\(\int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [388]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \] Output:

-a^(1/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(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
 

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f \sqrt {a+2 b+a \cos (2 e+2 f x)}}-\frac {\cot (e+f x) \left (23 a^2+40 a b+15 b^2-\left (11 a^2+21 a b+10 b^2\right ) \csc ^2(e+f x)+3 (a+b)^2 \csc ^4(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}{15 (a+b)^2 f} \] Input:

Integrate[Cot[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

-((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)
 

Rubi [A] (verified)

Time = 0.48 (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

\(\displaystyle \frac {\frac {1}{5} \left (\frac {-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}-15 \sqrt {a} (a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{3 (a+b)}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}\right )-\frac {1}{5} \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

Input:

Int[Cot[e + f*x]^6*Sqrt[a + b*Sec[e + f*x]^2],x]
 

Output:

(-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
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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])
 

rule 291

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]
 

rule 377

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]
 

rule 445

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]
 

rule 2075

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4629

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(547\) vs. \(2(149)=298\).

Time = 6.35 (sec) , antiderivative size = 548, normalized size of antiderivative = 3.28

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\right ) \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a^{3}+\sin \left (f x +e \right )^{3} \left (-30 \cos \left (f x +e \right )+30\right ) \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a^{2} b +\sin \left (f x +e \right )^{3} \left (-15 \cos \left (f x +e \right )+15\right ) \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a \,b^{2}+\left (23 \cos \left (f x +e \right )^{4}-35 \cos \left (f x +e \right )^{2}+15\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2}+\left (40 \cos \left (f x +e \right )^{4}-59 \cos \left (f x +e \right )^{2}+25\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b +\left (15 \cos \left (f x +e \right )^{4}-20 \cos \left (f x +e \right )^{2}+8\right ) \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}\right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}\, \cot \left (f x +e \right ) \csc \left (f x +e \right )^{4}}{15 f \left (a +b \right )^{2} \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\)	\(548\)

Input:

int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/f/(a+b)^2/(-a)^(1/2)*(sin(f*x+e)^3*(-15*cos(f*x+e)+15)*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)*a^3+sin(f*x+e)^3* 
(-30*cos(f*x+e)+30)*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)*a^2*b+sin(f*x+e)^3*(-15*cos(f*x+e)+15)*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)*a*b^2+(23*cos(f*x+e)^4- 
35*cos(f*x+e)^2+15)*(-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)^4-59*cos(f*x+e)^2+25)*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/( 
1+cos(f*x+e))^2)^(1/2)*a*b+(15*cos(f*x+e)^4-20*cos(f*x+e)^2+8)*(-a)^(1/2)* 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)*(a+b*sec(f*x+e)^2)^(1/2)/ 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cot(f*x+e)*csc(f*x+e)^4
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (149) = 298\).

Time = 2.05 (sec) , antiderivative size = 849, normalized size of antiderivative = 5.08 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[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...
 

Sympy [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \cot ^{6}{\left (e + f x \right )}\, dx \] Input:

integrate(cot(f*x+e)**6*(a+b*sec(f*x+e)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sec(e + f*x)**2)*cot(e + f*x)**6, x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*cot(f*x + e)^6, x)
 

Giac [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \] Input:

integrate(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e)^2 + a)*cot(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \] Input:

int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^(1/2),x)
 

Output:

int(cot(e + f*x)^6*(a + b/cos(e + f*x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {\sec \left (f x +e \right )^{2} b +a}\, \cot \left (f x +e \right )^{6}d x \] Input:

int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2)^(1/2),x)
 

Output:

int(sqrt(sec(e + f*x)**2*b + a)*cot(e + f*x)**6,x)