Integrand size = 23, antiderivative size = 65 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-a^2 x-\frac {a^2 \cot (e+f x)}{f}+\frac {\left (a^2-b^2\right ) \cot ^3(e+f x)}{3 f}-\frac {(a+b)^2 \cot ^5(e+f x)}{5 f} \] Output:
-a^2*x-a^2*cot(f*x+e)/f+1/3*(a^2-b^2)*cot(f*x+e)^3/f-1/5*(a+b)^2*cot(f*x+e )^5/f
Leaf count is larger than twice the leaf count of optimal. \(256\) vs. \(2(65)=130\).
Time = 2.46 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.94 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\csc (e) \csc ^5(e+f x) \left (-150 a^2 f x \cos (f x)+150 a^2 f x \cos (2 e+f x)+75 a^2 f x \cos (2 e+3 f x)-75 a^2 f x \cos (4 e+3 f x)-15 a^2 f x \cos (4 e+5 f x)+15 a^2 f x \cos (6 e+5 f x)+280 a^2 \sin (f x)+120 a b \sin (f x)+20 b^2 \sin (f x)+180 a^2 \sin (2 e+f x)-60 b^2 \sin (2 e+f x)-140 a^2 \sin (2 e+3 f x)+20 b^2 \sin (2 e+3 f x)-90 a^2 \sin (4 e+3 f x)-60 a b \sin (4 e+3 f x)+46 a^2 \sin (4 e+5 f x)+12 a b \sin (4 e+5 f x)-4 b^2 \sin (4 e+5 f x)\right )}{480 f} \] Input:
Integrate[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(Csc[e]*Csc[e + f*x]^5*(-150*a^2*f*x*Cos[f*x] + 150*a^2*f*x*Cos[2*e + f*x] + 75*a^2*f*x*Cos[2*e + 3*f*x] - 75*a^2*f*x*Cos[4*e + 3*f*x] - 15*a^2*f*x* Cos[4*e + 5*f*x] + 15*a^2*f*x*Cos[6*e + 5*f*x] + 280*a^2*Sin[f*x] + 120*a* b*Sin[f*x] + 20*b^2*Sin[f*x] + 180*a^2*Sin[2*e + f*x] - 60*b^2*Sin[2*e + f *x] - 140*a^2*Sin[2*e + 3*f*x] + 20*b^2*Sin[2*e + 3*f*x] - 90*a^2*Sin[4*e + 3*f*x] - 60*a*b*Sin[4*e + 3*f*x] + 46*a^2*Sin[4*e + 5*f*x] + 12*a*b*Sin[ 4*e + 5*f*x] - 4*b^2*Sin[4*e + 5*f*x]))/(480*f)
Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4629, 2075, 364, 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.
\(\displaystyle \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\tan (e+f x)^6}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^2}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^2}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left ((a+b)^2 \cot ^6(e+f x)+\left (b^2-a^2\right ) \cot ^4(e+f x)+a^2 \cot ^2(e+f x)-\frac {a^2}{\tan ^2(e+f x)+1}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a^2 \arctan (\tan (e+f x))+\frac {1}{3} \left (a^2-b^2\right ) \cot ^3(e+f x)-a^2 \cot (e+f x)-\frac {1}{5} (a+b)^2 \cot ^5(e+f x)}{f}\) |
Input:
Int[Cot[e + f*x]^6*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(-(a^2*ArcTan[Tan[e + f*x]]) - a^2*Cot[e + f*x] + ((a^2 - b^2)*Cot[e + f*x ]^3)/3 - ((a + b)^2*Cot[e + f*x]^5)/5)/f
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Time = 4.77 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cot \left (f x +e \right )^{5}}{5}+\frac {\cot \left (f x +e \right )^{3}}{3}-\cot \left (f x +e \right )-f x -e \right )-\frac {2 a b \cos \left (f x +e \right )^{5}}{5 \sin \left (f x +e \right )^{5}}+b^{2} \left (-\frac {\cos \left (f x +e \right )^{3}}{5 \sin \left (f x +e \right )^{5}}-\frac {2 \cos \left (f x +e \right )^{3}}{15 \sin \left (f x +e \right )^{3}}\right )}{f}\) | \(107\) |
default | \(\frac {a^{2} \left (-\frac {\cot \left (f x +e \right )^{5}}{5}+\frac {\cot \left (f x +e \right )^{3}}{3}-\cot \left (f x +e \right )-f x -e \right )-\frac {2 a b \cos \left (f x +e \right )^{5}}{5 \sin \left (f x +e \right )^{5}}+b^{2} \left (-\frac {\cos \left (f x +e \right )^{3}}{5 \sin \left (f x +e \right )^{5}}-\frac {2 \cos \left (f x +e \right )^{3}}{15 \sin \left (f x +e \right )^{3}}\right )}{f}\) | \(107\) |
risch | \(-a^{2} x -\frac {2 i \left (45 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+30 \,{\mathrm e}^{8 i \left (f x +e \right )} a b -90 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+30 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+140 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+60 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+10 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-70 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+10 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+23 a^{2}+6 a b -2 b^{2}\right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}\) | \(166\) |
Input:
int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(a^2*(-1/5*cot(f*x+e)^5+1/3*cot(f*x+e)^3-cot(f*x+e)-f*x-e)-2/5*a*b/sin (f*x+e)^5*cos(f*x+e)^5+b^2*(-1/5/sin(f*x+e)^5*cos(f*x+e)^3-2/15*cos(f*x+e) ^3/sin(f*x+e)^3))
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (61) = 122\).
Time = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.09 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {{\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 5 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \cos \left (f x + e\right ) + 15 \, {\left (a^{2} f x \cos \left (f x + e\right )^{4} - 2 \, a^{2} f x \cos \left (f x + e\right )^{2} + a^{2} f x\right )} \sin \left (f x + e\right )}{15 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )} \sin \left (f x + e\right )} \] Input:
integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
-1/15*((23*a^2 + 6*a*b - 2*b^2)*cos(f*x + e)^5 - 5*(7*a^2 - b^2)*cos(f*x + e)^3 + 15*a^2*cos(f*x + e) + 15*(a^2*f*x*cos(f*x + e)^4 - 2*a^2*f*x*cos(f *x + e)^2 + a^2*f*x)*sin(f*x + e))/((f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)*sin(f*x + e))
Timed out. \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)**6*(a+b*sec(f*x+e)**2)**2,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {15 \, {\left (f x + e\right )} a^{2} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} - 5 \, {\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{\tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:
integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
-1/15*(15*(f*x + e)*a^2 + (15*a^2*tan(f*x + e)^4 - 5*(a^2 - b^2)*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/tan(f*x + e)^5)/f
Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (61) = 122\).
Time = 0.18 (sec) , antiderivative size = 273, normalized size of antiderivative = 4.20 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 6 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 30 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 480 \, {\left (f x + e\right )} a^{2} + 330 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 60 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 30 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \frac {330 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 60 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 30 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 35 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 30 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{480 \, f} \] Input:
integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/480*(3*a^2*tan(1/2*f*x + 1/2*e)^5 + 6*a*b*tan(1/2*f*x + 1/2*e)^5 + 3*b^2 *tan(1/2*f*x + 1/2*e)^5 - 35*a^2*tan(1/2*f*x + 1/2*e)^3 - 30*a*b*tan(1/2*f *x + 1/2*e)^3 + 5*b^2*tan(1/2*f*x + 1/2*e)^3 - 480*(f*x + e)*a^2 + 330*a^2 *tan(1/2*f*x + 1/2*e) + 60*a*b*tan(1/2*f*x + 1/2*e) - 30*b^2*tan(1/2*f*x + 1/2*e) - (330*a^2*tan(1/2*f*x + 1/2*e)^4 + 60*a*b*tan(1/2*f*x + 1/2*e)^4 - 30*b^2*tan(1/2*f*x + 1/2*e)^4 - 35*a^2*tan(1/2*f*x + 1/2*e)^2 - 30*a*b*t an(1/2*f*x + 1/2*e)^2 + 5*b^2*tan(1/2*f*x + 1/2*e)^2 + 3*a^2 + 6*a*b + 3*b ^2)/tan(1/2*f*x + 1/2*e)^5)/f
Time = 16.44 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-a^2\,x-\frac {\frac {2\,a\,b}{5}+\frac {a^2}{5}+\frac {b^2}{5}-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a^2}{3}-\frac {b^2}{3}\right )+a^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5} \] Input:
int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^2,x)
Output:
- a^2*x - ((2*a*b)/5 + a^2/5 + b^2/5 - tan(e + f*x)^2*(a^2/3 - b^2/3) + a^ 2*tan(e + f*x)^4)/(f*tan(e + f*x)^5)
Time = 0.15 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.65 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {-23 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{2}-6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a b +2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b^{2}+11 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2}+12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a b +\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b^{2}-3 \cos \left (f x +e \right ) a^{2}-6 \cos \left (f x +e \right ) a b -3 \cos \left (f x +e \right ) b^{2}-15 \sin \left (f x +e \right )^{5} a^{2} f x}{15 \sin \left (f x +e \right )^{5} f} \] Input:
int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 23*cos(e + f*x)*sin(e + f*x)**4*a**2 - 6*cos(e + f*x)*sin(e + f*x)**4* a*b + 2*cos(e + f*x)*sin(e + f*x)**4*b**2 + 11*cos(e + f*x)*sin(e + f*x)** 2*a**2 + 12*cos(e + f*x)*sin(e + f*x)**2*a*b + cos(e + f*x)*sin(e + f*x)** 2*b**2 - 3*cos(e + f*x)*a**2 - 6*cos(e + f*x)*a*b - 3*cos(e + f*x)*b**2 - 15*sin(e + f*x)**5*a**2*f*x)/(15*sin(e + f*x)**5*f)