Integrand size = 21, antiderivative size = 53 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {b (2 a+b) \log (\cos (e+f x))}{f}+\frac {(a+b)^2 \log (\sin (e+f x))}{f}+\frac {b^2 \sec ^2(e+f x)}{2 f} \] Output:
-b*(2*a+b)*ln(cos(f*x+e))/f+(a+b)^2*ln(sin(f*x+e))/f+1/2*b^2*sec(f*x+e)^2/ f
Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.58 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {2 \left (b^2+2 \cos ^2(e+f x) \left (-b (2 a+b) \log (\cos (e+f x))+(a+b)^2 \log (\sin (e+f x))\right )\right ) (a \cos (e+f x)+b \sec (e+f x))^2}{f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:
Integrate[Cot[e + f*x]*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(2*(b^2 + 2*Cos[e + f*x]^2*(-(b*(2*a + b)*Log[Cos[e + f*x]]) + (a + b)^2*L og[Sin[e + f*x]]))*(a*Cos[e + f*x] + b*Sec[e + f*x])^2)/(f*(a + 2*b + a*Co s[2*(e + f*x)])^2)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04, 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, 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.
\(\displaystyle \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\tan (e+f x)}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (a \cos ^2(e+f x)+b\right )^2 \sec ^3(e+f x)}{1-\cos ^2(e+f x)}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\left (a \cos ^2(e+f x)+b\right )^2 \sec ^2(e+f x)}{1-\cos ^2(e+f x)}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {(a+b)^2}{\cos ^2(e+f x)-1}+b^2 \sec ^2(e+f x)+b (2 a+b) \sec (e+f x)\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b (2 a+b) \log \left (\cos ^2(e+f x)\right )-(a+b)^2 \log \left (1-\cos ^2(e+f x)\right )+b^2 (-\sec (e+f x))}{2 f}\) |
Input:
Int[Cot[e + f*x]*(a + b*Sec[e + f*x]^2)^2,x]
Output:
-1/2*(b*(2*a + b)*Log[Cos[e + f*x]^2] - (a + b)^2*Log[1 - Cos[e + f*x]^2] - b^2*Sec[e + f*x])/f
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[(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 = 1.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {a^{2} \ln \left (\sin \left (f x +e \right )\right )+2 a b \ln \left (\tan \left (f x +e \right )\right )+b^{2} \left (\frac {1}{2 \cos \left (f x +e \right )^{2}}+\ln \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(50\) |
default | \(\frac {a^{2} \ln \left (\sin \left (f x +e \right )\right )+2 a b \ln \left (\tan \left (f x +e \right )\right )+b^{2} \left (\frac {1}{2 \cos \left (f x +e \right )^{2}}+\ln \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(50\) |
risch | \(-i a^{2} x -\frac {2 i a^{2} e}{f}+\frac {2 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a b}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{f}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{f}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}\) | \(145\) |
Input:
int(cot(f*x+e)*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(a^2*ln(sin(f*x+e))+2*a*b*ln(tan(f*x+e))+b^2*(1/2/cos(f*x+e)^2+ln(tan( f*x+e))))
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.49 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {{\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\cos \left (f x + e\right )^{2}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\frac {1}{4} \, \cos \left (f x + e\right )^{2} + \frac {1}{4}\right ) - b^{2}}{2 \, f \cos \left (f x + e\right )^{2}} \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
-1/2*((2*a*b + b^2)*cos(f*x + e)^2*log(cos(f*x + e)^2) - (a^2 + 2*a*b + b^ 2)*cos(f*x + e)^2*log(-1/4*cos(f*x + e)^2 + 1/4) - b^2)/(f*cos(f*x + e)^2)
\[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cot {\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)**2)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*cot(e + f*x), x)
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {{\left (2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {b^{2}}{\sin \left (f x + e\right )^{2} - 1}}{2 \, f} \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
-1/2*((2*a*b + b^2)*log(sin(f*x + e)^2 - 1) - (a^2 + 2*a*b + b^2)*log(sin( f*x + e)^2) + b^2/(sin(f*x + e)^2 - 1))/f
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.25 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | \cos \left (f x + e\right )^{2} - 1 \right |}\right )}{2 \, f} - \frac {{\left (2 \, a b + b^{2}\right )} \log \left ({\left | \cos \left (f x + e\right ) \right |}\right )}{f} + \frac {b^{2}}{2 \, f \cos \left (f x + e\right )^{2}} \] Input:
integrate(cot(f*x+e)*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/2*(a^2 + 2*a*b + b^2)*log(abs(cos(f*x + e)^2 - 1))/f - (2*a*b + b^2)*log (abs(cos(f*x + e)))/f + 1/2*b^2/(f*cos(f*x + e)^2)
Time = 15.53 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+2\,a\,b+b^2\right )}{f}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,f} \] Input:
int(cot(e + f*x)*(a + b/cos(e + f*x)^2)^2,x)
Output:
(log(tan(e + f*x))*(2*a*b + a^2 + b^2))/f - (a^2*log(tan(e + f*x)^2 + 1))/ (2*f) + (b^2*tan(e + f*x)^2)/(2*f)
Time = 0.15 (sec) , antiderivative size = 353, normalized size of antiderivative = 6.66 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{2} a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2} a b -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2} b^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a b +2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) b^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2} a b -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2} b^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a b +2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a b +2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a b -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{2}-\sin \left (f x +e \right )^{2} b^{2}}{2 f \left (\sin \left (f x +e \right )^{2}-1\right )} \] Input:
int(cot(f*x+e)*(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 2*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**2 + 2*log(tan((e + f *x)/2)**2 + 1)*a**2 - 4*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a*b - 2* log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*b**2 + 4*log(tan((e + f*x)/2) - 1)*a*b + 2*log(tan((e + f*x)/2) - 1)*b**2 - 4*log(tan((e + f*x)/2) + 1)*si n(e + f*x)**2*a*b - 2*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*b**2 + 4*l og(tan((e + f*x)/2) + 1)*a*b + 2*log(tan((e + f*x)/2) + 1)*b**2 + 2*log(ta n((e + f*x)/2))*sin(e + f*x)**2*a**2 + 4*log(tan((e + f*x)/2))*sin(e + f*x )**2*a*b + 2*log(tan((e + f*x)/2))*sin(e + f*x)**2*b**2 - 2*log(tan((e + f *x)/2))*a**2 - 4*log(tan((e + f*x)/2))*a*b - 2*log(tan((e + f*x)/2))*b**2 - sin(e + f*x)**2*b**2)/(2*f*(sin(e + f*x)**2 - 1))