\(\int \frac {\tan ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [326]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}+\frac {\left (3 a^2+4 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 b^{5/2} f}-\frac {(3 a+4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 b f} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 6.24 (sec) , antiderivative size = 768, normalized size of antiderivative = 4.34 \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {-\frac {b \left (3 a^2+4 a b+4 b^2\right ) \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}+\frac {16 b^3 \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{4 b^2 f}+\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (-\frac {3 \sec (e+f x) (a \sin (e+f x)+2 b \sin (e+f x))}{8 b^2}+\frac {\sec ^2(e+f x) \tan (e+f x)}{4 b}\right )}{f} \] Input:

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

Output:

(-((b*(3*a^2 + 4*a*b + 4*b^2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + 
 Cos[2*(e + f*x)])]*Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e 
+ f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[ 
e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Co 
s[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(a*(a + b 
+ (a - b)*Cos[2*(e + f*x)]))) + (16*b^3*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(a 
 + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e 
 + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[( 
(a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*Ell 
ipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/ 
Sqrt[2]], 1]*Sin[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + 
(a - b)*Cos[2*(e + f*x)]]) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + 
 Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f 
*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(b/(a - b)), ArcSin[ 
Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*S 
in[e + f*x]^4)/(2*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)* 
Cos[2*(e + f*x)]])))/Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])/(4*b^2*f) + ( 
Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x 
)])]*((-3*Sec[e + f*x]*(a*Sin[e + f*x] + 2*b*Sin[e + f*x]))/(8*b^2) + (Sec 
[e + f*x]^2*Tan[e + f*x])/(4*b)))/f
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4153, 381, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 381

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

\(\Big \downarrow \) 444

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(4*b) - (-1/2*((-8*b^2*ArcTan 
[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[a - b] + ((3 
*a^2 + 4*a*b + 8*b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f* 
x]^2]])/Sqrt[b])/b + ((3*a + 4*b)*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2]) 
/(2*b))/(4*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 381

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
+ 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) 
Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 
2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q 
}, x] && NeQ[b*c - a*d, 0] && 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 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}+\frac {\tan \left (f x +e \right )^{3} \sqrt {a +b \tan \left (f x +e \right )^{2}}}{4 b}-\frac {3 a \left (\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}-\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}+\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}}{f}\)	\(248\)
default	\(\frac {\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}+\frac {\tan \left (f x +e \right )^{3} \sqrt {a +b \tan \left (f x +e \right )^{2}}}{4 b}-\frac {3 a \left (\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}-\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}+\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}}{f}\)	\(248\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 801, normalized size of antiderivative = 4.53 \[ \int \frac {\tan ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/16*(8*sqrt(-a + b)*b^3*log(-((a - 2*b)*tan(f*x + e)^2 + 2*sqrt(b*tan(f 
*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - (3*a 
^3 + a^2*b + 4*a*b^2 - 8*b^3)*sqrt(b)*log(2*b*tan(f*x + e)^2 + 2*sqrt(b*ta 
n(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) - 2*(2*(a*b^2 - b^3)*tan(f*x + 
 e)^3 - (3*a^2*b + a*b^2 - 4*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a) 
)/((a*b^3 - b^4)*f), -1/8*(4*sqrt(-a + b)*b^3*log(-((a - 2*b)*tan(f*x + e) 
^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x 
+ e)^2 + 1)) + (3*a^3 + a^2*b + 4*a*b^2 - 8*b^3)*sqrt(-b)*arctan(sqrt(-b)* 
tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) - (2*(a*b^2 - b^3)*tan(f*x + e)^3 
 - (3*a^2*b + a*b^2 - 4*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a 
*b^3 - b^4)*f), -1/16*(16*sqrt(a - b)*b^3*arctan(sqrt(a - b)*tan(f*x + e)/ 
sqrt(b*tan(f*x + e)^2 + a)) - (3*a^3 + a^2*b + 4*a*b^2 - 8*b^3)*sqrt(b)*lo 
g(2*b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + 
 a) - 2*(2*(a*b^2 - b^3)*tan(f*x + e)^3 - (3*a^2*b + a*b^2 - 4*b^3)*tan(f* 
x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a*b^3 - b^4)*f), -1/8*(8*sqrt(a - b) 
*b^3*arctan(sqrt(a - b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) + (3*a^3 
+ a^2*b + 4*a*b^2 - 8*b^3)*sqrt(-b)*arctan(sqrt(-b)*tan(f*x + e)/sqrt(b*ta 
n(f*x + e)^2 + a)) - (2*(a*b^2 - b^3)*tan(f*x + e)^3 - (3*a^2*b + a*b^2 - 
4*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a*b^3 - b^4)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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