Integrand size = 23, antiderivative size = 120 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {x}{a}+\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a (a+b)^{7/2} f}-\frac {\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{(a+b)^3 f}+\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f} \] Output:
-x/a+b^(7/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a/(a+b)^(7/2)/f-(a^2+3 *a*b+3*b^2)*cot(f*x+e)/(a+b)^3/f+1/3*(a+2*b)*cot(f*x+e)^3/(a+b)^2/f-1/5*co t(f*x+e)^5/(a+b)/f
Result contains complex when optimal does not.
Time = 2.57 (sec) , antiderivative size = 671, normalized size of antiderivative = 5.59 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (-\frac {480 b^4 \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\csc (e) \csc ^5(e+f x) \left (-150 (a+b)^3 f x \cos (f x)+150 (a+b)^3 f x \cos (2 e+f x)+75 a^3 f x \cos (2 e+3 f x)+225 a^2 b f x \cos (2 e+3 f x)+225 a b^2 f x \cos (2 e+3 f x)+75 b^3 f x \cos (2 e+3 f x)-75 a^3 f x \cos (4 e+3 f x)-225 a^2 b f x \cos (4 e+3 f x)-225 a b^2 f x \cos (4 e+3 f x)-75 b^3 f x \cos (4 e+3 f x)-15 a^3 f x \cos (4 e+5 f x)-45 a^2 b f x \cos (4 e+5 f x)-45 a b^2 f x \cos (4 e+5 f x)-15 b^3 f x \cos (4 e+5 f x)+15 a^3 f x \cos (6 e+5 f x)+45 a^2 b f x \cos (6 e+5 f x)+45 a b^2 f x \cos (6 e+5 f x)+15 b^3 f x \cos (6 e+5 f x)+280 a^3 \sin (f x)+780 a^2 b \sin (f x)+680 a b^2 \sin (f x)+180 a^3 \sin (2 e+f x)+540 a^2 b \sin (2 e+f x)+480 a b^2 \sin (2 e+f x)-140 a^3 \sin (2 e+3 f x)-420 a^2 b \sin (2 e+3 f x)-400 a b^2 \sin (2 e+3 f x)-90 a^3 \sin (4 e+3 f x)-240 a^2 b \sin (4 e+3 f x)-180 a b^2 \sin (4 e+3 f x)+46 a^3 \sin (4 e+5 f x)+132 a^2 b \sin (4 e+5 f x)+116 a b^2 \sin (4 e+5 f x)\right )\right )}{960 a (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )} \] Input:
Integrate[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*((-480*b^4*ArcTan[(Sec[f*x] *(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sq rt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + Csc[e]*Csc[e + f*x]^5*(-150*(a + b) ^3*f*x*Cos[f*x] + 150*(a + b)^3*f*x*Cos[2*e + f*x] + 75*a^3*f*x*Cos[2*e + 3*f*x] + 225*a^2*b*f*x*Cos[2*e + 3*f*x] + 225*a*b^2*f*x*Cos[2*e + 3*f*x] + 75*b^3*f*x*Cos[2*e + 3*f*x] - 75*a^3*f*x*Cos[4*e + 3*f*x] - 225*a^2*b*f*x *Cos[4*e + 3*f*x] - 225*a*b^2*f*x*Cos[4*e + 3*f*x] - 75*b^3*f*x*Cos[4*e + 3*f*x] - 15*a^3*f*x*Cos[4*e + 5*f*x] - 45*a^2*b*f*x*Cos[4*e + 5*f*x] - 45* a*b^2*f*x*Cos[4*e + 5*f*x] - 15*b^3*f*x*Cos[4*e + 5*f*x] + 15*a^3*f*x*Cos[ 6*e + 5*f*x] + 45*a^2*b*f*x*Cos[6*e + 5*f*x] + 45*a*b^2*f*x*Cos[6*e + 5*f* x] + 15*b^3*f*x*Cos[6*e + 5*f*x] + 280*a^3*Sin[f*x] + 780*a^2*b*Sin[f*x] + 680*a*b^2*Sin[f*x] + 180*a^3*Sin[2*e + f*x] + 540*a^2*b*Sin[2*e + f*x] + 480*a*b^2*Sin[2*e + f*x] - 140*a^3*Sin[2*e + 3*f*x] - 420*a^2*b*Sin[2*e + 3*f*x] - 400*a*b^2*Sin[2*e + 3*f*x] - 90*a^3*Sin[4*e + 3*f*x] - 240*a^2*b* Sin[4*e + 3*f*x] - 180*a*b^2*Sin[4*e + 3*f*x] + 46*a^3*Sin[4*e + 5*f*x] + 132*a^2*b*Sin[4*e + 5*f*x] + 116*a*b^2*Sin[4*e + 5*f*x])))/(960*a*(a + b)^ 3*f*(a + b*Sec[e + f*x]^2))
Time = 0.49 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.22, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4629, 2075, 382, 27, 445, 27, 445, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^6 \left (a+b \sec (e+f x)^2\right )}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 382 |
| \(\displaystyle \frac {\frac {\int -\frac {5 \cot ^4(e+f x) \left (b \tan ^2(e+f x)+a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\cot ^4(e+f x) \left (b \tan ^2(e+f x)+a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {3 \cot ^2(e+f x) \left (a^2+3 b a+3 b^2+b (a+2 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{3 (a+b)}-\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)}}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (a^2+3 b a+3 b^2+b (a+2 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)}}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\int \frac {b \left (a^2+3 b a+3 b^2\right ) \tan ^2(e+f x)+(a+2 b) \left (a^2+2 b a+2 b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{a+b}}{a+b}-\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)}}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\frac {(a+b)^3 \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {b^4 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a+b}-\frac {\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{a+b}}{a+b}-\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)}}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\frac {(a+b)^3 \arctan (\tan (e+f x))}{a}-\frac {b^4 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a+b}-\frac {\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{a+b}}{a+b}-\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)}}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{a+b}-\frac {\frac {(a+b)^3 \arctan (\tan (e+f x))}{a}-\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a+b}}{a+b}-\frac {(a+2 b) \cot ^3(e+f x)}{3 (a+b)}}{a+b}-\frac {\cot ^5(e+f x)}{5 (a+b)}}{f}\) |
Input:
Int[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
Output:
(-1/5*Cot[e + f*x]^5/(a + b) - (-1/3*((a + 2*b)*Cot[e + f*x]^3)/(a + b) - (-((((a + b)^3*ArcTan[Tan[e + f*x]])/a - (b^(7/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(a + b)) - ((a^2 + 3*a*b + 3*b^2)*Cot [e + f*x])/(a + b))/(a + b))/(a + b))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ (a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b* x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m + 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
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 = 2.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{5 \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {-a -2 b}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2}+3 a b +3 b^{2}}{\left (a +b \right )^{3} \tan \left (f x +e \right )}+\frac {b^{4} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3} a \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\) | \(118\) |
| default | \(\frac {-\frac {1}{5 \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {-a -2 b}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2}+3 a b +3 b^{2}}{\left (a +b \right )^{3} \tan \left (f x +e \right )}+\frac {b^{4} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3} a \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\) | \(118\) |
| risch | \(-\frac {x}{a}-\frac {2 i \left (45 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+120 \,{\mathrm e}^{8 i \left (f x +e \right )} a b +90 \,{\mathrm e}^{8 i \left (f x +e \right )} b^{2}-90 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-270 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-240 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+140 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+390 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+340 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-70 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-210 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-200 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+23 a^{2}+66 a b +58 b^{2}\right )}{15 f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{4} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{4} f a}\) | \(324\) |
Input:
int(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/5/(a+b)/tan(f*x+e)^5-1/3*(-a-2*b)/(a+b)^2/tan(f*x+e)^3-(a^2+3*a*b+ 3*b^2)/(a+b)^3/tan(f*x+e)+1/(a+b)^3*b^4/a/((a+b)*b)^(1/2)*arctan(b*tan(f*x +e)/((a+b)*b)^(1/2))-1/a*arctan(tan(f*x+e)))
Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (108) = 216\).
Time = 0.13 (sec) , antiderivative size = 833, normalized size of antiderivative = 6.94 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:
integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
[-1/60*(4*(23*a^3 + 66*a^2*b + 58*a*b^2)*cos(f*x + e)^5 - 20*(7*a^3 + 21*a ^2*b + 20*a*b^2)*cos(f*x + e)^3 - 15*(b^3*cos(f*x + e)^4 - 2*b^3*cos(f*x + e)^2 + b^3)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos (f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 60*(a^3 + 3*a^2* b + 3*a*b^2)*cos(f*x + e) + 60*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x*cos(f* x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x*cos(f*x + e)^2 + (a^3 + 3 *a^2*b + 3*a*b^2 + b^3)*f*x)*sin(f*x + e))/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f)*sin(f*x + e)), -1/30*(2*(2 3*a^3 + 66*a^2*b + 58*a*b^2)*cos(f*x + e)^5 - 10*(7*a^3 + 21*a^2*b + 20*a* b^2)*cos(f*x + e)^3 + 15*(b^3*cos(f*x + e)^4 - 2*b^3*cos(f*x + e)^2 + b^3) *sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b)) /(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*(a^3 + 3*a^2*b + 3*a*b^2 )*cos(f*x + e) + 30*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3* a*b^2 + b^3)*f*x)*sin(f*x + e))/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*co s(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a ^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f)*sin(f*x + e))]
\[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(cot(f*x+e)**6/(a+b*sec(f*x+e)**2),x)
Output:
Integral(cot(e + f*x)**6/(a + b*sec(e + f*x)**2), x)
Time = 0.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.34 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, b^{4} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {15 \, {\left (f x + e\right )}}{a} - \frac {15 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 5 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:
integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/15*(15*b^4*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^4 + 3*a^3*b + 3*a^ 2*b^2 + a*b^3)*sqrt((a + b)*b)) - 15*(f*x + e)/a - (15*(a^2 + 3*a*b + 3*b^ 2)*tan(f*x + e)^4 - 5*(a^2 + 3*a*b + 2*b^2)*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x + e)^5))/f
Time = 0.18 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.77 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt {a b + b^{2}}} - \frac {15 \, {\left (f x + e\right )}}{a} - \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 45 \, a b \tan \left (f x + e\right )^{4} + 45 \, b^{2} \tan \left (f x + e\right )^{4} - 5 \, a^{2} \tan \left (f x + e\right )^{2} - 15 \, a b \tan \left (f x + e\right )^{2} - 10 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:
integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/15*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt (a*b + b^2)))*b^4/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sqrt(a*b + b^2)) - 15*(f*x + e)/a - (15*a^2*tan(f*x + e)^4 + 45*a*b*tan(f*x + e)^4 + 45*b^2*t an(f*x + e)^4 - 5*a^2*tan(f*x + e)^2 - 15*a*b*tan(f*x + e)^2 - 10*b^2*tan( f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f *x + e)^5))/f
Time = 20.83 (sec) , antiderivative size = 4324, normalized size of antiderivative = 36.03 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \] Input:
int(cot(e + f*x)^6/(a + b/cos(e + f*x)^2),x)
Output:
(atan(((((tan(e + f*x)*(48*a*b^17 + 4*b^18 + 282*a^2*b^16 + 1078*a^3*b^15 + 2982*a^4*b^14 + 6258*a^5*b^13 + 10178*a^6*b^12 + 12942*a^7*b^11 + 12888* a^8*b^10 + 10012*a^9*b^9 + 6006*a^10*b^8 + 2730*a^11*b^7 + 910*a^12*b^6 + 210*a^13*b^5 + 30*a^14*b^4 + 2*a^15*b^3))/2 - ((-b^7*(a + b)^7)^(1/2)*(8*a ^2*b^17 + 108*a^3*b^16 + 680*a^4*b^15 + 2650*a^5*b^14 + 7152*a^6*b^13 + 14 168*a^7*b^12 + 21296*a^8*b^11 + 24750*a^9*b^10 + 22440*a^10*b^9 + 15884*a^ 11*b^8 + 8712*a^12*b^7 + 3638*a^13*b^6 + 1120*a^14*b^5 + 240*a^15*b^4 + 32 *a^16*b^3 + 2*a^17*b^2 - (tan(e + f*x)*(-b^7*(a + b)^7)^(1/2)*(16*a^2*b^18 + 248*a^3*b^17 + 1800*a^4*b^16 + 8120*a^5*b^15 + 25480*a^6*b^14 + 58968*a ^7*b^13 + 104104*a^8*b^12 + 143000*a^9*b^11 + 154440*a^10*b^10 + 131560*a^ 11*b^9 + 88088*a^12*b^8 + 45864*a^13*b^7 + 18200*a^14*b^6 + 5320*a^15*b^5 + 1080*a^16*b^4 + 136*a^17*b^3 + 8*a^18*b^2))/(4*a*(a + b)^7)))/(2*a*(a + b)^7))*(-b^7*(a + b)^7)^(1/2)*1i)/(a*(a + b)^7) + (((tan(e + f*x)*(48*a*b^ 17 + 4*b^18 + 282*a^2*b^16 + 1078*a^3*b^15 + 2982*a^4*b^14 + 6258*a^5*b^13 + 10178*a^6*b^12 + 12942*a^7*b^11 + 12888*a^8*b^10 + 10012*a^9*b^9 + 6006 *a^10*b^8 + 2730*a^11*b^7 + 910*a^12*b^6 + 210*a^13*b^5 + 30*a^14*b^4 + 2* a^15*b^3))/2 + ((-b^7*(a + b)^7)^(1/2)*(8*a^2*b^17 + 108*a^3*b^16 + 680*a^ 4*b^15 + 2650*a^5*b^14 + 7152*a^6*b^13 + 14168*a^7*b^12 + 21296*a^8*b^11 + 24750*a^9*b^10 + 22440*a^10*b^9 + 15884*a^11*b^8 + 8712*a^12*b^7 + 3638*a ^13*b^6 + 1120*a^14*b^5 + 240*a^15*b^4 + 32*a^16*b^3 + 2*a^17*b^2 + (ta...
Time = 0.19 (sec) , antiderivative size = 423, normalized size of antiderivative = 3.52 \[ \int \frac {\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {15 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{5} b^{3}+15 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{5} b^{3}-23 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{4}-89 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{3} b -124 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{2} b^{2}-58 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a \,b^{3}+11 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{4}+38 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{3} b +43 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2} b^{2}+16 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a \,b^{3}-3 \cos \left (f x +e \right ) a^{4}-9 \cos \left (f x +e \right ) a^{3} b -9 \cos \left (f x +e \right ) a^{2} b^{2}-3 \cos \left (f x +e \right ) a \,b^{3}-15 \sin \left (f x +e \right )^{5} a^{4} f x -60 \sin \left (f x +e \right )^{5} a^{3} b f x -90 \sin \left (f x +e \right )^{5} a^{2} b^{2} f x -60 \sin \left (f x +e \right )^{5} a \,b^{3} f x -15 \sin \left (f x +e \right )^{5} b^{4} f x}{15 \sin \left (f x +e \right )^{5} a f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )} \] Input:
int(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x)
Output:
(15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt (b))*sin(e + f*x)**5*b**3 + 15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan(( e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**5*b**3 - 23*cos(e + f*x)*sin (e + f*x)**4*a**4 - 89*cos(e + f*x)*sin(e + f*x)**4*a**3*b - 124*cos(e + f *x)*sin(e + f*x)**4*a**2*b**2 - 58*cos(e + f*x)*sin(e + f*x)**4*a*b**3 + 1 1*cos(e + f*x)*sin(e + f*x)**2*a**4 + 38*cos(e + f*x)*sin(e + f*x)**2*a**3 *b + 43*cos(e + f*x)*sin(e + f*x)**2*a**2*b**2 + 16*cos(e + f*x)*sin(e + f *x)**2*a*b**3 - 3*cos(e + f*x)*a**4 - 9*cos(e + f*x)*a**3*b - 9*cos(e + f* x)*a**2*b**2 - 3*cos(e + f*x)*a*b**3 - 15*sin(e + f*x)**5*a**4*f*x - 60*si n(e + f*x)**5*a**3*b*f*x - 90*sin(e + f*x)**5*a**2*b**2*f*x - 60*sin(e + f *x)**5*a*b**3*f*x - 15*sin(e + f*x)**5*b**4*f*x)/(15*sin(e + f*x)**5*a*f*( a**4 + 4*a**3*b + 6*a**2*b**2 + 4*a*b**3 + b**4))