\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx\) [736]

Optimal result

Integrand size = 41, antiderivative size = 99 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {5 A x}{16 a^3 c^3}+\frac {5 A \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac {5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac {\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f} \] Output:

5/16*A*x/a^3/c^3+5/16*A*cos(f*x+e)*sin(f*x+e)/a^3/c^3/f+5/24*A*cos(f*x+e)^ 
3*sin(f*x+e)/a^3/c^3/f-1/6*cos(f*x+e)^6*(B-A*tan(f*x+e))/a^3/c^3/f
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.64 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {-32 B \cos ^6(e+f x)+A (60 e+60 f x+45 \sin (2 (e+f x))+9 \sin (4 (e+f x))+\sin (6 (e+f x)))}{192 a^3 c^3 f} \] Input:

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + 
f*x])^3),x]
 

Output:

(-32*B*Cos[e + f*x]^6 + A*(60*e + 60*f*x + 45*Sin[2*(e + f*x)] + 9*Sin[4*( 
e + f*x)] + Sin[6*(e + f*x)]))/(192*a^3*c^3*f)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3042, 4071, 27, 82, 454, 215, 215, 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.

\(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^4 c^4 (1-i \tan (e+f x))^4 (i \tan (e+f x)+1)^4}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {A+B \tan (e+f x)}{(1-i \tan (e+f x))^4 (i \tan (e+f x)+1)^4}d\tan (e+f x)}{a^3 c^3 f}\)

\(\Big \downarrow \) 82

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

\(\Big \downarrow \) 454

\(\displaystyle \frac {\frac {5}{6} A \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)-\frac {B-A \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{a^3 c^3 f}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {5}{6} A \left (\frac {3}{4} \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {B-A \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{a^3 c^3 f}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {5}{6} A \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {B-A \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{a^3 c^3 f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {5}{6} A \left (\frac {3}{4} \left (\frac {1}{2} \arctan (\tan (e+f x))+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {\tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )-\frac {B-A \tan (e+f x)}{6 \left (\tan ^2(e+f x)+1\right )^3}}{a^3 c^3 f}\)

Input:

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^ 
3),x]
 

Output:

(-1/6*(B - A*Tan[e + f*x])/(1 + Tan[e + f*x]^2)^3 + (5*A*(Tan[e + f*x]/(4* 
(1 + Tan[e + f*x]^2)^2) + (3*(ArcTan[Tan[e + f*x]]/2 + Tan[e + f*x]/(2*(1 
+ Tan[e + f*x]^2))))/4))/6)/(a^3*c^3*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 82

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) 
)^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, 
 e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
 

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 454

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d 
 - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a 
*(p + 1)))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L 
tQ[p, -1] && NeQ[p, -3/2]
 

rule 3042

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

rule 4071

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a*(c/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x 
, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c 
+ a*d, 0] && EqQ[a^2 + b^2, 0]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.39

method	result	size
risch	\(\frac {5 A x}{16 a^{3} c^{3}}-\frac {B \cos \left (6 f x +6 e \right )}{192 f \,c^{3} a^{3}}+\frac {A \sin \left (6 f x +6 e \right )}{192 f \,c^{3} a^{3}}-\frac {B \cos \left (4 f x +4 e \right )}{32 f \,c^{3} a^{3}}+\frac {3 A \sin \left (4 f x +4 e \right )}{64 f \,c^{3} a^{3}}-\frac {5 B \cos \left (2 f x +2 e \right )}{64 f \,c^{3} a^{3}}+\frac {15 A \sin \left (2 f x +2 e \right )}{64 f \,c^{3} a^{3}}\)	\(138\)
norman	\(\frac {\frac {5 A x}{16 a c}-\frac {B}{6 a c f}+\frac {11 A \tan \left (f x +e \right )}{16 a c f}+\frac {5 A \tan \left (f x +e \right )^{3}}{6 a c f}+\frac {5 A \tan \left (f x +e \right )^{5}}{16 a c f}+\frac {15 A x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {15 A x \tan \left (f x +e \right )^{4}}{16 a c}+\frac {5 A x \tan \left (f x +e \right )^{6}}{16 a c}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a^{2} c^{2}}\)	\(155\)
derivativedivides	\(\frac {i B}{48 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{3} c^{3}}-\frac {A}{48 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B}{32 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{16 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {5 A}{32 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {i A}{16 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i B}{48 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {i B}{32 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {A}{48 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {5 A}{32 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}\)	\(303\)
default	\(\frac {i B}{48 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{3} c^{3}}-\frac {A}{48 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B}{32 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{16 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {5 A}{32 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )}+\frac {i A}{16 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i B}{48 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {i B}{32 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {A}{48 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {5 A}{32 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{3} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}\)	\(303\)

Input:

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^3,x,method=_R 
ETURNVERBOSE)
 

Output:

5/16*A*x/a^3/c^3-1/192*B/f/c^3/a^3*cos(6*f*x+6*e)+1/192*A/f/c^3/a^3*sin(6* 
f*x+6*e)-1/32*B/f/c^3/a^3*cos(4*f*x+4*e)+3/64*A/f/c^3/a^3*sin(4*f*x+4*e)-5 
/64*B/f/c^3/a^3*cos(2*f*x+2*e)+15/64*A/f/c^3/a^3*sin(2*f*x+2*e)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.27 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (120 \, A f x e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-i \, A - B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} - 3 \, {\left (3 i \, A + 2 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 15 \, {\left (3 i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 15 \, {\left (-3 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-3 i \, A + 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{3} f} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^3,x, al 
gorithm="fricas")
 

Output:

1/384*(120*A*f*x*e^(6*I*f*x + 6*I*e) + (-I*A - B)*e^(12*I*f*x + 12*I*e) - 
3*(3*I*A + 2*B)*e^(10*I*f*x + 10*I*e) - 15*(3*I*A + B)*e^(8*I*f*x + 8*I*e) 
 - 15*(-3*I*A + B)*e^(4*I*f*x + 4*I*e) - 3*(-3*I*A + 2*B)*e^(2*I*f*x + 2*I 
*e) + I*A - B)*e^(-6*I*f*x - 6*I*e)/(a^3*c^3*f)
 

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 508, normalized size of antiderivative = 5.13 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {5 A x}{16 a^{3} c^{3}} + \begin {cases} \frac {\left (\left (103079215104 i A a^{15} c^{15} f^{5} e^{6 i e} - 103079215104 B a^{15} c^{15} f^{5} e^{6 i e}\right ) e^{- 6 i f x} + \left (927712935936 i A a^{15} c^{15} f^{5} e^{8 i e} - 618475290624 B a^{15} c^{15} f^{5} e^{8 i e}\right ) e^{- 4 i f x} + \left (4638564679680 i A a^{15} c^{15} f^{5} e^{10 i e} - 1546188226560 B a^{15} c^{15} f^{5} e^{10 i e}\right ) e^{- 2 i f x} + \left (- 4638564679680 i A a^{15} c^{15} f^{5} e^{14 i e} - 1546188226560 B a^{15} c^{15} f^{5} e^{14 i e}\right ) e^{2 i f x} + \left (- 927712935936 i A a^{15} c^{15} f^{5} e^{16 i e} - 618475290624 B a^{15} c^{15} f^{5} e^{16 i e}\right ) e^{4 i f x} + \left (- 103079215104 i A a^{15} c^{15} f^{5} e^{18 i e} - 103079215104 B a^{15} c^{15} f^{5} e^{18 i e}\right ) e^{6 i f x}\right ) e^{- 12 i e}}{39582418599936 a^{18} c^{18} f^{6}} & \text {for}\: a^{18} c^{18} f^{6} e^{12 i e} \neq 0 \\x \left (- \frac {5 A}{16 a^{3} c^{3}} + \frac {\left (A e^{12 i e} + 6 A e^{10 i e} + 15 A e^{8 i e} + 20 A e^{6 i e} + 15 A e^{4 i e} + 6 A e^{2 i e} + A - i B e^{12 i e} - 4 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{4 i e} + 4 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{64 a^{3} c^{3}}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**3,x)
 

Output:

5*A*x/(16*a**3*c**3) + Piecewise((((103079215104*I*A*a**15*c**15*f**5*exp( 
6*I*e) - 103079215104*B*a**15*c**15*f**5*exp(6*I*e))*exp(-6*I*f*x) + (9277 
12935936*I*A*a**15*c**15*f**5*exp(8*I*e) - 618475290624*B*a**15*c**15*f**5 
*exp(8*I*e))*exp(-4*I*f*x) + (4638564679680*I*A*a**15*c**15*f**5*exp(10*I* 
e) - 1546188226560*B*a**15*c**15*f**5*exp(10*I*e))*exp(-2*I*f*x) + (-46385 
64679680*I*A*a**15*c**15*f**5*exp(14*I*e) - 1546188226560*B*a**15*c**15*f* 
*5*exp(14*I*e))*exp(2*I*f*x) + (-927712935936*I*A*a**15*c**15*f**5*exp(16* 
I*e) - 618475290624*B*a**15*c**15*f**5*exp(16*I*e))*exp(4*I*f*x) + (-10307 
9215104*I*A*a**15*c**15*f**5*exp(18*I*e) - 103079215104*B*a**15*c**15*f**5 
*exp(18*I*e))*exp(6*I*f*x))*exp(-12*I*e)/(39582418599936*a**18*c**18*f**6) 
, Ne(a**18*c**18*f**6*exp(12*I*e), 0)), (x*(-5*A/(16*a**3*c**3) + (A*exp(1 
2*I*e) + 6*A*exp(10*I*e) + 15*A*exp(8*I*e) + 20*A*exp(6*I*e) + 15*A*exp(4* 
I*e) + 6*A*exp(2*I*e) + A - I*B*exp(12*I*e) - 4*I*B*exp(10*I*e) - 5*I*B*ex 
p(8*I*e) + 5*I*B*exp(4*I*e) + 4*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(64*a**3 
*c**3)), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^3,x, al 
gorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {5 \, {\left (f x + e\right )} A}{16 \, a^{3} c^{3} f} + \frac {15 \, A \tan \left (f x + e\right )^{5} + 40 \, A \tan \left (f x + e\right )^{3} + 33 \, A \tan \left (f x + e\right ) - 8 \, B}{48 \, {\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3} c^{3} f} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^3,x, al 
gorithm="giac")
 

Output:

5/16*(f*x + e)*A/(a^3*c^3*f) + 1/48*(15*A*tan(f*x + e)^5 + 40*A*tan(f*x + 
e)^3 + 33*A*tan(f*x + e) - 8*B)/((tan(f*x + e)^2 + 1)^3*a^3*c^3*f)
 

Mupad [B] (verification not implemented)

Time = 5.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {5\,A\,x}{16\,a^3\,c^3}+\frac {{\cos \left (e+f\,x\right )}^6\,\left (\frac {5\,A\,{\mathrm {tan}\left (e+f\,x\right )}^5}{16}+\frac {5\,A\,{\mathrm {tan}\left (e+f\,x\right )}^3}{6}+\frac {11\,A\,\mathrm {tan}\left (e+f\,x\right )}{16}-\frac {B}{6}\right )}{a^3\,c^3\,f} \] Input:

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i 
)^3),x)
 

Output:

(5*A*x)/(16*a^3*c^3) + (cos(e + f*x)^6*((11*A*tan(e + f*x))/16 - B/6 + (5* 
A*tan(e + f*x)^3)/6 + (5*A*tan(e + f*x)^5)/16))/(a^3*c^3*f)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.54 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx=\frac {15 \tan \left (f x +e \right )^{6} a f x +8 \tan \left (f x +e \right )^{6} b +15 \tan \left (f x +e \right )^{5} a +45 \tan \left (f x +e \right )^{4} a f x +24 \tan \left (f x +e \right )^{4} b +40 \tan \left (f x +e \right )^{3} a +45 \tan \left (f x +e \right )^{2} a f x +24 \tan \left (f x +e \right )^{2} b +33 \tan \left (f x +e \right ) a +15 a f x}{48 a^{3} c^{3} f \left (\tan \left (f x +e \right )^{6}+3 \tan \left (f x +e \right )^{4}+3 \tan \left (f x +e \right )^{2}+1\right )} \] Input:

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^3,x)
 

Output:

(15*tan(e + f*x)**6*a*f*x + 8*tan(e + f*x)**6*b + 15*tan(e + f*x)**5*a + 4 
5*tan(e + f*x)**4*a*f*x + 24*tan(e + f*x)**4*b + 40*tan(e + f*x)**3*a + 45 
*tan(e + f*x)**2*a*f*x + 24*tan(e + f*x)**2*b + 33*tan(e + f*x)*a + 15*a*f 
*x)/(48*a**3*c**3*f*(tan(e + f*x)**6 + 3*tan(e + f*x)**4 + 3*tan(e + f*x)* 
*2 + 1))