Integrand size = 19, antiderivative size = 54 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {(a+b) \log (1-\cos (e+f x))}{2 f}+\frac {(a-b) \log (1+\cos (e+f x))}{2 f}+\frac {b \sec (e+f x)}{f} \] Output:
1/2*(a+b)*ln(1-cos(f*x+e))/f+1/2*(a-b)*ln(1+cos(f*x+e))/f+b*sec(f*x+e)/f
Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=-\frac {b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {a \log (\sin (e+f x))}{f}+\frac {b \sec (e+f x)}{f} \] Input:
Integrate[Cot[e + f*x]*(a + b*Sec[e + f*x]^3),x]
Output:
-((b*Log[Cos[(e + f*x)/2]])/f) + (b*Log[Sin[(e + f*x)/2]])/f + (a*Log[Sin[ e + f*x]])/f + (b*Sec[e + f*x])/f
Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 4626, 2333, 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.
\(\displaystyle \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (e+f x)^3}{\tan (e+f x)}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (a \cos ^3(e+f x)+b\right ) \sec ^2(e+f x)}{1-\cos ^2(e+f x)}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle -\frac {\int \left (b \sec ^2(e+f x)+\frac {-a-b}{2 (\cos (e+f x)-1)}+\frac {b-a}{2 (\cos (e+f x)+1)}\right )d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{2} (a+b) \log (1-\cos (e+f x))-\frac {1}{2} (a-b) \log (\cos (e+f x)+1)-b \sec (e+f x)}{f}\) |
Input:
Int[Cot[e + f*x]*(a + b*Sec[e + f*x]^3),x]
Output:
-((-1/2*((a + b)*Log[1 - Cos[e + f*x]]) - ((a - b)*Log[1 + Cos[e + f*x]])/ 2 - b*Sec[e + f*x])/f)
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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 = 0.41 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {a \ln \left (\sin \left (f x +e \right )\right )+b \left (\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\) | \(42\) |
default | \(\frac {a \ln \left (\sin \left (f x +e \right )\right )+b \left (\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\) | \(42\) |
risch | \(-i a x -\frac {2 i a e}{f}+\frac {2 b \,{\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{f}\) | \(112\) |
Input:
int(cot(f*x+e)*(a+b*sec(f*x+e)^3),x,method=_RETURNVERBOSE)
Output:
1/f*(a*ln(sin(f*x+e))+b*(1/cos(f*x+e)+ln(csc(f*x+e)-cot(f*x+e))))
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {{\left (a - b\right )} \cos \left (f x + e\right ) \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left (a + b\right )} \cos \left (f x + e\right ) \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 2 \, b}{2 \, f \cos \left (f x + e\right )} \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)^3),x, algorithm="fricas")
Output:
1/2*((a - b)*cos(f*x + e)*log(1/2*cos(f*x + e) + 1/2) + (a + b)*cos(f*x + e)*log(-1/2*cos(f*x + e) + 1/2) + 2*b)/(f*cos(f*x + e))
\[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\int \left (a + b \sec ^{3}{\left (e + f x \right )}\right ) \cot {\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)**3),x)
Output:
Integral((a + b*sec(e + f*x)**3)*cot(e + f*x), x)
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {{\left (a - b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) + {\left (a + b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) + \frac {2 \, b}{\cos \left (f x + e\right )}}{2 \, f} \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)^3),x, algorithm="maxima")
Output:
1/2*((a - b)*log(cos(f*x + e) + 1) + (a + b)*log(cos(f*x + e) - 1) + 2*b/c os(f*x + e))/f
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {{\left (a - b\right )} \log \left ({\left | \cos \left (f x + e\right ) + 1 \right |}\right )}{2 \, f} + \frac {{\left (a + b\right )} \log \left ({\left | \cos \left (f x + e\right ) - 1 \right |}\right )}{2 \, f} + \frac {b}{f \cos \left (f x + e\right )} \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)^3),x, algorithm="giac")
Output:
1/2*(a - b)*log(abs(cos(f*x + e) + 1))/f + 1/2*(a + b)*log(abs(cos(f*x + e ) - 1))/f + b/(f*cos(f*x + e))
Time = 15.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.33 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}{f}-\frac {2\,b}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f} \] Input:
int(cot(e + f*x)*(a + b/cos(e + f*x)^3),x)
Output:
(a*log(tan(e/2 + (f*x)/2)))/f - (a*log(tan(e/2 + (f*x)/2)^2 + 1))/f - (2*b )/(f*(tan(e/2 + (f*x)/2)^2 - 1)) + (b*log(tan(e/2 + (f*x)/2)))/f
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.52 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {-\cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a +\cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +\cos \left (f x +e \right ) \mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b -\cos \left (f x +e \right ) b +b}{\cos \left (f x +e \right ) f} \] Input:
int(cot(f*x+e)*(a+b*sec(f*x+e)^3),x)
Output:
( - cos(e + f*x)*log(tan((e + f*x)/2)**2 + 1)*a + cos(e + f*x)*log(tan((e + f*x)/2))*a + cos(e + f*x)*log(tan((e + f*x)/2))*b - cos(e + f*x)*b + b)/ (cos(e + f*x)*f)