\(\int \sqrt {a+b \sec ^2(e+f x)} \tan ^6(e+f x) \, dx\) [382]

Optimal result

Integrand size = 25, antiderivative size = 219 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^6(e+f x) \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 b^{5/2} f}-\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 b^2 f}+\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{24 b f}+\frac {\tan ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{6 f} \] Output:

-a^(1/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f+1/16*(a^3 
+5*a^2*b+15*a*b^2-5*b^3)*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-b)*(a+5*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/b 
^2/f+1/24*(a-5*b)*tan(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/b/f+1/6*tan(f*x+ 
e)^5*(a+b+b*tan(f*x+e)^2)^(1/2)/f
 

Mathematica [A] (verified)

Time = 2.46 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.20 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^6(e+f x) \, dx=-\frac {\left (16 \sqrt {a} b^2 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )-\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{8 \sqrt {2} b^2 f \sqrt {a+2 b+a \cos (2 e+2 f x)}}-\frac {\left (9 a^2+34 a b-59 b^2+4 \left (3 a^2+12 a b-7 b^2\right ) \cos (2 (e+f x))+\left (3 a^2+14 a b-33 b^2\right ) \cos (4 (e+f x))\right ) \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \tan (e+f x)}{384 b^2 f} \] Input:

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

Output:

-1/8*((16*Sqrt[a]*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + 
 f*x]^2]] - ((a^3 + 5*a^2*b + 15*a*b^2 - 5*b^3)*ArcTanh[(Sqrt[b]*Sin[e + f 
*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*Cos[e + f*x]*Sqrt[a + b*Sec 
[e + f*x]^2])/(Sqrt[2]*b^2*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]) - ((9*a^2 
 + 34*a*b - 59*b^2 + 4*(3*a^2 + 12*a*b - 7*b^2)*Cos[2*(e + f*x)] + (3*a^2 
+ 14*a*b - 33*b^2)*Cos[4*(e + f*x)])*Sec[e + f*x]^4*Sqrt[a + b*Sec[e + f*x 
]^2]*Tan[e + f*x])/(384*b^2*f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4629, 2075, 380, 444, 27, 444, 398, 224, 219, 291, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \sqrt {b \tan ^2(e+f x)+a+b}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 380

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

\(\Big \downarrow \) 444

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\int \frac {\left (a^3+5 b a^2+15 b^2 a-5 b^3\right ) \tan ^2(e+f x)+(a+5 b) \left (a^2-b^2\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-16 a b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-16 a b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-16 a b^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}\right )}{4 b}\right )+\frac {1}{6} \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-16 a b^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(a-5 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {3 \left (\frac {(a-b) (a+5 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\frac {\left (a^3+5 a^2 b+15 a b^2-5 b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b}}-16 \sqrt {a} b^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b}\right )}{4 b}\right )+\frac {1}{6} \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f}\)

Input:

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

Output:

((Tan[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/6 + (((a - 5*b)*Tan[e + f 
*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*b) - (3*(-1/2*(-16*Sqrt[a]*b^2*Ar 
cTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]] + ((a^3 + 5*a^ 
2*b + 15*a*b^2 - 5*b^3)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[ 
e + f*x]^2]])/Sqrt[b])/b + ((a - b)*(a + 5*b)*Tan[e + f*x]*Sqrt[a + b + b* 
Tan[e + f*x]^2])/(2*b)))/(4*b))/6)/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

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 291

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

rule 380

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

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 444

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

rule 2075

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

rule 3042

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

rule 4629

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]}, Sim 
p[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] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1256\) vs. \(2(193)=386\).

Time = 49.50 (sec) , antiderivative size = 1257, normalized size of antiderivative = 5.74

method	result	size
default	\(\text {Expression too large to display}\)	\(1257\)

Input:

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

Output:

-1/96/f/(-a)^(1/2)/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)*(-a)^(1/2)*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*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))*a^3 
*b^3-15*cos(f*x+e)*(-a)^(1/2)*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))*a^2*b^4-45*cos(f*x+e)*(-a)^(1/2)*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 
))*a*b^5+15*cos(f*x+e)*(-a)^(1/2)*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))*b^6-3*cos(f*x+e)*(-a)^(1/2)*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 
))*a^3*b^3-15*cos(f*x+e)*(-a)^(1/2)*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))*a^2*b^4-45*cos(f*x+e)*(-a)^(1/ 
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))*a*b^5+15*cos(f*x+e)*(-a)^(1/2)*ln(-4*(b^(1/2)*((b+a*cos(f*x+...
 

Fricas [A] (verification not implemented)

Time = 3.75 (sec) , antiderivative size = 1775, normalized size of antiderivative = 8.11 \[ \int \sqrt {a+b \sec ^2(e+f x)} \tan ^6(e+f x) \, dx=\text {Too large to display} \] Input:

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

Output:

[1/192*(24*sqrt(-a)*b^3*cos(f*x + e)^5*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)) - 3*(a^3 + 5*a^2*b + 15*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)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f* 
x + e)^4) - 4*((3*a^2*b + 14*a*b^2 - 33*b^3)*cos(f*x + e)^4 - 8*b^3 - 2*(a 
*b^2 - 13*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), 1/96*(12*sqrt(-a)*b^3*cos(f*x + e)^ 
5*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 - 2 
8*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)) + 
3*(a^3 + 5*a^2*b + 15*a*b^2 - 5*b^3)*sqrt(-b)*arctan(-1/2*((a - b)*cos(...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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