\(\int (a+b \sec ^2(e+f x))^2 \tan ^6(e+f x) \, dx\) [330]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-a^2 x+\frac {a^2 \tan (e+f x)}{f}-\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan ^5(e+f x)}{5 f}+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \] Output:

-a^2*x+a^2*tan(f*x+e)/f-1/3*a^2*tan(f*x+e)^3/f+1/5*a^2*tan(f*x+e)^5/f+1/7* 
b*(2*a+b)*tan(f*x+e)^7/f+1/9*b^2*tan(f*x+e)^9/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(275\) vs. \(2(95)=190\).

Time = 2.31 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.89 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {4 \left (b+a \cos ^2(e+f x)\right )^2 \sec ^9(e+f x) \left (315 a^2 f x \cos ^9(e+f x)-35 b^2 \sec (e) \sin (f x)-5 (18 a-19 b) b \cos ^2(e+f x) \sec (e) \sin (f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \cos ^4(e+f x) \sec (e) \sin (f x)+\left (231 a^2-270 a b+5 b^2\right ) \cos ^6(e+f x) \sec (e) \sin (f x)-\left (483 a^2-90 a b-10 b^2\right ) \cos ^8(e+f x) \sec (e) \sin (f x)-35 b^2 \cos (e+f x) \tan (e)-5 (18 a-19 b) b \cos ^3(e+f x) \tan (e)-3 \left (21 a^2-90 a b+25 b^2\right ) \cos ^5(e+f x) \tan (e)+\left (231 a^2-270 a b+5 b^2\right ) \cos ^7(e+f x) \tan (e)\right )}{315 f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:

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

Output:

(-4*(b + a*Cos[e + f*x]^2)^2*Sec[e + f*x]^9*(315*a^2*f*x*Cos[e + f*x]^9 - 
35*b^2*Sec[e]*Sin[f*x] - 5*(18*a - 19*b)*b*Cos[e + f*x]^2*Sec[e]*Sin[f*x] 
- 3*(21*a^2 - 90*a*b + 25*b^2)*Cos[e + f*x]^4*Sec[e]*Sin[f*x] + (231*a^2 - 
 270*a*b + 5*b^2)*Cos[e + f*x]^6*Sec[e]*Sin[f*x] - (483*a^2 - 90*a*b - 10* 
b^2)*Cos[e + f*x]^8*Sec[e]*Sin[f*x] - 35*b^2*Cos[e + f*x]*Tan[e] - 5*(18*a 
 - 19*b)*b*Cos[e + f*x]^3*Tan[e] - 3*(21*a^2 - 90*a*b + 25*b^2)*Cos[e + f* 
x]^5*Tan[e] + (231*a^2 - 270*a*b + 5*b^2)*Cos[e + f*x]^7*Tan[e]))/(315*f*( 
a + 2*b + a*Cos[2*(e + f*x)])^2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4629, 2075, 364, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 364

\(\displaystyle \frac {\int \left (b^2 \tan ^8(e+f x)+b (2 a+b) \tan ^6(e+f x)+a^2 \tan ^4(e+f x)-a^2 \tan ^2(e+f x)+a^2-\frac {a^2}{\tan ^2(e+f x)+1}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-a^2 \arctan (\tan (e+f x))+\frac {1}{5} a^2 \tan ^5(e+f x)-\frac {1}{3} a^2 \tan ^3(e+f x)+a^2 \tan (e+f x)+\frac {1}{7} b (2 a+b) \tan ^7(e+f x)+\frac {1}{9} b^2 \tan ^9(e+f x)}{f}\)

Input:

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

Output:

(-(a^2*ArcTan[Tan[e + f*x]]) + a^2*Tan[e + f*x] - (a^2*Tan[e + f*x]^3)/3 + 
 (a^2*Tan[e + f*x]^5)/5 + (b*(2*a + b)*Tan[e + f*x]^7)/7 + (b^2*Tan[e + f* 
x]^9)/9)/f
 

Defintions of rubi rules used

rule 364

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), 
x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x 
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In 
tegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 = 7.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93

method	result	size
parts	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{9}}{9}+\frac {\tan \left (f x +e \right )^{7}}{7}\right )}{f}+\frac {2 a b \tan \left (f x +e \right )^{7}}{7 f}\)	\(88\)
derivativedivides	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{7}}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{7}}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \sin \left (f x +e \right )^{7}}{63 \cos \left (f x +e \right )^{7}}\right )}{f}\)	\(105\)
default	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{7}}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{7}}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \sin \left (f x +e \right )^{7}}{63 \cos \left (f x +e \right )^{7}}\right )}{f}\)	\(105\)
risch	\(-a^{2} x -\frac {2 i \left (10 b^{2}-483 a^{2}+90 a b -28350 \,{\mathrm e}^{10 i \left (f x +e \right )} a^{2}+3150 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}-32508 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}-1890 \,{\mathrm e}^{8 i \left (f x +e \right )} b^{2}+1890 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-11718 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-270 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-3402 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+90 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-24402 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+6300 \,{\mathrm e}^{10 i \left (f x +e \right )} a b -5670 a^{2} {\mathrm e}^{14 i \left (f x +e \right )}-16170 \,{\mathrm e}^{12 i \left (f x +e \right )} a^{2}-1050 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}+5040 \,{\mathrm e}^{8 i \left (f x +e \right )} a b +3780 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+1980 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+180 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+630 a b \,{\mathrm e}^{16 i \left (f x +e \right )}+630 b^{2} {\mathrm e}^{14 i \left (f x +e \right )}-945 a^{2} {\mathrm e}^{16 i \left (f x +e \right )}+1260 a b \,{\mathrm e}^{14 i \left (f x +e \right )}+3780 a b \,{\mathrm e}^{12 i \left (f x +e \right )}\right )}{315 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{9}}\)	\(356\)

Input:

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

Output:

a^2/f*(1/5*tan(f*x+e)^5-1/3*tan(f*x+e)^3+tan(f*x+e)-arctan(tan(f*x+e)))+b^ 
2/f*(1/9*tan(f*x+e)^9+1/7*tan(f*x+e)^7)+2/7*a*b/f*tan(f*x+e)^7
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.44 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {315 \, a^{2} f x \cos \left (f x + e\right )^{9} - {\left ({\left (483 \, a^{2} - 90 \, a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{8} - {\left (231 \, a^{2} - 270 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (21 \, a^{2} - 90 \, a b + 25 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \, {\left (18 \, a b - 19 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \] Input:

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

Output:

-1/315*(315*a^2*f*x*cos(f*x + e)^9 - ((483*a^2 - 90*a*b - 10*b^2)*cos(f*x 
+ e)^8 - (231*a^2 - 270*a*b + 5*b^2)*cos(f*x + e)^6 + 3*(21*a^2 - 90*a*b + 
 25*b^2)*cos(f*x + e)^4 + 5*(18*a*b - 19*b^2)*cos(f*x + e)^2 + 35*b^2)*sin 
(f*x + e))/(f*cos(f*x + e)^9)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 45 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \] Input:

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

Output:

1/315*(35*b^2*tan(f*x + e)^9 + 45*(2*a*b + b^2)*tan(f*x + e)^7 + 63*a^2*ta 
n(f*x + e)^5 - 105*a^2*tan(f*x + e)^3 - 315*(f*x + e)*a^2 + 315*a^2*tan(f* 
x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.19 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {{\left (f x + e\right )} a^{2}}{f} + \frac {35 \, b^{2} f^{8} \tan \left (f x + e\right )^{9} + 90 \, a b f^{8} \tan \left (f x + e\right )^{7} + 45 \, b^{2} f^{8} \tan \left (f x + e\right )^{7} + 63 \, a^{2} f^{8} \tan \left (f x + e\right )^{5} - 105 \, a^{2} f^{8} \tan \left (f x + e\right )^{3} + 315 \, a^{2} f^{8} \tan \left (f x + e\right )}{315 \, f^{9}} \] Input:

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

Output:

-(f*x + e)*a^2/f + 1/315*(35*b^2*f^8*tan(f*x + e)^9 + 90*a*b*f^8*tan(f*x + 
 e)^7 + 45*b^2*f^8*tan(f*x + e)^7 + 63*a^2*f^8*tan(f*x + e)^5 - 105*a^2*f^ 
8*tan(f*x + e)^3 + 315*a^2*f^8*tan(f*x + e))/f^9
 

Mupad [B] (verification not implemented)

Time = 15.37 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {{\left (a+b\right )}^2}{3}+\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {{\left (a+b\right )}^2}{5}+\frac {b^2}{5}-\frac {2\,b\,\left (a+b\right )}{5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (\frac {b^2}{7}-\frac {2\,b\,\left (a+b\right )}{7}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9}{9}-a^2\,f\,x}{f} \] Input:

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

Output:

(tan(e + f*x)*((a + b)^2 + b^2 - 2*b*(a + b)) - tan(e + f*x)^3*((a + b)^2/ 
3 + b^2/3 - (2*b*(a + b))/3) + tan(e + f*x)^5*((a + b)^2/5 + b^2/5 - (2*b* 
(a + b))/5) - tan(e + f*x)^7*(b^2/7 - (2*b*(a + b))/7) + (b^2*tan(e + f*x) 
^9)/9 - a^2*f*x)/f
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 574, normalized size of antiderivative = 6.04 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {63 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{8} \tan \left (f x +e \right )^{5} a^{2}-105 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{8} \tan \left (f x +e \right )^{3} a^{2}+315 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{8} \tan \left (f x +e \right ) a^{2}-315 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{8} a^{2} f x -252 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} \tan \left (f x +e \right )^{5} a^{2}+420 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} \tan \left (f x +e \right )^{3} a^{2}-1260 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} \tan \left (f x +e \right ) a^{2}+1260 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} a^{2} f x +378 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right )^{5} a^{2}-630 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right )^{3} a^{2}+1890 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right ) a^{2}-1890 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{2} f x -252 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )^{5} a^{2}+420 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )^{3} a^{2}-1260 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right ) a^{2}+1260 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2} f x +63 \cos \left (f x +e \right ) \tan \left (f x +e \right )^{5} a^{2}-105 \cos \left (f x +e \right ) \tan \left (f x +e \right )^{3} a^{2}+315 \cos \left (f x +e \right ) \tan \left (f x +e \right ) a^{2}-315 \cos \left (f x +e \right ) a^{2} f x -90 \sin \left (f x +e \right )^{9} a b -10 \sin \left (f x +e \right )^{9} b^{2}+90 \sin \left (f x +e \right )^{7} a b +45 \sin \left (f x +e \right )^{7} b^{2}}{315 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{8}-4 \sin \left (f x +e \right )^{6}+6 \sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{2}+1\right )} \] Input:

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

Output:

(63*cos(e + f*x)*sin(e + f*x)**8*tan(e + f*x)**5*a**2 - 105*cos(e + f*x)*s 
in(e + f*x)**8*tan(e + f*x)**3*a**2 + 315*cos(e + f*x)*sin(e + f*x)**8*tan 
(e + f*x)*a**2 - 315*cos(e + f*x)*sin(e + f*x)**8*a**2*f*x - 252*cos(e + f 
*x)*sin(e + f*x)**6*tan(e + f*x)**5*a**2 + 420*cos(e + f*x)*sin(e + f*x)** 
6*tan(e + f*x)**3*a**2 - 1260*cos(e + f*x)*sin(e + f*x)**6*tan(e + f*x)*a* 
*2 + 1260*cos(e + f*x)*sin(e + f*x)**6*a**2*f*x + 378*cos(e + f*x)*sin(e + 
 f*x)**4*tan(e + f*x)**5*a**2 - 630*cos(e + f*x)*sin(e + f*x)**4*tan(e + f 
*x)**3*a**2 + 1890*cos(e + f*x)*sin(e + f*x)**4*tan(e + f*x)*a**2 - 1890*c 
os(e + f*x)*sin(e + f*x)**4*a**2*f*x - 252*cos(e + f*x)*sin(e + f*x)**2*ta 
n(e + f*x)**5*a**2 + 420*cos(e + f*x)*sin(e + f*x)**2*tan(e + f*x)**3*a**2 
 - 1260*cos(e + f*x)*sin(e + f*x)**2*tan(e + f*x)*a**2 + 1260*cos(e + f*x) 
*sin(e + f*x)**2*a**2*f*x + 63*cos(e + f*x)*tan(e + f*x)**5*a**2 - 105*cos 
(e + f*x)*tan(e + f*x)**3*a**2 + 315*cos(e + f*x)*tan(e + f*x)*a**2 - 315* 
cos(e + f*x)*a**2*f*x - 90*sin(e + f*x)**9*a*b - 10*sin(e + f*x)**9*b**2 + 
 90*sin(e + f*x)**7*a*b + 45*sin(e + f*x)**7*b**2)/(315*cos(e + f*x)*f*(si 
n(e + f*x)**8 - 4*sin(e + f*x)**6 + 6*sin(e + f*x)**4 - 4*sin(e + f*x)**2 
+ 1))