\(\int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [263]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{5/2} f}-\frac {3 (a-b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f} \]

[Out]

1/8*(3*a^2-2*a*b+3*b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f-3/8*(a-b)*(a+b+b*tan(
f*x+e)^2)^(1/2)*tan(f*x+e)/b^2/f+1/4*sec(f*x+e)^2*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4231, 427, 396, 223, 212} \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 b^{5/2} f}-\frac {3 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 b^2 f}+\frac {\tan (e+f x) \sec ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b f} \]

[In]

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

[Out]

((3*a^2 - 2*a*b + 3*b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*b^(5/2)*f) - (3*(a
- b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(8*b^2*f) + (Sec[e + f*x]^2*Tan[e + f*x]*Sqrt[a + b + b*Tan
[e + f*x]^2])/(4*b*f)

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 396

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

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\sec ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f}+\frac {\text {Subst}\left (\int \frac {-a+3 b-3 (a-b) x^2}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{4 b f} \\ & = -\frac {3 (a-b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f}+\frac {\left (3 a^2-2 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 b^2 f} \\ & = -\frac {3 (a-b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f}+\frac {\left (3 a^2-2 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^2 f} \\ & = \frac {\left (3 a^2-2 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{5/2} f}-\frac {3 (a-b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.55 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.38 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \sqrt {a+2 b+a \cos (2 e+2 f x)} \left (-\frac {i \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (-3 a \left (1+e^{2 i (e+f x)}\right )^2+b \left (3+14 e^{2 i (e+f x)}+3 e^{4 i (e+f x)}\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^4}-\frac {\left (3 a^2-2 a b+3 b^2\right ) \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sec (e+f x)}{8 \sqrt {2} b^{5/2} f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Sqrt[a + 2*b + a*Cos[2*e + 2*
f*x]]*(((-I)*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*(-3*a*(1 + E^((2*I)*(e + f*x)))^2 + b*(3 + 14*E^((2*I)*(e + f*
x)) + 3*E^((4*I)*(e + f*x)))))/(1 + E^((2*I)*(e + f*x)))^4 - ((3*a^2 - 2*a*b + 3*b^2)*Log[(-4*Sqrt[b]*(-1 + E^
((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e
+ f*x)))])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*Sec[e + f*x])/(8*Sqrt[2]*b^(5/2)*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. \(1611\) vs. \(2(121)=242\).

Time = 16.83 (sec) , antiderivative size = 1612, normalized size of antiderivative = 11.77

method	result	size
default	\(\text {Expression too large to display}\)	\(1612\)

[In]

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

[Out]

1/16/f/b^(9/2)/(a+b*sec(f*x+e)^2)^(1/2)*(6*b^(7/2)*a*tan(f*x+e)+6*b^(9/2)*tan(f*x+e)*sec(f*x+e)^2-6*b^(5/2)*a^
2*tan(f*x+e)-2*b^(7/2)*a*tan(f*x+e)*sec(f*x+e)^2+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin
(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a^2*b^2-2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2
)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+
a+b)/(sin(f*x+e)-1))*a*b^3+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*
x+e)-1))*b^4+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b
^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2*b^
2-2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(
f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a*b^3+3*((b+a*cos
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2
)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^4+4*b^(9/2)*tan(f*x+e)*sec(f
*x+e)^4+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/
2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a^2*b^2*se
c(f*x+e)-2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1
/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a*b^3*sec
(f*x+e)+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/
2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*b^4*sec(f*
x+e)+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*c
os(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2*b^2*sec(f*
x+e)-2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*c
os(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a*b^3*sec(f*x+
e)+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos
(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^4*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.89 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \sqrt {b} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left (3 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, b^{3} f \cos \left (f x + e\right )^{3}}, \frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, b^{3} f \cos \left (f x + e\right )^{3}}\right ] \]

[In]

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

[Out]

[1/32*((3*a^2 - 2*a*b + 3*b^2)*sqrt(b)*cos(f*x + e)^3*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*
cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e
)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 4*(3*(a*b - b^2)*cos(f*x + e)^2 - 2*b^2)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e)^3), 1/16*((3*a^2 - 2*a*b + 3*b^2)*sqrt(-b)*arctan(-1/2*(
(a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x
+ e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e)^3 - 2*(3*(a*b - b^2)*cos(f*x + e)^2 - 2*b^2)*sqrt((a*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e)^3)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{3}}{b} + \frac {3 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {3 \, {\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {8 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {8 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )}{b}}{8 \, f} \]

[In]

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

[Out]

1/8*(2*sqrt(b*tan(f*x + e)^2 + a + b)*tan(f*x + e)^3/b + 3*(a + b)*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b
^(5/2) + 3*(a + b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 8*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b
))/b^(3/2) - 3*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e)/b^2 + 8*sqrt(b*tan(f*x + e)^2 + a + b)*tan(
f*x + e)/b)/f

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

[In]

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

[Out]

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