\(\int \frac {\tan ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [421]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f} \] Output:

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

Mathematica [A] (verified)

Time = 6.12 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (\frac {2 b^2 \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {a}}+\frac {a (3 a+5 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{4 \sqrt {2} a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 a^2+6 a b+2 b^2+\left (3 a^2+4 a b+2 b^2\right ) \cos (2 (e+f x))\right ) \sec ^4(e+f x) \tan (e+f x)}{8 a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

-1/4*(((2*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2] 
])/Sqrt[a] + (a*(3*a + 5*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a* 
Sin[e + f*x]^2]])/Sqrt[b])*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f* 
x]^3)/(Sqrt[2]*a*b^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) + ((a + 2*b + a*Cos[2 
*(e + f*x)])*(3*a^2 + 6*a*b + 2*b^2 + (3*a^2 + 4*a*b + 2*b^2)*Cos[2*(e + f 
*x)])*Sec[e + f*x]^4*Tan[e + f*x])/(8*a*b^2*f*(a + b*Sec[e + f*x]^2)^(3/2) 
)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4629, 2075, 372, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 444

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && 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. \(1295\) vs. \(2(152)=304\).

Time = 13.58 (sec) , antiderivative size = 1296, normalized size of antiderivative = 7.53

method	result	size
default	\(\text {Expression too large to display}\)	\(1296\)

Input:

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

Output:

1/f*((cos(f*x+e)^4+2*cos(f*x+e)^2+1/2)*tan(f*x+e)*sec(f*x+e)^4*a*(-a)^(1/2 
)*b^(7/2)+(2*sec(f*x+e)^2*tan(f*x+e)+2*tan(f*x+e))*a^2*(-a)^(1/2)*b^(5/2)+ 
3/2*b^(3/2)*(-a)^(1/2)*a^3*tan(f*x+e)+b^(9/2)*(-a)^(1/2)*tan(f*x+e)*sec(f* 
x+e)^2-1/4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-a)^(1/2)*a^3*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))*(3+3*sec(f*x+e))-1/4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-a) 
^(1/2)*a^2*b^2*ln(4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(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))*(5+5*sec(f*x+e)+3*sec(f*x+e)^2+3*sec(f*x+e)^3)-1/4*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(-a)^(1/2)*a*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*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))*(5*sec( 
f*x+e)^2+5*sec(f*x+e)^3)-1/4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*( 
-a)^(1/2)*a^3*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))*(3+3*sec(f*x+e))-1/4*((b+a*cos(f*x+e)^2)/(1+cos(f* 
x+e))^2)^(1/2)*(-a)^(1/2)*a^2*b^2*ln(-4*(b^(1/2)*((b+a*cos(f*x+e)^2)/(1+co 
s(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))*(5+5*sec(f*x+e)+3*sec(f*x+e)^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (152) = 304\).

Time = 1.49 (sec) , antiderivative size = 1895, normalized size of antiderivative = 11.02 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/8*((a*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(-a)*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)*co 
s(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^4 + 5*a^3*b) 
*cos(f*x + e)^3 + (3*a^3*b + 5*a^2*b^2)*cos(f*x + e))*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)*co 
s(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*(a^2*b^2 + (3*a^3*b + 
 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
+ e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)), 
 -1/8*(2*((3*a^4 + 5*a^3*b)*cos(f*x + e)^3 + (3*a^3*b + 5*a^2*b^2)*cos(f*x 
 + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sq 
rt(-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))) + (a*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(-a)*l 
og(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*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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