\(\int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [414]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f} \]

[Out]

-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/a^(1/2)-1/15*(15*a^2+40*a*b+33*b^2)*cot(f*x+e)*(a+b+b
*tan(f*x+e)^2)^(1/2)/(a+b)^3/f+1/15*(5*a+9*b)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^2/f-1/5*cot(f*x+e)
^5*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)/f

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4226, 2000, 491, 597, 12, 385, 209} \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)^3}-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a} f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 f (a+b)}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2} \]

[In]

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

[Out]

-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(Sqrt[a]*f)) - ((15*a^2 + 40*a*b + 33*b^2)*Cot
[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^3*f) + ((5*a + 9*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e
+ f*x]^2])/(15*(a + b)^2*f) - (Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(5*(a + b)*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 491

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 597

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 2000

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4226

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]}, Dist[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] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^6 \left (1+x^2\right ) \sqrt {a+b \left (1+x^2\right )}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x^6 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac {\text {Subst}\left (\int \frac {-5 a-9 b-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f} \\ & = \frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}-\frac {\text {Subst}\left (\int \frac {-15 a^2-40 a b-33 b^2-2 b (5 a+9 b) x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f} \\ & = -\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac {\text {Subst}\left (\int -\frac {15 (a+b)^3}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^3 f} \\ & = -\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}-\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f}-\frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f} \\ & = -\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{\sqrt {2} \sqrt {a} f \sqrt {a+b \sec ^2(e+f x)}}-\frac {(a+2 b+a \cos (2 (e+f x))) \csc (e+f x) \left (23 a^2+60 a b+45 b^2-\left (11 a^2+26 a b+15 b^2\right ) \csc ^2(e+f x)+3 (a+b)^2 \csc ^4(e+f x)\right ) \sec (e+f x)}{30 (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

-((ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*
x])/(Sqrt[2]*Sqrt[a]*f*Sqrt[a + b*Sec[e + f*x]^2])) - ((a + 2*b + a*Cos[2*(e + f*x)])*Csc[e + f*x]*(23*a^2 + 6
0*a*b + 45*b^2 - (11*a^2 + 26*a*b + 15*b^2)*Csc[e + f*x]^2 + 3*(a + b)^2*Csc[e + f*x]^4)*Sec[e + f*x])/(30*(a
+ b)^3*f*Sqrt[a + b*Sec[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1278\) vs. \(2(154)=308\).

Time = 6.78 (sec) , antiderivative size = 1279, normalized size of antiderivative = 7.44

method	result	size
default	\(\text {Expression too large to display}\)	\(1279\)

[In]

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

[Out]

-1/15/f/(a+b)^3/(-a)^(1/2)*(15*sin(f*x+e)^5*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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e
))^2)^(1/2)*a^3*cos(f*x+e)+45*sin(f*x+e)^5*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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)^2)^(1/2)*a^2*b*cos(f*x+e)+45*sin(f*x+e)^5*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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e
))^2)^(1/2)*a*b^2*cos(f*x+e)+15*sin(f*x+e)^5*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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*b^3*cos(f*x+e)+23*cos(f*x+e)^6*(-a)^(1/2)*a^3+60*cos(f*x+e)^6*(-a)^(1/2)*a^2*b+45*cos(f*x+e)^6*(-
a)^(1/2)*a*b^2+15*sin(f*x+e)^5*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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a
^3+45*sin(f*x+e)^5*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*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b+45*sin(
f*x+e)^5*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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^2+15*sin(f*x+e)^5*l
n(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)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3-35*cos(f*x+e)^4*(-a)^(1/2)*a^
3-71*(-a)^(1/2)*a^2*b*cos(f*x+e)^4-15*cos(f*x+e)^4*(-a)^(1/2)*a*b^2+45*cos(f*x+e)^4*(-a)^(1/2)*b^3+15*cos(f*x+
e)^2*(-a)^(1/2)*a^3+5*(-a)^(1/2)*a^2*b*cos(f*x+e)^2-61*(-a)^(1/2)*a*b^2*cos(f*x+e)^2-75*cos(f*x+e)^2*(-a)^(1/2
)*b^3+15*(-a)^(1/2)*a^2*b+40*(-a)^(1/2)*a*b^2+33*(-a)^(1/2)*b^3)/(a+b*sec(f*x+e)^2)^(1/2)/(-1+cos(f*x+e))^2/(1
+cos(f*x+e))^2*sec(f*x+e)*csc(f*x+e)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (154) = 308\).

Time = 1.91 (sec) , antiderivative size = 987, normalized size of antiderivative = 5.74 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/120*(15*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*cos(f*x + e)^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^3 + 60*a^2*b + 45*a*b^2)*cos(f*x + e)^5 - (
35*a^3 + 94*a^2*b + 75*a*b^2)*cos(f*x + e)^3 + (15*a^3 + 40*a^2*b + 33*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x +
e)^2 + b)/cos(f*x + e)^2))/(((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)), 1/60*(15*((a^3 + 3*a^2*b
+ 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x +
e)^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^3 + 60*a^2*b + 45*a*b^2)*cos(f*x + e)^5 - (35*a^3 +
94*a^2*b + 75*a*b^2)*cos(f*x + e)^3 + (15*a^3 + 40*a^2*b + 33*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)
/cos(f*x + e)^2))/(((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))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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