\(\int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [343]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {x}{a}+\frac {(a+b)^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a b^{5/2} f}-\frac {(a+2 b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.45 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.76 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (-\frac {3 x}{a}-\frac {3 (a+b)^{5/2} \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{a b^2 f \sqrt {b (\cos (e)-i \sin (e))^4}}-\frac {(3 a+7 b) \sec (e) \sec (e+f x) \sin (f x)}{b^2 f}+\frac {\sec (e) \sec ^3(e+f x) \sin (f x)}{b f}+\frac {\sec ^2(e+f x) \tan (e)}{b f}\right )}{6 \left (a+b \sec ^2(e+f x)\right )} \] Input:

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

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*((-3*x)/a - (3*(a + b)^(5/2 
)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[ 
2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I* 
Sin[2*e]))/(a*b^2*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) - ((3*a + 7*b)*Sec[e]*S 
ec[e + f*x]*Sin[f*x])/(b^2*f) + (Sec[e]*Sec[e + f*x]^3*Sin[f*x])/(b*f) + ( 
Sec[e + f*x]^2*Tan[e])/(b*f)))/(6*(a + b*Sec[e + f*x]^2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4629, 2075, 381, 27, 444, 397, 216, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 381

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 444

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(Tan[e + f*x]^3/(3*b) - (-((-((b^2*ArcTan[Tan[e + f*x]])/a) + ((a + b)^(5/ 
2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[b]))/b) + ((a + 2*b 
)*Tan[e + f*x])/b)/b)/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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(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 [A] (verified)

Time = 1.62 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {b \tan \left (f x +e \right )^{3}}{3}+a \tan \left (f x +e \right )+2 b \tan \left (f x +e \right )}{b^{2}}+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b^{2} a \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\)	\(101\)
default	\(\frac {-\frac {-\frac {b \tan \left (f x +e \right )^{3}}{3}+a \tan \left (f x +e \right )+2 b \tan \left (f x +e \right )}{b^{2}}+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b^{2} a \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}}{f}\)	\(101\)
risch	\(-\frac {x}{a}-\frac {2 i \left (3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+9 b \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a \,{\mathrm e}^{2 i \left (f x +e \right )}+12 b \,{\mathrm e}^{2 i \left (f x +e \right )}+3 a +7 b \right )}{3 f \,b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}-\frac {\sqrt {-\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b^{3} f}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{b^{2} f}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f a}+\frac {\sqrt {-\left (a +b \right ) b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b^{3} f}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{b^{2} f}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f a}\)	\(383\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (73) = 146\).

Time = 0.12 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.49 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {12 \, b^{2} f x \cos \left (f x + e\right )^{3} - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a + b}{b}} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sin \left (f x + e\right )}{12 \, a b^{2} f \cos \left (f x + e\right )^{3}}, -\frac {6 \, b^{2} f x \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} + 2 \, {\left ({\left (3 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sin \left (f x + e\right )}{6 \, a b^{2} f \cos \left (f x + e\right )^{3}}\right ] \] Input:

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

Output:

[-1/12*(12*b^2*f*x*cos(f*x + e)^3 - 3*(a^2 + 2*a*b + b^2)*sqrt(-(a + b)/b) 
*cos(f*x + e)^3*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b 
^2)*cos(f*x + e)^2 - 4*((a*b + 2*b^2)*cos(f*x + e)^3 - b^2*cos(f*x + e))*s 
qrt(-(a + b)/b)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + 
e)^2 + b^2)) + 4*((3*a^2 + 7*a*b)*cos(f*x + e)^2 - a*b)*sin(f*x + e))/(a*b 
^2*f*cos(f*x + e)^3), -1/6*(6*b^2*f*x*cos(f*x + e)^3 + 3*(a^2 + 2*a*b + b^ 
2)*sqrt((a + b)/b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a + b)/ 
b)/((a + b)*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e)^3 + 2*((3*a^2 + 7*a*b 
)*cos(f*x + e)^2 - a*b)*sin(f*x + e))/(a*b^2*f*cos(f*x + e)^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {3 \, {\left (f x + e\right )}}{a} - \frac {b \tan \left (f x + e\right )^{3} - 3 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )}{b^{2}} - \frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a b^{2}}}{3 \, f} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.46 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {f x + e}{a f} + \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{\sqrt {a b + b^{2}} a b^{2} f} + \frac {b^{2} f^{2} \tan \left (f x + e\right )^{3} - 3 \, a b f^{2} \tan \left (f x + e\right ) - 6 \, b^{2} f^{2} \tan \left (f x + e\right )}{3 \, b^{3} f^{3}} \] Input:

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

Output:

-(f*x + e)/(a*f) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*tan(f*x + e)/s 
qrt(a*b + b^2))/(sqrt(a*b + b^2)*a*b^2*f) + 1/3*(b^2*f^2*tan(f*x + e)^3 - 
3*a*b*f^2*tan(f*x + e) - 6*b^2*f^2*tan(f*x + e))/(b^3*f^3)
 

Mupad [B] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 1109, normalized size of antiderivative = 13.36 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

tan(e + f*x)^3/(3*b*f) - atan((40*a^2*tan(e + f*x))/(30*a*b + 40*a^2 + 10* 
b^2 + (30*a^3)/b + (12*a^4)/b^2 + (2*a^5)/b^3) + (30*a^3*tan(e + f*x))/(30 
*a*b^2 + 40*a^2*b + 30*a^3 + 10*b^3 + (12*a^4)/b + (2*a^5)/b^2) + (12*a^4* 
tan(e + f*x))/(30*a*b^3 + 30*a^3*b + 12*a^4 + 10*b^4 + 40*a^2*b^2 + (2*a^5 
)/b) + (2*a^5*tan(e + f*x))/(30*a*b^4 + 12*a^4*b + 2*a^5 + 10*b^5 + 40*a^2 
*b^3 + 30*a^3*b^2) + (10*b^2*tan(e + f*x))/(30*a*b + 40*a^2 + 10*b^2 + (30 
*a^3)/b + (12*a^4)/b^2 + (2*a^5)/b^3) + (30*a*b*tan(e + f*x))/(30*a*b + 40 
*a^2 + 10*b^2 + (30*a^3)/b + (12*a^4)/b^2 + (2*a^5)/b^3))/(a*f) - (tan(e + 
 f*x)*(a + 2*b))/(b^2*f) - (atan((((-b^5*(a + b)^5)^(1/2)*((2*tan(e + f*x) 
*(6*a*b^5 + 6*a^5*b + a^6 + 2*b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2)) 
/b^3 + ((-b^5*(a + b)^5)^(1/2)*((8*a^2*b^5 + 12*a^3*b^4 + 4*a^4*b^3)/b^3 + 
 (tan(e + f*x)*(8*a^2*b^6 + 4*a^3*b^5)*(-b^5*(a + b)^5)^(1/2))/(a*b^8)))/( 
2*a*b^5))*1i)/(2*a*b^5) + ((-b^5*(a + b)^5)^(1/2)*((2*tan(e + f*x)*(6*a*b^ 
5 + 6*a^5*b + a^6 + 2*b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2))/b^3 - ( 
(-b^5*(a + b)^5)^(1/2)*((8*a^2*b^5 + 12*a^3*b^4 + 4*a^4*b^3)/b^3 - (tan(e 
+ f*x)*(8*a^2*b^6 + 4*a^3*b^5)*(-b^5*(a + b)^5)^(1/2))/(a*b^8)))/(2*a*b^5) 
)*1i)/(2*a*b^5))/((2*(12*a*b^4 + 6*a^4*b + a^5 + 3*b^5 + 19*a^2*b^3 + 15*a 
^3*b^2))/b^3 - ((-b^5*(a + b)^5)^(1/2)*((2*tan(e + f*x)*(6*a*b^5 + 6*a^5*b 
 + a^6 + 2*b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2))/b^3 + ((-b^5*(a + 
b)^5)^(1/2)*((8*a^2*b^5 + 12*a^3*b^4 + 4*a^4*b^3)/b^3 + (tan(e + f*x)*(...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 654, normalized size of antiderivative = 7.88 \[ \int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*cos(e + f*x)*sin(e + f*x)**2*a**2 + 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a 
 + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e + f*x)*sin(e + f*x)**2*a* 
b + 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sq 
rt(b))*cos(e + f*x)*sin(e + f*x)**2*b**2 - 3*sqrt(b)*sqrt(a + b)*atan((sqr 
t(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e + f*x)*a**2 - 6*sqrt(b 
)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e 
 + f*x)*a*b - 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - s 
qrt(a))/sqrt(b))*cos(e + f*x)*b**2 + 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + 
b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*sin(e + f*x)**2*a**2 
+ 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt 
(b))*cos(e + f*x)*sin(e + f*x)**2*a*b + 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a 
 + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*sin(e + f*x)**2*b* 
*2 - 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/s 
qrt(b))*cos(e + f*x)*a**2 - 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
 + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*a*b - 3*sqrt(b)*sqrt(a + b)*at 
an((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*b**2 - 3 
*cos(e + f*x)*sin(e + f*x)**2*b**3*f*x + 3*cos(e + f*x)*b**3*f*x - 3*sin(e 
 + f*x)**3*a**2*b - 7*sin(e + f*x)**3*a*b**2 + 3*sin(e + f*x)*a**2*b + 6*s 
in(e + f*x)*a*b**2)/(3*cos(e + f*x)*a*b**3*f*(sin(e + f*x)**2 - 1))