\(\int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [234]

Optimal result

Integrand size = 25, antiderivative size = 178 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(a+b) \left (a^2-2 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 b^{5/2} f}+\frac {\left (a^2-2 a b+5 b^2\right ) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 b^2 f}-\frac {(a-3 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{8 b^2 f}+\frac {\tan ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{6 b f} \] Output:

1/16*(a+b)*(a^2-2*a*b+5*b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^ 
2)^(1/2))/b^(5/2)/f+1/16*(a^2-2*a*b+5*b^2)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2) 
^(1/2)/b^2/f-1/8*(a-3*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/b^2/f+1/6*t 
an(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(3/2)/b/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 12.23 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.43 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {i e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (\sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} \left (-3 a^2 \left (1+e^{2 i (e+f x)}\right )^4+4 a b \left (1+e^{2 i (e+f x)}\right )^2 \left (1+4 e^{2 i (e+f x)}+e^{4 i (e+f x)}\right )+b^2 \left (15+100 e^{2 i (e+f x)}+298 e^{4 i (e+f x)}+100 e^{6 i (e+f x)}+15 e^{8 i (e+f x)}\right )\right )+3 \left (a^3-a^2 b+3 a b^2+5 b^3\right ) \left (1+e^{2 i (e+f x)}\right )^6 \arctan \left (\frac {\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right )\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{24 \sqrt {2} b^{5/2} \left (1+e^{2 i (e+f x)}\right )^6 \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \] Input:

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

Output:

((-1/24*I)*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^(( 
2*I)*(e + f*x))]*(Sqrt[b]*(-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]*(-3*a^2*(1 + E^((2*I)*(e + f*x))) 
^4 + 4*a*b*(1 + E^((2*I)*(e + f*x)))^2*(1 + 4*E^((2*I)*(e + f*x)) + E^((4* 
I)*(e + f*x))) + b^2*(15 + 100*E^((2*I)*(e + f*x)) + 298*E^((4*I)*(e + f*x 
)) + 100*E^((6*I)*(e + f*x)) + 15*E^((8*I)*(e + f*x)))) + 3*(a^3 - a^2*b + 
 3*a*b^2 + 5*b^3)*(1 + E^((2*I)*(e + f*x)))^6*ArcTan[(Sqrt[b]*(-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]])*Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(Sqrt[2]*b^(5/2)*(1 + E^((2 
*I)*(e + f*x)))^6*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x) 
))^2]*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4634, 318, 25, 299, 211, 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.

\(\displaystyle \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^6 \sqrt {a+b \sec (e+f x)^2}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 318

\(\displaystyle \frac {\frac {\int -\left (\left ((3 a-5 b) \tan ^2(e+f x)+a-5 b\right ) \sqrt {b \tan ^2(e+f x)+a+b}\right )d\tan (e+f x)}{6 b}+\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b}-\frac {\int \left ((3 a-5 b) \tan ^2(e+f x)+a-5 b\right ) \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{6 b}}{f}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b}-\frac {\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {3 \left (a^2-2 a b+5 b^2\right ) \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{4 b}}{6 b}}{f}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b}-\frac {\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {3 \left (a^2-2 a b+5 b^2\right ) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{6 b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b}-\frac {\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {3 \left (a^2-2 a b+5 b^2\right ) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{6 b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{6 b}-\frac {\frac {(3 a-5 b) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {3 \left (a^2-2 a b+5 b^2\right ) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{6 b}}{f}\)

Input:

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

Output:

((Tan[e + f*x]*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^(3/2))/(6*b 
) - (((3*a - 5*b)*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*b) - ( 
3*(a^2 - 2*a*b + 5*b^2)*(((a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + 
b + b*Tan[e + f*x]^2]])/(2*Sqrt[b]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + 
 f*x]^2])/2))/(4*b))/(6*b))/f
 

Defintions of rubi rules used

rule 25

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

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 299

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

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 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. \(1105\) vs. \(2(158)=316\).

Time = 47.89 (sec) , antiderivative size = 1106, normalized size of antiderivative = 6.21

method	result	size
default	\(\text {Expression too large to display}\)	\(1106\)

Input:

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

Output:

1/96/f/b^(11/2)*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2 
)/(1+cos(f*x+e))^2)^(1/2)*(3*cos(f*x+e)*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))*a^3*b^3-3*cos(f*x+e)*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)) 
*a^2*b^4+9*cos(f*x+e)*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))*a*b^5+15*cos(f*x+e)*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))*b^6+3*cos(f*x+e)*l 
n(-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))*a^3*b^3-3*cos(f*x+e)*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))*a^2*b^4+9*cos(f*x+e)*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))*a*b^5+15* 
cos(f*x+e)*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...
 

Fricas [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.63 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} \sqrt {b} \cos \left (f x + e\right )^{5} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (3 \, a^{2} b - 4 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, b^{3} - 2 \, {\left (a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, b^{3} f \cos \left (f x + e\right )^{5}}, \frac {3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - 2 \, {\left ({\left (3 \, a^{2} b - 4 \, a b^{2} - 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, b^{3} - 2 \, {\left (a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{96 \, b^{3} f \cos \left (f x + e\right )^{5}}\right ] \] Input:

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

Output:

[1/192*(3*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*sqrt(b)*cos(f*x + e)^5*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)/co 
s(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 4*((3*a^2*b - 4*a*b^ 
2 - 15*b^3)*cos(f*x + e)^4 - 8*b^3 - 2*(a*b^2 + 5*b^3)*cos(f*x + e)^2)*sqr 
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e) 
^5), 1/96*(3*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*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)/c 
os(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e)^5 - 
 2*((3*a^2*b - 4*a*b^2 - 15*b^3)*cos(f*x + e)^4 - 8*b^3 - 2*(a*b^2 + 5*b^3 
)*cos(f*x + e)^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)^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (158) = 316\).

Time = 0.04 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.78 \[ \int \sec ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {\frac {8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{3}}{b} + \frac {3 \, {\left (a + b\right )}^{2} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {12 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {12 \, {\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \frac {24 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 24 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 24 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right ) - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b^{2}} + \frac {24 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )}{b} - \frac {12 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b}}{48 \, f} \] Input:

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

Output:

1/48*(8*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e)^3/b + 3*(a + b)^2*a* 
arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(5/2) + 3*(a + b)^2*arcsinh(b*ta 
n(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 12*(a + b)*a*arcsinh(b*tan(f*x + e)/ 
sqrt((a + b)*b))/b^(3/2) - 12*(a + b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)* 
b))/sqrt(b) + 24*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 24*sq 
rt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 24*sqrt(b*tan(f*x + e)^2 + 
 a + b)*tan(f*x + e) - 6*(b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)*tan(f*x 
+ e)/b^2 + 3*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2*tan(f*x + e)/b^2 + 2 
4*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e)/b - 12*sqrt(b*tan(f*x + e) 
^2 + a + b)*(a + b)*tan(f*x + e)/b)/f
 

Giac [F]

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

integrate(sec(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)*sec(f*x + e)^6, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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