\(\int \frac {\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [289]

Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}+\frac {a^2 \tan (e+f x)}{3 b^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 a (2 a+3 b) \tan (e+f x)}{3 b^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:

arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+1/3*a^2*t 
an(f*x+e)/b^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-2/3*a*(2*a+3*b)*tan(f*x+e 
)/b^2/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 9.89 (sec) , antiderivative size = 592, normalized size of antiderivative = 4.66 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^6(e+f x) \sqrt {1-\frac {2 a \sin ^2(e+f x)}{2 a+2 b}} \tan (e+f x) \left (-\frac {24 b \cos ^2(e+f x) \, _3F_2\left (2,2,2;1,\frac {9}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^2(e+f x) \left (-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}\right )^{5/2}}{a+b}-\frac {\sec ^6(e+f x) \left (24 b^3 \operatorname {Hypergeometric2F1}\left (2,2,\frac {9}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \sin ^6(e+f x) \left (4 b^2+a b \left (8-7 \sin ^2(e+f x)\right )+a^2 \left (4-7 \sin ^2(e+f x)+3 \sin ^4(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}+35 (a+b) \cos ^2(e+f x) \left (15 b^2+10 a b \left (3-2 \sin ^2(e+f x)\right )+a^2 \left (15-20 \sin ^2(e+f x)+8 \sin ^4(e+f x)\right )\right ) \left (-3 \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \left (a+b-a \sin ^2(e+f x)\right )^2-(a+b) \cos ^2(e+f x) \left (-3 a \cos ^2(e+f x)-b \left (3+\sin ^2(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}\right )\right )}{(a+b)^5}\right )}{315 (2 a+2 b)^2 f \left (a+b \sec ^2(e+f x)\right )^{5/2} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \sqrt {2 a+2 b-2 a \sin ^2(e+f x)} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right )^{3/2} \left (-\frac {b \tan ^2(e+f x)}{a+b}\right )^{5/2}} \] Input:

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^6*Sqrt[1 - (2*a*Sin[e + 
 f*x]^2)/(2*a + 2*b)]*Tan[e + f*x]*((-24*b*Cos[e + f*x]^2*HypergeometricPF 
Q[{2, 2, 2}, {1, 9/2}, -((b*Tan[e + f*x]^2)/(a + b))]*Sin[e + f*x]^2*(-((b 
*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2))^(5/ 
2))/(a + b) - (Sec[e + f*x]^6*(24*b^3*Hypergeometric2F1[2, 2, 9/2, -((b*Ta 
n[e + f*x]^2)/(a + b))]*Sin[e + f*x]^6*(4*b^2 + a*b*(8 - 7*Sin[e + f*x]^2) 
 + a^2*(4 - 7*Sin[e + f*x]^2 + 3*Sin[e + f*x]^4))*Sqrt[-((b*Sec[e + f*x]^2 
*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2)] + 35*(a + b)*Cos[e 
 + f*x]^2*(15*b^2 + 10*a*b*(3 - 2*Sin[e + f*x]^2) + a^2*(15 - 20*Sin[e + f 
*x]^2 + 8*Sin[e + f*x]^4))*(-3*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]] 
*(a + b - a*Sin[e + f*x]^2)^2 - (a + b)*Cos[e + f*x]^2*(-3*a*Cos[e + f*x]^ 
2 - b*(3 + Sin[e + f*x]^2))*Sqrt[-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f* 
x]^2)*Tan[e + f*x]^2)/(a + b)^2)])))/(a + b)^5))/(315*(2*a + 2*b)^2*f*(a + 
 b*Sec[e + f*x]^2)^(5/2)*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqrt[2*a + 2 
*b - 2*a*Sin[e + f*x]^2]*(1 - (a*Sin[e + f*x]^2)/(a + b))^(3/2)*(-((b*Tan[ 
e + f*x]^2)/(a + b)))^(5/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4634, 315, 298, 224, 219}

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.

\(\displaystyle \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4634

\(\displaystyle \frac {\int \frac {\left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {\frac {3 (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{b}-\frac {a (3 a+5 b) \tan (e+f x)}{b (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {3 (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{b}-\frac {a (3 a+5 b) \tan (e+f x)}{b (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {3 (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{b^{3/2}}-\frac {a (3 a+5 b) \tan (e+f x)}{b (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 b (a+b)}-\frac {a \tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 b (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\)

Input:

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(5/2),x]
 

Output:

(-1/3*(a*Tan[e + f*x]*(1 + Tan[e + f*x]^2))/(b*(a + b)*(a + b + b*Tan[e + 
f*x]^2)^(3/2)) + ((3*(a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b 
*Tan[e + f*x]^2]])/b^(3/2) - (a*(3*a + 5*b)*Tan[e + f*x])/(b*(a + b)*Sqrt[ 
a + b + b*Tan[e + f*x]^2]))/(3*b*(a + b)))/f
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1636\) vs. \(2(113)=226\).

Time = 21.56 (sec) , antiderivative size = 1637, normalized size of antiderivative = 12.89

method	result	size
default	\(\text {Expression too large to display}\)	\(1637\)

Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6/f/(a+b)^2/b^(7/2)*(cos(f*x+e)^4*(3*cos(f*x+e)+3)*((b+a*cos(f*x+e)^2)/( 
1+cos(f*x+e))^2)^(1/2)*a^4*b*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(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))+cos(f*x+e)^2*(6*cos(f*x+e)^3+6*cos(f*x 
+e)^2+6*cos(f*x+e)+6)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b^2* 
ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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))+(3*cos(f*x+e)^5+3*cos(f*x+e)^4+12*cos(f*x+e)^3+12*cos(f*x+e)^2+3*co 
s(f*x+e)+3)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2*b^3*ln(4*(b^(1 
/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*c 
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))+(6* 
cos(f*x+e)^3+6*cos(f*x+e)^2+6*cos(f*x+e)+6)*((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)*a*b^4*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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))+(3*cos(f*x+e)+3)*((b+a*cos(f*x+e)^2)/(1+cos(f*x 
+e))^2)^(1/2)*b^5*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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))+cos(f*x+e)^4*(3*cos(f*x+e)+3)*((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*a^4*b*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f 
*x+e))^2)^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (113) = 226\).

Time = 0.32 (sec) , antiderivative size = 688, normalized size of antiderivative = 5.42 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b 
^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sqrt(b)*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^3*b + 5*a^2*b^2)*c 
os(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) 
^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*c 
os(f*x + e)^4 + 2*(a^3*b^4 + 2*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^ 
5 + 2*a*b^6 + b^7)*f), 1/6*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + 
a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sq 
rt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sq 
rt((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))) - 2*((3*a^3*b + 5*a^2*b^2)*cos(f*x + e)^3 + 2*(2*a^2*b^2 + 3*a* 
b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e 
))/((a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^3*b^4 + 2*a^2* 
b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^2*b^5 + 2*a*b^6 + b^7)*f)]
 

Sympy [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**(5/2),x)
 

Output:

Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (113) = 226\).

Time = 0.04 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (\frac {3 \, \tan \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b^{2}} + \frac {2}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}\right )} \tan \left (f x + e\right ) - \frac {3 \, \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} - \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} + \frac {3 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} b^{2}} - \frac {2 \, a \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b^{2}} + \frac {2 \, \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} - \frac {4 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate(sec(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^6\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \] Input:

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(5/2)),x)
 

Output:

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{6}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**6)/(sec(e + f*x)**6*b**3 + 
3*sec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)