\(\int (a+b \sec ^2(e+f x))^2 \tan ^4(e+f x) \, dx\) [331]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=a^2 x-\frac {a^2 \tan (e+f x)}{f}+\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(395\) vs. \(2(77)=154\).

Time = 1.47 (sec) , antiderivative size = 395, normalized size of antiderivative = 5.13 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {\sec (e) \sec ^7(e+f x) \left (3675 a^2 f x \cos (f x)+3675 a^2 f x \cos (2 e+f x)+2205 a^2 f x \cos (2 e+3 f x)+2205 a^2 f x \cos (4 e+3 f x)+735 a^2 f x \cos (4 e+5 f x)+735 a^2 f x \cos (6 e+5 f x)+105 a^2 f x \cos (6 e+7 f x)+105 a^2 f x \cos (8 e+7 f x)-5320 a^2 \sin (f x)+1680 a b \sin (f x)+840 b^2 \sin (f x)+4480 a^2 \sin (2 e+f x)-1260 a b \sin (2 e+f x)+420 b^2 \sin (2 e+f x)-3780 a^2 \sin (2 e+3 f x)+924 a b \sin (2 e+3 f x)-168 b^2 \sin (2 e+3 f x)+2100 a^2 \sin (4 e+3 f x)-840 a b \sin (4 e+3 f x)-420 b^2 \sin (4 e+3 f x)-1540 a^2 \sin (4 e+5 f x)+168 a b \sin (4 e+5 f x)+84 b^2 \sin (4 e+5 f x)+420 a^2 \sin (6 e+5 f x)-420 a b \sin (6 e+5 f x)-280 a^2 \sin (6 e+7 f x)+84 a b \sin (6 e+7 f x)+12 b^2 \sin (6 e+7 f x)\right )}{13440 f} \] Input:

Integrate[(a + b*Sec[e + f*x]^2)^2*Tan[e + f*x]^4,x]
 

Output:

(Sec[e]*Sec[e + f*x]^7*(3675*a^2*f*x*Cos[f*x] + 3675*a^2*f*x*Cos[2*e + f*x 
] + 2205*a^2*f*x*Cos[2*e + 3*f*x] + 2205*a^2*f*x*Cos[4*e + 3*f*x] + 735*a^ 
2*f*x*Cos[4*e + 5*f*x] + 735*a^2*f*x*Cos[6*e + 5*f*x] + 105*a^2*f*x*Cos[6* 
e + 7*f*x] + 105*a^2*f*x*Cos[8*e + 7*f*x] - 5320*a^2*Sin[f*x] + 1680*a*b*S 
in[f*x] + 840*b^2*Sin[f*x] + 4480*a^2*Sin[2*e + f*x] - 1260*a*b*Sin[2*e + 
f*x] + 420*b^2*Sin[2*e + f*x] - 3780*a^2*Sin[2*e + 3*f*x] + 924*a*b*Sin[2* 
e + 3*f*x] - 168*b^2*Sin[2*e + 3*f*x] + 2100*a^2*Sin[4*e + 3*f*x] - 840*a* 
b*Sin[4*e + 3*f*x] - 420*b^2*Sin[4*e + 3*f*x] - 1540*a^2*Sin[4*e + 5*f*x] 
+ 168*a*b*Sin[4*e + 5*f*x] + 84*b^2*Sin[4*e + 5*f*x] + 420*a^2*Sin[6*e + 5 
*f*x] - 420*a*b*Sin[6*e + 5*f*x] - 280*a^2*Sin[6*e + 7*f*x] + 84*a*b*Sin[6 
*e + 7*f*x] + 12*b^2*Sin[6*e + 7*f*x]))/(13440*f)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97, 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.

\(\displaystyle \int \tan ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^2dx\)

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 364

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*Sec[e + f*x]^2)^2*Tan[e + f*x]^4,x]
 

Output:

(a^2*ArcTan[Tan[e + f*x]] - a^2*Tan[e + f*x] + (a^2*Tan[e + f*x]^3)/3 + (b 
*(2*a + b)*Tan[e + f*x]^5)/5 + (b^2*Tan[e + f*x]^7)/7)/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 = 4.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01

method	result	size
parts	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{7}}{7}+\frac {\tan \left (f x +e \right )^{5}}{5}\right )}{f}+\frac {2 a b \tan \left (f x +e \right )^{5}}{5 f}\)	\(78\)
derivativedivides	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {2 a b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{5}}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \sin \left (f x +e \right )^{5}}{35 \cos \left (f x +e \right )^{5}}\right )}{f}\)	\(94\)
default	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {2 a b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{5}}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \sin \left (f x +e \right )^{5}}{35 \cos \left (f x +e \right )^{5}}\right )}{f}\)	\(94\)
risch	\(a^{2} x +\frac {4 i \left (-105 \,{\mathrm e}^{12 i \left (f x +e \right )} a^{2}+105 a b \,{\mathrm e}^{12 i \left (f x +e \right )}-525 \,{\mathrm e}^{10 i \left (f x +e \right )} a^{2}+210 \,{\mathrm e}^{10 i \left (f x +e \right )} a b +105 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}-1120 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+315 \,{\mathrm e}^{8 i \left (f x +e \right )} a b -105 \,{\mathrm e}^{8 i \left (f x +e \right )} b^{2}-1330 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+420 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+210 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-945 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+231 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-42 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-385 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+42 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+21 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-70 a^{2}+21 a b +3 b^{2}\right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\)	\(273\)

Input:

int((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.47 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {105 \, a^{2} f x \cos \left (f x + e\right )^{7} - {\left (2 \, {\left (70 \, a^{2} - 21 \, a b - 3 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - {\left (35 \, a^{2} - 84 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 6 \, {\left (7 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, b^{2}\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \] Input:

integrate((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^4,x, algorithm="fricas")
 

Output:

1/105*(105*a^2*f*x*cos(f*x + e)^7 - (2*(70*a^2 - 21*a*b - 3*b^2)*cos(f*x + 
 e)^6 - (35*a^2 - 84*a*b + 3*b^2)*cos(f*x + e)^4 - 6*(7*a*b - 4*b^2)*cos(f 
*x + e)^2 - 15*b^2)*sin(f*x + e))/(f*cos(f*x + e)^7)
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate((a+b*sec(f*x+e)**2)**2*tan(f*x+e)**4,x)
 

Output:

Integral((a + b*sec(e + f*x)**2)**2*tan(e + f*x)**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 21 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, a^{2} \tan \left (f x + e\right )^{3} + 105 \, {\left (f x + e\right )} a^{2} - 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \] Input:

integrate((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^4,x, algorithm="maxima")
 

Output:

1/105*(15*b^2*tan(f*x + e)^7 + 21*(2*a*b + b^2)*tan(f*x + e)^5 + 35*a^2*ta 
n(f*x + e)^3 + 105*(f*x + e)*a^2 - 105*a^2*tan(f*x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {{\left (f x + e\right )} a^{2}}{f} + \frac {15 \, b^{2} f^{6} \tan \left (f x + e\right )^{7} + 42 \, a b f^{6} \tan \left (f x + e\right )^{5} + 21 \, b^{2} f^{6} \tan \left (f x + e\right )^{5} + 35 \, a^{2} f^{6} \tan \left (f x + e\right )^{3} - 105 \, a^{2} f^{6} \tan \left (f x + e\right )}{105 \, f^{7}} \] Input:

integrate((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^4,x, algorithm="giac")
 

Output:

(f*x + e)*a^2/f + 1/105*(15*b^2*f^6*tan(f*x + e)^7 + 42*a*b*f^6*tan(f*x + 
e)^5 + 21*b^2*f^6*tan(f*x + e)^5 + 35*a^2*f^6*tan(f*x + e)^3 - 105*a^2*f^6 
*tan(f*x + e))/f^7
 

Mupad [B] (verification not implemented)

Time = 15.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.26 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {{\left (a+b\right )}^2}{3}+\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {b^2}{5}-\frac {2\,b\,\left (a+b\right )}{5}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^7}{7}+a^2\,f\,x}{f} \] Input:

int(tan(e + f*x)^4*(a + b/cos(e + f*x)^2)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 364, normalized size of antiderivative = 4.73 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {35 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} \tan \left (f x +e \right )^{3} a^{2}-105 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} \tan \left (f x +e \right ) a^{2}+105 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} a^{2} f x -105 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right )^{3} a^{2}+315 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right ) a^{2}-315 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{2} f x +105 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )^{3} a^{2}-315 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right ) a^{2}+315 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2} f x -35 \cos \left (f x +e \right ) \tan \left (f x +e \right )^{3} a^{2}+105 \cos \left (f x +e \right ) \tan \left (f x +e \right ) a^{2}-105 \cos \left (f x +e \right ) a^{2} f x +42 \sin \left (f x +e \right )^{7} a b +6 \sin \left (f x +e \right )^{7} b^{2}-42 \sin \left (f x +e \right )^{5} a b -21 \sin \left (f x +e \right )^{5} b^{2}}{105 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1\right )} \] Input:

int((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^4,x)
 

Output:

(35*cos(e + f*x)*sin(e + f*x)**6*tan(e + f*x)**3*a**2 - 105*cos(e + f*x)*s 
in(e + f*x)**6*tan(e + f*x)*a**2 + 105*cos(e + f*x)*sin(e + f*x)**6*a**2*f 
*x - 105*cos(e + f*x)*sin(e + f*x)**4*tan(e + f*x)**3*a**2 + 315*cos(e + f 
*x)*sin(e + f*x)**4*tan(e + f*x)*a**2 - 315*cos(e + f*x)*sin(e + f*x)**4*a 
**2*f*x + 105*cos(e + f*x)*sin(e + f*x)**2*tan(e + f*x)**3*a**2 - 315*cos( 
e + f*x)*sin(e + f*x)**2*tan(e + f*x)*a**2 + 315*cos(e + f*x)*sin(e + f*x) 
**2*a**2*f*x - 35*cos(e + f*x)*tan(e + f*x)**3*a**2 + 105*cos(e + f*x)*tan 
(e + f*x)*a**2 - 105*cos(e + f*x)*a**2*f*x + 42*sin(e + f*x)**7*a*b + 6*si 
n(e + f*x)**7*b**2 - 42*sin(e + f*x)**5*a*b - 21*sin(e + f*x)**5*b**2)/(10 
5*cos(e + f*x)*f*(sin(e + f*x)**6 - 3*sin(e + f*x)**4 + 3*sin(e + f*x)**2 
- 1))