Integrand size = 21, antiderivative size = 46 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {b \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b) f}+\frac {\log (\sin (e+f x))}{(a+b) f} \] Output:
1/2*b*ln(b+a*cos(f*x+e)^2)/a/(a+b)/f+ln(sin(f*x+e))/(a+b)/f
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {2 a \log (\sin (e+f x))+b \log \left (a+b-a \sin ^2(e+f x)\right )}{2 a^2 f+2 a b f} \] Input:
Integrate[Cot[e + f*x]/(a + b*Sec[e + f*x]^2),x]
Output:
(2*a*Log[Sin[e + f*x]] + b*Log[a + b - a*Sin[e + f*x]^2])/(2*a^2*f + 2*a*b *f)
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4626, 354, 86, 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.
\(\displaystyle \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x) \left (a+b \sec (e+f x)^2\right )}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\cos ^3(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\cos ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {\int \left (\frac {1}{(-a-b) \left (\cos ^2(e+f x)-1\right )}-\frac {b}{(a+b) \left (a \cos ^2(e+f x)+b\right )}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\log \left (1-\cos ^2(e+f x)\right )}{a+b}-\frac {b \log \left (a \cos ^2(e+f x)+b\right )}{a (a+b)}}{2 f}\) |
Input:
Int[Cot[e + f*x]/(a + b*Sec[e + f*x]^2),x]
Output:
-1/2*(-(Log[1 - Cos[e + f*x]^2]/(a + b)) - (b*Log[b + a*Cos[e + f*x]^2])/( a*(a + b)))/f
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 0.64 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 a +2 b}+\frac {b \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right ) a}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 a +2 b}}{f}\) | \(68\) |
default | \(\frac {\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 a +2 b}+\frac {b \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right ) a}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 a +2 b}}{f}\) | \(68\) |
risch | \(\frac {i x}{a}-\frac {2 i x}{a +b}-\frac {2 i e}{f \left (a +b \right )}-\frac {2 i b x}{a \left (a +b \right )}-\frac {2 i b e}{a f \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{f \left (a +b \right )}+\frac {b \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f \left (a +b \right )}\) | \(125\) |
Input:
int(cot(f*x+e)/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/(2*a+2*b)*ln(-1+cos(f*x+e))+1/2/(a+b)*b/a*ln(b+a*cos(f*x+e)^2)+1/(2 *a+2*b)*ln(1+cos(f*x+e)))
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {b \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 2 \, a \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{2 \, {\left (a^{2} + a b\right )} f} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/2*(b*log(a*cos(f*x + e)^2 + b) + 2*a*log(1/2*sin(f*x + e)))/((a^2 + a*b) *f)
\[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot {\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)**2),x)
Output:
Integral(cot(e + f*x)/(a + b*sec(e + f*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {b \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{2} + a b} + \frac {\log \left (\sin \left (f x + e\right )^{2}\right )}{a + b}}{2 \, f} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/2*(b*log(a*sin(f*x + e)^2 - a - b)/(a^2 + a*b) + log(sin(f*x + e)^2)/(a + b))/f
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {b \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, {\left (a^{2} f + a b f\right )}} + \frac {\log \left ({\left | \cos \left (f x + e\right )^{2} - 1 \right |}\right )}{2 \, {\left (a f + b f\right )}} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/2*b*log(abs(a*cos(f*x + e)^2 + b))/(a^2*f + a*b*f) + 1/2*log(abs(cos(f*x + e)^2 - 1))/(a*f + b*f)
Time = 16.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.41 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{f\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}+\frac {b\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,f\,\left (a^2+b\,a\right )} \] Input:
int(cot(e + f*x)/(a + b/cos(e + f*x)^2),x)
Output:
log(tan(e + f*x))/(f*(a + b)) - log(tan(e + f*x)^2 + 1)/(2*a*f) + (b*log(a + b + b*tan(e + f*x)^2))/(2*f*(a*b + a^2))
Time = 0.15 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.93 \[ \int \frac {\cot (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) b +\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b +\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b +2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a}{2 a f \left (a +b \right )} \] Input:
int(cot(f*x+e)/(a+b*sec(f*x+e)^2),x)
Output:
( - 2*log(tan((e + f*x)/2)**2 + 1)*a - 2*log(tan((e + f*x)/2)**2 + 1)*b + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x )/2))*b + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*ta n((e + f*x)/2))*b + 2*log(tan((e + f*x)/2))*a)/(2*a*f*(a + b))