \ln \left (\sin \left (d x +c \right )\right )+b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\) | \(33\) |
default | \(\frac {a \ln \left (\sin \left (d x +c \right )\right )+b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\) | \(33\) |
risch | \(-i a x -\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{d}\) | \(84\) |
Input:
int(cot(d*x+c)*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*ln(sin(d*x+c))+b*ln(csc(d*x+c)-cot(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\left (a - b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a + b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, d} \] Input:
integrate(cot(d*x+c)*(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
1/2*((a - b)*log(1/2*cos(d*x + c) + 1/2) + (a + b)*log(-1/2*cos(d*x + c) + 1/2))/d
\[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)*(a+b*sec(d*x+c)),x)
Output:
Integral((a + b*sec(c + d*x))*cot(c + d*x), x)
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\left (a - b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (a + b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{2 \, d} \] Input:
integrate(cot(d*x+c)*(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
1/2*((a - b)*log(cos(d*x + c) + 1) + (a + b)*log(cos(d*x + c) - 1))/d
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\left (a - b\right )} \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, d} + \frac {{\left (a + b\right )} \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \] Input:
integrate(cot(d*x+c)*(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
1/2*(a - b)*log(abs(cos(d*x + c) + 1))/d + 1/2*(a + b)*log(abs(cos(d*x + c ) - 1))/d
Time = 10.85 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:
int(cot(c + d*x)*(a + b/cos(c + d*x)),x)
Output:
(a*log(tan(c/2 + (d*x)/2)))/d - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d + (b*l og(tan(c/2 + (d*x)/2)))/d
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \cot (c+d x) (a+b \sec (c+d x)) \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d} \] Input:
int(cot(d*x+c)*(a+b*sec(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2)**2 + 1)*a + log(tan((c + d*x)/2))*a + log(tan((c + d*x)/2))*b)/d