\(\int \frac {\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [212]

Optimal result

Integrand size = 23, antiderivative size = 138 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\left (3 a^2+8 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} (a+b)^{5/2} f}+\frac {a^2 \tan (e+f x)}{4 b^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {a (5 a+8 b) \tan (e+f x)}{8 b^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {\left (3 a^2+8 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {a \sqrt {b} \left (3 a^2+16 a b+16 b^2+3 a (a+2 b) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2}}{8 b^{5/2} f} \] Input:

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(((3*a^2 + 8*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + 
 b)^(5/2) - (a*Sqrt[b]*(3*a^2 + 16*a*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + 
 f*x)])*Sin[2*(e + f*x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))/(8 
*b^(5/2)*f)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4634, 315, 298, 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.

\(\displaystyle \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \frac {\left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 315

\(\displaystyle \frac {\frac {\int \frac {(3 a+4 b) \tan ^2(e+f x)+a+4 b}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{4 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {\frac {\left (3 a^2+8 a b+8 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{2 b (a+b)}-\frac {3 a (a+2 b) \tan (e+f x)}{2 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{4 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\left (3 a^2+8 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 b^{3/2} (a+b)^{3/2}}-\frac {3 a (a+2 b) \tan (e+f x)}{2 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{4 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

Input:

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(-1/4*(a*Tan[e + f*x]*(1 + Tan[e + f*x]^2))/(b*(a + b)*(a + b + b*Tan[e + 
f*x]^2)^2) + (((3*a^2 + 8*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[ 
a + b]])/(2*b^(3/2)*(a + b)^(3/2)) - (3*a*(a + 2*b)*Tan[e + f*x])/(2*b*(a 
+ b)*(a + b + b*Tan[e + f*x]^2)))/(4*b*(a + b)))/f
 

Defintions of rubi rules used

rule 218

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]
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 3042

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

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
 

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {-\frac {a \left (5 a +8 b \right ) \tan \left (f x +e \right )^{3}}{8 b \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (3 a +8 b \right ) a \tan \left (f x +e \right )}{8 b^{2} \left (a +b \right )}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+8 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) b^{2} \sqrt {\left (a +b \right ) b}}}{f}\)	\(137\)
default	\(\frac {\frac {-\frac {a \left (5 a +8 b \right ) \tan \left (f x +e \right )^{3}}{8 b \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (3 a +8 b \right ) a \tan \left (f x +e \right )}{8 b^{2} \left (a +b \right )}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+8 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) b^{2} \sqrt {\left (a +b \right ) b}}}{f}\)	\(137\)
risch	\(-\frac {i \left (3 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+8 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+8 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+9 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+42 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+72 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+9 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+40 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b +40 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}+3 a^{3}+6 a^{2} b \right )}{4 \left (a +b \right )^{2} b^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) a^{2}}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f \,b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) a}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f b}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) a^{2}}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) a}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f b}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f}\)	\(773\)

Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (124) = 248\).

Time = 0.12 (sec) , antiderivative size = 722, normalized size of antiderivative = 5.23 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[-1/32*(((3*a^4 + 8*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 8*a*b^ 
3 + 8*b^4 + 2*(3*a^3*b + 8*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)*sqrt(-a*b - 
b^2)*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*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*s 
in(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + 4* 
(3*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*cos(f*x + e)^3 + (5*a^3*b^2 + 13*a^2*b^ 
3 + 8*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^5*b^3 + 3*a^4*b^4 + 3*a^3*b^5 
 + a^2*b^6)*f*cos(f*x + e)^4 + 2*(a^4*b^4 + 3*a^3*b^5 + 3*a^2*b^6 + a*b^7) 
*f*cos(f*x + e)^2 + (a^3*b^5 + 3*a^2*b^6 + 3*a*b^7 + b^8)*f), -1/16*(((3*a 
^4 + 8*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 8*a*b^3 + 8*b^4 + 2 
*(3*a^3*b + 8*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)*sqrt(a*b + b^2)*arctan(1/ 
2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e 
))) + 2*(3*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*cos(f*x + e)^3 + (5*a^3*b^2 + 1 
3*a^2*b^3 + 8*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^5*b^3 + 3*a^4*b^4 + 3 
*a^3*b^5 + a^2*b^6)*f*cos(f*x + e)^4 + 2*(a^4*b^4 + 3*a^3*b^5 + 3*a^2*b^6 
+ a*b^7)*f*cos(f*x + e)^2 + (a^3*b^5 + 3*a^2*b^6 + 3*a*b^7 + b^8)*f)]
 

Sympy [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \] Input:

integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {{\left (5 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{3} + 11 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )}{a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6} + {\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

1/8*((3*a^2 + 8*a*b + 8*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^2* 
b^2 + 2*a*b^3 + b^4)*sqrt((a + b)*b)) - ((5*a^2*b + 8*a*b^2)*tan(f*x + e)^ 
3 + (3*a^3 + 11*a^2*b + 8*a*b^2)*tan(f*x + e))/(a^4*b^2 + 4*a^3*b^3 + 6*a^ 
2*b^4 + 4*a*b^5 + b^6 + (a^2*b^4 + 2*a*b^5 + b^6)*tan(f*x + e)^4 + 2*(a^3* 
b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*tan(f*x + e)^2))/f
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )}}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \sqrt {a b + b^{2}}} - \frac {5 \, a^{2} b \tan \left (f x + e\right )^{3} + 8 \, a b^{2} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 11 \, a^{2} b \tan \left (f x + e\right ) + 8 \, a b^{2} \tan \left (f x + e\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}}}{8 \, f} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

1/8*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b 
 + b^2)))*(3*a^2 + 8*a*b + 8*b^2)/((a^2*b^2 + 2*a*b^3 + b^4)*sqrt(a*b + b^ 
2)) - (5*a^2*b*tan(f*x + e)^3 + 8*a*b^2*tan(f*x + e)^3 + 3*a^3*tan(f*x + e 
) + 11*a^2*b*tan(f*x + e) + 8*a*b^2*tan(f*x + e))/((a^2*b^2 + 2*a*b^3 + b^ 
4)*(b*tan(f*x + e)^2 + a + b)^2))/f
 

Mupad [B] (verification not implemented)

Time = 17.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{8\,b^{5/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+8\,b\,a\right )}{8\,b\,{\left (a+b\right )}^2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a^2+8\,b\,a\right )}{8\,b^2\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )} \] Input:

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^3),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1296, normalized size of antiderivative = 9.39 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

(3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*sin(e + f*x)**4*a**4 + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
+ f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b + 8*sqrt(b)*sqrt(a + 
b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4* 
a**2*b**2 - 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqr 
t(a))/sqrt(b))*sin(e + f*x)**2*a**4 - 22*sqrt(b)*sqrt(a + b)*atan((sqrt(a 
+ b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3*b - 32*sqrt 
(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin 
(e + f*x)**2*a**2*b**2 - 16*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + 
 f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**3 + 3*sqrt(b)*sqrt(a + b 
)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 + 14*sqrt(b) 
*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3*b 
 + 27*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sq 
rt(b))*a**2*b**2 + 24*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/ 
2) - sqrt(a))/sqrt(b))*a*b**3 + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*ta 
n((e + f*x)/2) - sqrt(a))/sqrt(b))*b**4 + 3*sqrt(b)*sqrt(a + b)*atan((sqrt 
(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**4 + 8*sqrt 
(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin 
(e + f*x)**4*a**3*b + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x 
)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b**2 - 6*sqrt(b)*sqrt(a +...