Integrand size = 23, antiderivative size = 51 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a (a+b) \csc ^2(e+f x)}{f}-\frac {(a+b)^2 \csc ^4(e+f x)}{4 f}+\frac {a^2 \log (\sin (e+f x))}{f} \] Output:
a*(a+b)*csc(f*x+e)^2/f-1/4*(a+b)^2*csc(f*x+e)^4/f+a^2*ln(sin(f*x+e))/f
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.51 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {\left (b+a \cos ^2(e+f x)\right )^2 \left (-4 a (a+b) \csc ^2(e+f x)+(a+b)^2 \csc ^4(e+f x)-4 a^2 \log (\sin (e+f x))\right )}{f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:
Integrate[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]
Output:
-(((b + a*Cos[e + f*x]^2)^2*(-4*a*(a + b)*Csc[e + f*x]^2 + (a + b)^2*Csc[e + f*x]^4 - 4*a^2*Log[Sin[e + f*x]]))/(f*(a + 2*b + a*Cos[2*(e + f*x)])^2) )
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.35, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 353, 49, 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.
\(\displaystyle \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\tan (e+f x)^5}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\cos (e+f x) \left (a \cos ^2(e+f x)+b\right )^2}{\left (1-\cos ^2(e+f x)\right )^3}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {\int \frac {\left (a \cos ^2(e+f x)+b\right )^2}{\left (1-\cos ^2(e+f x)\right )^3}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\int \left (-\frac {a^2}{\cos ^2(e+f x)-1}-\frac {2 (a+b) a}{\left (\cos ^2(e+f x)-1\right )^2}-\frac {(a+b)^2}{\left (\cos ^2(e+f x)-1\right )^3}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \left (-\log \left (1-\cos ^2(e+f x)\right )\right )-\frac {2 a (a+b)}{1-\cos ^2(e+f x)}+\frac {(a+b)^2}{2 \left (1-\cos ^2(e+f x)\right )^2}}{2 f}\) |
Input:
Int[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]
Output:
-1/2*((a + b)^2/(2*(1 - Cos[e + f*x]^2)^2) - (2*a*(a + b))/(1 - Cos[e + f* x]^2) - a^2*Log[1 - Cos[e + f*x]^2])/f
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
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 = 3.46 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {a b \cos \left (f x +e \right )^{4}}{2 \sin \left (f x +e \right )^{4}}-\frac {b^{2}}{4 \sin \left (f x +e \right )^{4}}}{f}\) | \(71\) |
default | \(\frac {a^{2} \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {a b \cos \left (f x +e \right )^{4}}{2 \sin \left (f x +e \right )^{4}}-\frac {b^{2}}{4 \sin \left (f x +e \right )^{4}}}{f}\) | \(71\) |
risch | \(-i a^{2} x -\frac {2 i a^{2} e}{f}-\frac {4 \left (a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+a b \,{\mathrm e}^{6 i \left (f x +e \right )}-a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+a b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f}\) | \(134\) |
Input:
int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(a^2*(-1/4*cot(f*x+e)^4+1/2*cot(f*x+e)^2+ln(sin(f*x+e)))-1/2*a*b/sin(f *x+e)^4*cos(f*x+e)^4-1/4*b^2/sin(f*x+e)^4)
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.90 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 2 \, a b + b^{2} - 4 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}} \] Input:
integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
-1/4*(4*(a^2 + a*b)*cos(f*x + e)^2 - 3*a^2 - 2*a*b + b^2 - 4*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^2)*log(1/2*sin(f*x + e)))/(f*cos(f*x + e )^4 - 2*f*cos(f*x + e)^2 + f)
\[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cot ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**5*(a+b*sec(f*x+e)**2)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*cot(e + f*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {2 \, a^{2} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {4 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\sin \left (f x + e\right )^{4}}}{4 \, f} \] Input:
integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
1/4*(2*a^2*log(sin(f*x + e)^2) + (4*(a^2 + a*b)*sin(f*x + e)^2 - a^2 - 2*a *b - b^2)/sin(f*x + e)^4)/f
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^{2} \log \left ({\left | \cos \left (f x + e\right )^{2} - 1 \right |}\right )}{2 \, f} - \frac {4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 2 \, a b + b^{2}}{4 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )}^{2} f} \] Input:
integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/2*a^2*log(abs(cos(f*x + e)^2 - 1))/f - 1/4*(4*(a^2 + a*b)*cos(f*x + e)^2 - 3*a^2 - 2*a*b + b^2)/((cos(f*x + e)^2 - 1)^2*f)
Time = 15.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{f}-\frac {\frac {a\,b}{2}+\frac {a^2}{4}+\frac {b^2}{4}-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )}{f\,{\mathrm {tan}\left (e+f\,x\right )}^4}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f} \] Input:
int(cot(e + f*x)^5*(a + b/cos(e + f*x)^2)^2,x)
Output:
(a^2*log(tan(e + f*x)))/f - ((a*b)/2 + a^2/4 + b^2/4 - tan(e + f*x)^2*(a^2 /2 - b^2/2))/(f*tan(e + f*x)^4) - (a^2*log(tan(e + f*x)^2 + 1))/(2*f)
Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.76 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {-32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{4} a^{2}+32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} a^{2}-13 \sin \left (f x +e \right )^{4} a^{2}-10 \sin \left (f x +e \right )^{4} a b +3 \sin \left (f x +e \right )^{4} b^{2}+32 \sin \left (f x +e \right )^{2} a^{2}+32 \sin \left (f x +e \right )^{2} a b -8 a^{2}-16 a b -8 b^{2}}{32 \sin \left (f x +e \right )^{4} f} \] Input:
int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 32*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**2 + 32*log(tan((e + f*x)/2))*sin(e + f*x)**4*a**2 - 13*sin(e + f*x)**4*a**2 - 10*sin(e + f*x) **4*a*b + 3*sin(e + f*x)**4*b**2 + 32*sin(e + f*x)**2*a**2 + 32*sin(e + f* x)**2*a*b - 8*a**2 - 16*a*b - 8*b**2)/(32*sin(e + f*x)**4*f)