Integrand size = 21, antiderivative size = 51 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(2 a+b) \csc ^2(e+f x)}{2 f}-\frac {(a+b) \csc ^4(e+f x)}{4 f}+\frac {a \log (\sin (e+f x))}{f} \] Output:
1/2*(2*a+b)*csc(f*x+e)^2/f-1/4*(a+b)*csc(f*x+e)^4/f+a*ln(sin(f*x+e))/f
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {b \cot ^4(e+f x)}{4 f}+\frac {a \csc ^2(e+f x)}{f}-\frac {a \csc ^4(e+f x)}{4 f}+\frac {a \log (\sin (e+f x))}{f} \] Input:
Integrate[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2),x]
Output:
-1/4*(b*Cot[e + f*x]^4)/f + (a*Csc[e + f*x]^2)/f - (a*Csc[e + f*x]^4)/(4*f ) + (a*Log[Sin[e + f*x]])/f
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.29, 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.
\(\displaystyle \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (e+f x)^2}{\tan (e+f x)^5}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\cos ^3(e+f x) \left (a \cos ^2(e+f x)+b\right )}{\left (1-\cos ^2(e+f x)\right )^3}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\cos ^2(e+f x) \left (a \cos ^2(e+f x)+b\right )}{\left (1-\cos ^2(e+f x)\right )^3}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {\int \left (-\frac {a}{\cos ^2(e+f x)-1}+\frac {-2 a-b}{\left (\cos ^2(e+f x)-1\right )^2}+\frac {-a-b}{\left (\cos ^2(e+f x)-1\right )^3}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {a+b}{2 \left (1-\cos ^2(e+f x)\right )^2}-\frac {2 a+b}{1-\cos ^2(e+f x)}-a \log \left (1-\cos ^2(e+f x)\right )}{2 f}\) |
Input:
Int[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2),x]
Output:
-1/2*((a + b)/(2*(1 - Cos[e + f*x]^2)^2) - (2*a + b)/(1 - Cos[e + f*x]^2) - a*Log[1 - Cos[e + f*x]^2])/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 = 0.96 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {b \cos \left (f x +e \right )^{4}}{4 \sin \left (f x +e \right )^{4}}}{f}\) | \(55\) |
default | \(\frac {a \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {b \cos \left (f x +e \right )^{4}}{4 \sin \left (f x +e \right )^{4}}}{f}\) | \(55\) |
risch | \(-i a x -\frac {2 i a e}{f}-\frac {2 \left (2 a \,{\mathrm e}^{6 i \left (f x +e \right )}+b \,{\mathrm e}^{6 i \left (f x +e \right )}-2 a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f}\) | \(109\) |
Input:
int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(a*(-1/4*cot(f*x+e)^4+1/2*cot(f*x+e)^2+ln(sin(f*x+e)))-1/4*b/sin(f*x+e )^4*cos(f*x+e)^4)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {2 \, {\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (a \cos \left (f x + e\right )^{4} - 2 \, a \cos \left (f x + e\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right ) - 3 \, a - b}{4 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}} \] Input:
integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
-1/4*(2*(2*a + b)*cos(f*x + e)^2 - 4*(a*cos(f*x + e)^4 - 2*a*cos(f*x + e)^ 2 + a)*log(1/2*sin(f*x + e)) - 3*a - b)/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)
\[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \cot ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**5*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*cot(e + f*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {2 \, a \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {2 \, {\left (2 \, a + b\right )} \sin \left (f x + e\right )^{2} - a - b}{\sin \left (f x + e\right )^{4}}}{4 \, f} \] Input:
integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/4*(2*a*log(sin(f*x + e)^2) + (2*(2*a + b)*sin(f*x + e)^2 - a - b)/sin(f* x + e)^4)/f
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \log \left ({\left | \cos \left (f x + e\right )^{2} - 1 \right |}\right )}{2 \, f} - \frac {2 \, {\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - b}{4 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )}^{2} f} \] Input:
integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/2*a*log(abs(cos(f*x + e)^2 - 1))/f - 1/4*(2*(2*a + b)*cos(f*x + e)^2 - 3 *a - b)/((cos(f*x + e)^2 - 1)^2*f)
Time = 15.64 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{f}-\frac {a\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}-\frac {-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {a}{4}+\frac {b}{4}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^4} \] Input:
int(cot(e + f*x)^5*(a + b/cos(e + f*x)^2),x)
Output:
(a*log(tan(e + f*x)))/f - (a*log(tan(e + f*x)^2 + 1))/(2*f) - (a/4 + b/4 - (a*tan(e + f*x)^2)/2)/(f*tan(e + f*x)^4)
Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.16 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {-32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{4} a +32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} a -13 \sin \left (f x +e \right )^{4} a -5 \sin \left (f x +e \right )^{4} b +32 \sin \left (f x +e \right )^{2} a +16 \sin \left (f x +e \right )^{2} b -8 a -8 b}{32 \sin \left (f x +e \right )^{4} f} \] Input:
int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2),x)
Output:
( - 32*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a + 32*log(tan((e + f* x)/2))*sin(e + f*x)**4*a - 13*sin(e + f*x)**4*a - 5*sin(e + f*x)**4*b + 32 *sin(e + f*x)**2*a + 16*sin(e + f*x)**2*b - 8*a - 8*b)/(32*sin(e + f*x)**4 *f)