\(\int \frac {\tan ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [369]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {x}{a^3}+\frac {\sqrt {a+b} \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.12 (sec) , antiderivative size = 523, normalized size of antiderivative = 3.56 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (\frac {2 \left (3 a^3-a^2 b+4 a b^2+8 b^3\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\sec (2 e) \left (8 b^2 \left (3 a^2+8 a b+8 b^2\right ) f x \cos (2 e)+16 a b^2 (a+2 b) f x \cos (2 f x)+4 a^2 b^2 f x \cos (2 (e+2 f x))+16 a^2 b^2 f x \cos (4 e+2 f x)+32 a b^3 f x \cos (4 e+2 f x)+4 a^2 b^2 f x \cos (6 e+4 f x)-9 a^4 \sin (2 e)-15 a^3 b \sin (2 e)+18 a^2 b^2 \sin (2 e)+72 a b^3 \sin (2 e)+48 b^4 \sin (2 e)+9 a^4 \sin (2 f x)+13 a^3 b \sin (2 f x)-28 a^2 b^2 \sin (2 f x)-32 a b^3 \sin (2 f x)+3 a^4 \sin (2 (e+2 f x))-3 a^3 b \sin (2 (e+2 f x))-6 a^2 b^2 \sin (2 (e+2 f x))-3 a^4 \sin (4 e+2 f x)+a^3 b \sin (4 e+2 f x)+20 a^2 b^2 \sin (4 e+2 f x)+16 a b^3 \sin (4 e+2 f x)\right )\right )}{128 a^3 b^2 f \left (a+b \sec ^2(e+f x)\right )^3} \] Input:

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

Output:

-1/128*((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*((2*(3*a^3 - a^2*b + 
 4*a*b^2 + 8*b^3)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Si 
n[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) 
]*(a + 2*b + a*Cos[2*(e + f*x)])^2*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*S 
qrt[b*(Cos[e] - I*Sin[e])^4]) + Sec[2*e]*(8*b^2*(3*a^2 + 8*a*b + 8*b^2)*f* 
x*Cos[2*e] + 16*a*b^2*(a + 2*b)*f*x*Cos[2*f*x] + 4*a^2*b^2*f*x*Cos[2*(e + 
2*f*x)] + 16*a^2*b^2*f*x*Cos[4*e + 2*f*x] + 32*a*b^3*f*x*Cos[4*e + 2*f*x] 
+ 4*a^2*b^2*f*x*Cos[6*e + 4*f*x] - 9*a^4*Sin[2*e] - 15*a^3*b*Sin[2*e] + 18 
*a^2*b^2*Sin[2*e] + 72*a*b^3*Sin[2*e] + 48*b^4*Sin[2*e] + 9*a^4*Sin[2*f*x] 
 + 13*a^3*b*Sin[2*f*x] - 28*a^2*b^2*Sin[2*f*x] - 32*a*b^3*Sin[2*f*x] + 3*a 
^4*Sin[2*(e + 2*f*x)] - 3*a^3*b*Sin[2*(e + 2*f*x)] - 6*a^2*b^2*Sin[2*(e + 
2*f*x)] - 3*a^4*Sin[4*e + 2*f*x] + a^3*b*Sin[4*e + 2*f*x] + 20*a^2*b^2*Sin 
[4*e + 2*f*x] + 16*a*b^3*Sin[4*e + 2*f*x])))/(a^3*b^2*f*(a + b*Sec[e + f*x 
]^2)^3)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.16, 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, 372, 440, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 440

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 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 440

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + 
 b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ 
g^2/(2*b*(b*c - a*d)*(p + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + 
d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c 
*f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && 
 LtQ[p, -1] && 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 = 15.48 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\frac {\left (a +b \right ) \left (\frac {-\frac {a \left (5 a -4 b \right ) \tan \left (f x +e \right )^{3}}{8 b}-\frac {a \left (3 a^{2}-a b -4 b^{2}\right ) \tan \left (f x +e \right )}{8 b^{2}}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b^{2} \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\)	\(134\)
default	\(\frac {\frac {\left (a +b \right ) \left (\frac {-\frac {a \left (5 a -4 b \right ) \tan \left (f x +e \right )^{3}}{8 b}-\frac {a \left (3 a^{2}-a b -4 b^{2}\right ) \tan \left (f x +e \right )}{8 b^{2}}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b^{2} \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\)	\(134\)
risch	\(-\frac {x}{a^{3}}+\frac {i \left (-3 a^{4} {\mathrm e}^{6 i \left (f x +e \right )}+a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+20 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-9 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}-15 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+18 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+72 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}-13 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}+28 a^{2} b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+32 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-3 a^{4}+3 b \,a^{3}+6 a^{2} b^{2}\right )}{4 a^{3} f \,b^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 b^{3} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 b^{2} f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 b^{3} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 b^{2} f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f \,a^{3}}\)	\(584\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (133) = 266\).

Time = 0.13 (sec) , antiderivative size = 664, normalized size of antiderivative = 4.52 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/32*(32*a^2*b^2*f*x*cos(f*x + e)^4 + 64*a*b^3*f*x*cos(f*x + e)^2 + 32*b 
^4*f*x - ((3*a^4 - 4*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 - 4*a*b 
^3 + 8*b^4 + 2*(3*a^3*b - 4*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)*sqrt(-(a + 
b)/b)*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))*sqrt(-(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^4 - a^3*b - 2*a^2*b^2)*cos(f*x + e)^3 + (5*a^3*b + a^2*b^2 - 
4*a*b^3)*cos(f*x + e))*sin(f*x + e))/(a^5*b^2*f*cos(f*x + e)^4 + 2*a^4*b^3 
*f*cos(f*x + e)^2 + a^3*b^4*f), -1/16*(16*a^2*b^2*f*x*cos(f*x + e)^4 + 32* 
a*b^3*f*x*cos(f*x + e)^2 + 16*b^4*f*x + ((3*a^4 - 4*a^3*b + 8*a^2*b^2)*cos 
(f*x + e)^4 + 3*a^2*b^2 - 4*a*b^3 + 8*b^4 + 2*(3*a^3*b - 4*a^2*b^2 + 8*a*b 
^3)*cos(f*x + e)^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))) + 2*(3*(a^4 - a^3 
*b - 2*a^2*b^2)*cos(f*x + e)^3 + (5*a^3*b + a^2*b^2 - 4*a*b^3)*cos(f*x + e 
))*sin(f*x + e))/(a^5*b^2*f*cos(f*x + e)^4 + 2*a^4*b^3*f*cos(f*x + e)^2 + 
a^3*b^4*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (5 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{3} + 2 \, a^{2} b - 5 \, a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{2} b^{4} \tan \left (f x + e\right )^{4} + a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4} + 2 \, {\left (a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}} + \frac {8 \, {\left (f x + e\right )}}{a^{3}} - \frac {{\left (3 \, a^{3} - a^{2} b + 4 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3} b^{2}}}{8 \, f} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {f x + e}{a^{3} f} + \frac {{\left (3 \, a^{3} - a^{2} b + 4 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{8 \, \sqrt {a b + b^{2}} a^{3} b^{2} f} - \frac {5 \, a^{2} b \tan \left (f x + e\right )^{3} + a b^{2} \tan \left (f x + e\right )^{3} - 4 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 2 \, a^{2} b \tan \left (f x + e\right ) - 5 \, a b^{2} \tan \left (f x + e\right ) - 4 \, b^{3} \tan \left (f x + e\right )}{8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{2} b^{2} f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 15.69 (sec) , antiderivative size = 615, normalized size of antiderivative = 4.18 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

- atan((25*tan(e + f*x))/(32*((5*b)/(4*a) - (3*a)/(16*b) + (9*a^2)/(32*b^2 
) + 25/32)) - (3*tan(e + f*x))/(16*((9*a)/(32*b) + (25*b)/(32*a) + (5*b^2) 
/(4*a^2) - 3/16)) + (9*tan(e + f*x))/(32*((25*b^2)/(32*a^2) - (3*b)/(16*a) 
 + (5*b^3)/(4*a^3) + 9/32)) + (5*tan(e + f*x))/(4*((25*a)/(32*b) - (3*a^2) 
/(16*b^2) + (9*a^3)/(32*b^3) + 5/4)))/(a^3*f) - ((tan(e + f*x)^3*(a*b + 5* 
a^2 - 4*b^2))/(8*a^2*b) - (tan(e + f*x)*(a + b)*(a*b - 3*a^2 + 4*b^2))/(8* 
a^2*b^2))/(f*(2*a*b + a^2 + b^2 + tan(e + f*x)^2*(2*a*b + 2*b^2) + b^2*tan 
(e + f*x)^4)) - (atanh((27*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(256*((27*a 
*b^2)/256 - (27*b^3)/128 + (171*b^4)/(256*a) - (7*b^5)/(64*a^2) + (5*b^6)/ 
(32*a^3) + (5*b^7)/(4*a^4))) - (81*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(25 
6*((27*a^2*b)/256 - (27*a*b^2)/128 + (171*b^3)/256 - (7*b^4)/(64*a) + (5*b 
^5)/(32*a^2) + (5*b^6)/(4*a^3))) - (35*tan(e + f*x)*(- a*b^5 - b^6)^(1/2)) 
/(32*((171*a^2*b)/256 - (7*a*b^2)/64 - (27*a^3)/128 + (5*b^3)/32 + (5*b^4) 
/(4*a) + (27*a^4)/(256*b))) + (5*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(4*(( 
5*a*b^2)/32 - (7*a^2*b)/64 + (171*a^3)/256 + (5*b^3)/4 - (27*a^4)/(128*b) 
+ (27*a^5)/(256*b^2))) + (63*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(64*((171 
*a*b^2)/256 - (27*a^2*b)/128 + (27*a^3)/256 - (7*b^3)/64 + (5*b^4)/(32*a) 
+ (5*b^5)/(4*a^2))))*(-b^5*(a + b))^(1/2)*(3*a^2 - 4*a*b + 8*b^2))/(8*a^3* 
b^5*f)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1259, normalized size of antiderivative = 8.56 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*sin(e + f*x)**4*a**4 - 4*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
+ f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b + 8*sqrt(b)*sqrt(a + 
b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4* 
a**2*b**2 - 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqr 
t(a))/sqrt(b))*sin(e + f*x)**2*a**4 + 2*sqrt(b)*sqrt(a + b)*atan((sqrt(a + 
 b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3*b - 8*sqrt(b 
)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e 
 + f*x)**2*a**2*b**2 - 16*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f 
*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**3 + 3*sqrt(b)*sqrt(a + b)* 
atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 + 2*sqrt(b)*sq 
rt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3*b + 
3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b 
))*a**2*b**2 + 12*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - 
 sqrt(a))/sqrt(b))*a*b**3 + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
 + f*x)/2) - sqrt(a))/sqrt(b))*b**4 + 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + 
 b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**4 - 4*sqrt(b)* 
sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + 
 f*x)**4*a**3*b + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) 
 + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b**2 - 6*sqrt(b)*sqrt(a + b)*...