Integrand size = 25, antiderivative size = 167 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \]
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*a^(1/2)/f-1/15*(15* a^2+25*a*b+8*b^2)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^2/f-1/15*(-4 *b-5*a)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)/f-1/5*cot(f*x+e)^5*( a+b+b*tan(f*x+e)^2)^(1/2)/f
Time = 2.88 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f \sqrt {a+2 b+a \cos (2 e+2 f x)}}-\frac {\cot (e+f x) \left (23 a^2+40 a b+15 b^2-\left (11 a^2+21 a b+10 b^2\right ) \csc ^2(e+f x)+3 (a+b)^2 \csc ^4(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}{15 (a+b)^2 f} \]
-((Sqrt[2]*Sqrt[a]*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f* x]^2]]*Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(f*Sqrt[a + 2*b + a*Cos[2* e + 2*f*x]])) - (Cot[e + f*x]*(23*a^2 + 40*a*b + 15*b^2 - (11*a^2 + 21*a*b + 10*b^2)*Csc[e + f*x]^2 + 3*(a + b)^2*Csc[e + f*x]^4)*Sqrt[a + b*Sec[e + f*x]^2])/(15*(a + b)^2*f)
Time = 0.45 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4629, 2075, 377, 445, 25, 445, 27, 291, 216}
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.
\(\displaystyle \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sec (e+f x)^2}}{\tan (e+f x)^6}dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \sqrt {a+b \left (\tan ^2(e+f x)+1\right )}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \sqrt {b \tan ^2(e+f x)+a+b}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 377 |
\(\displaystyle \frac {\frac {1}{5} \int \frac {\cot ^4(e+f x) \left (-4 b \tan ^2(e+f x)+b-5 (a+b)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 445 |
\(\displaystyle \frac {\frac {1}{5} \left (-\frac {\int -\frac {\cot ^2(e+f x) \left (15 a^2+25 b a+8 b^2-2 b (b-5 (a+b)) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {\int \frac {\cot ^2(e+f x) \left (15 a^2+25 b a+8 b^2-2 b (b-5 (a+b)) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 445 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {-\frac {\int \frac {15 a (a+b)^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {-15 a (a+b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {-15 a (a+b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}-15 \sqrt {a} (a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\) |
(-1/5*(Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2]) + (-1/3*((b - 5*(a + b))*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b) + (-15*Sqrt[a] *(a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] - ( (15*a^2 + 25*a*b + 8*b^2)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b))/(3*(a + b)))/5)/f
3.4.88.3.1 Defintions of rubi rules used
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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
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/(a*e*( m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[b*c*(m + 1) + 2*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + 2*b*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b *c - a*d, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m , 2, p, q, 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])
Leaf count of result is larger than twice the leaf count of optimal. \(1015\) vs. \(2(149)=298\).
Time = 5.90 (sec) , antiderivative size = 1016, normalized size of antiderivative = 6.08
1/15/f/(a+b)^2/(-a)^(1/2)*(15*sin(f*x+e)^3*a^3*ln(4*(-a)^(1/2)*((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^ 2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*cos(f*x+e)+30*sin(f*x+e)^3*a^2* b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4 *(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*co s(f*x+e)+15*sin(f*x+e)^3*a*b^2*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) )^2)^(1/2)-4*sin(f*x+e)*a)*cos(f*x+e)-23*cos(f*x+e)^4*(-a)^(1/2)*((b+a*cos (f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2-40*cos(f*x+e)^4*(-a)^(1/2)*((b+a*co s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-15*cos(f*x+e)^4*(-a)^(1/2)*((b+a*c os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2-15*sin(f*x+e)^3*a^3*ln(4*(-a)^(1/ 2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b +a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)-30*sin(f*x+e)^3*a ^2*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e )+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a) -15*sin(f*x+e)^3*a*b^2*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 2)-4*sin(f*x+e)*a)+35*cos(f*x+e)^2*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f *x+e))^2)^(1/2)*a^2+59*cos(f*x+e)^2*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( f*x+e))^2)^(1/2)*a*b+20*cos(f*x+e)^2*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+...
Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (149) = 298\).
Time = 2.30 (sec) , antiderivative size = 849, normalized size of antiderivative = 5.08 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 8 \, {\left ({\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (35 \, a^{2} + 59 \, a b + 20 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 25 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (35 \, a^{2} + 59 \, a b + 20 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 25 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/120*(15*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos (f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256 *(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - ( a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) ^2 + b)/cos(f*x + e)^2)*sin(f*x + e))*sin(f*x + e) - 8*((23*a^2 + 40*a*b + 15*b^2)*cos(f*x + e)^5 - (35*a^2 + 59*a*b + 20*b^2)*cos(f*x + e)^3 + (15* a^2 + 25*a*b + 8*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*co s(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f)*sin(f*x + e)), 1/60*(15*((a^2 + 2*a* b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a *b + b^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^ 2*b)*cos(f*x + e)^2)*sin(f*x + e)))*sin(f*x + e) - 4*((23*a^2 + 40*a*b + 1 5*b^2)*cos(f*x + e)^5 - (35*a^2 + 59*a*b + 20*b^2)*cos(f*x + e)^3 + (15*a^ 2 + 25*a*b + 8*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) ^2))/(((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*c...
\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \cot ^{6}{\left (e + f x \right )}\, dx \]
\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \]
\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]