\(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) [763]

Optimal result

Integrand size = 43, antiderivative size = 142 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {8 a^3 (i A+B)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {8 a^3 (i A+2 B)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^3 (i A+5 B)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B}{c^3 f \sqrt {c-i c \tan (e+f x)}} \] Output:

-8/7*a^3*(I*A+B)/f/(c-I*c*tan(f*x+e))^(7/2)+8/5*a^3*(I*A+2*B)/c/f/(c-I*c*t 
an(f*x+e))^(5/2)-2/3*a^3*(I*A+5*B)/c^2/f/(c-I*c*tan(f*x+e))^(3/2)+2*a^3*B/ 
c^3/f/(c-I*c*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 5.85 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.67 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\frac {2 a^3 \left (-11 A-38 i B+(-14 i A-133 B) \tan (e+f x)+35 (A+4 i B) \tan ^2(e+f x)+105 B \tan ^3(e+f x)\right )}{105 c^3 f (i+\tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + 
 f*x])^(7/2),x]
 

Output:

(2*a^3*(-11*A - (38*I)*B + ((-14*I)*A - 133*B)*Tan[e + f*x] + 35*(A + (4*I 
)*B)*Tan[e + f*x]^2 + 105*B*Tan[e + f*x]^3))/(105*c^3*f*(I + Tan[e + f*x]) 
^3*Sqrt[c - I*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3042, 4071, 27, 86, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {a^2 (i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^3 c \int \frac {(i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {a^3 c \int \left (\frac {4 (A-i B)}{(c-i c \tan (e+f x))^{9/2}}+\frac {i B}{c^3 (c-i c \tan (e+f x))^{3/2}}+\frac {A-5 i B}{c^2 (c-i c \tan (e+f x))^{5/2}}-\frac {4 (A-2 i B)}{c (c-i c \tan (e+f x))^{7/2}}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 c \left (-\frac {2 (5 B+i A)}{3 c^3 (c-i c \tan (e+f x))^{3/2}}+\frac {8 (2 B+i A)}{5 c^2 (c-i c \tan (e+f x))^{5/2}}-\frac {8 (B+i A)}{7 c (c-i c \tan (e+f x))^{7/2}}+\frac {2 B}{c^4 \sqrt {c-i c \tan (e+f x)}}\right )}{f}\)

Input:

Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x]) 
^(7/2),x]
 

Output:

(a^3*c*((-8*(I*A + B))/(7*c*(c - I*c*Tan[e + f*x])^(7/2)) + (8*(I*A + 2*B) 
)/(5*c^2*(c - I*c*Tan[e + f*x])^(5/2)) - (2*(I*A + 5*B))/(3*c^3*(c - I*c*T 
an[e + f*x])^(3/2)) + (2*B)/(c^4*Sqrt[c - I*c*Tan[e + f*x]])))/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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 2009

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

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 = 0.67 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (-\frac {c \left (-5 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (-2 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {4 c^{3} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\right )}{f \,c^{3}}\)	\(105\)
default	\(\frac {2 i a^{3} \left (-\frac {c \left (-5 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (-2 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {4 c^{3} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\right )}{f \,c^{3}}\)	\(105\)
risch	\(-\frac {a^{3} \left (15 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+15 B \,{\mathrm e}^{6 i \left (f x +e \right )}+3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-39 B \,{\mathrm e}^{4 i \left (f x +e \right )}-4 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+52 B \,{\mathrm e}^{2 i \left (f x +e \right )}+8 i A -104 B \right ) \sqrt {2}}{210 c^{3} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(115\)
parts	\(\frac {2 i A \,a^{3} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {9}{2}}}-\frac {1}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{20 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {1}{14 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\right )}{f}+\frac {a^{3} \left (3 i A +B \right ) \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 c^{\frac {7}{2}}}-\frac {1}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {1}{8 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{10 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}+\frac {2 B \,a^{3} \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}+\frac {15}{16 \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {17 c}{24 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {7 c^{2}}{20 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {c^{3}}{14 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\right )}{f \,c^{3}}-\frac {6 i a^{3} \left (-i B +A \right ) \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {5}{2}}}-\frac {3}{20 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{16 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{24 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {c}{14 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\right )}{f c}-\frac {2 a^{3} \left (i A +3 B \right ) \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {3}{2}}}+\frac {7}{24 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{16 c \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {c}{4 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {c^{2}}{14 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}\right )}{f \,c^{2}}\)	\(597\)

Input:

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

2*I/f*a^3/c^3*(-1/3*c*(A-5*I*B)/(c-I*c*tan(f*x+e))^(3/2)+4/5*c^2*(A-2*I*B) 
/(c-I*c*tan(f*x+e))^(5/2)-I*B/(c-I*c*tan(f*x+e))^(1/2)-4/7*c^3*(A-I*B)/(c- 
I*c*tan(f*x+e))^(7/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {\sqrt {2} {\left (15 \, {\left (i \, A + B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} + 6 \, {\left (3 i \, A - 4 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (i \, A - 13 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, {\left (i \, A - 13 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, A - 13 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{210 \, c^{4} f} \] Input:

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x 
, algorithm="fricas")
 

Output:

-1/210*sqrt(2)*(15*(I*A + B)*a^3*e^(8*I*f*x + 8*I*e) + 6*(3*I*A - 4*B)*a^3 
*e^(6*I*f*x + 6*I*e) - (I*A - 13*B)*a^3*e^(4*I*f*x + 4*I*e) + 4*(I*A - 13* 
B)*a^3*e^(2*I*f*x + 2*I*e) + 8*(I*A - 13*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I* 
e) + 1))/(c^4*f)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx =\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(7/2) 
,x)
 

Output:

-I*a**3*(Integral(I*A/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 
- 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c* 
tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + I 
ntegral(-3*A*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) 
**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(- 
I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) 
 + Integral(A*tan(e + f*x)**3/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + 
f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sq 
rt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)) 
, x) + Integral(-3*B*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*t 
an(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I* 
c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x 
) + c)), x) + Integral(B*tan(e + f*x)**4/(I*c**3*sqrt(-I*c*tan(e + f*x) + 
c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 
3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + 
 f*x) + c)), x) + Integral(-3*I*A*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e 
+ f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f 
*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I* 
c*tan(e + f*x) + c)), x) + Integral(I*B*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan 
(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {2 i \, {\left (105 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} B a^{3} + 35 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (A - 5 i \, B\right )} a^{3} c - 84 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 2 i \, B\right )} a^{3} c^{2} + 60 \, {\left (A - i \, B\right )} a^{3} c^{3}\right )}}{105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} c^{3} f} \] Input:

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x 
, algorithm="maxima")
 

Output:

-2/105*I*(105*I*(-I*c*tan(f*x + e) + c)^3*B*a^3 + 35*(-I*c*tan(f*x + e) + 
c)^2*(A - 5*I*B)*a^3*c - 84*(-I*c*tan(f*x + e) + c)*(A - 2*I*B)*a^3*c^2 + 
60*(A - I*B)*a^3*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*c^3*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x 
, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 6.98 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.13 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (\frac {a^3\,\left (A+B\,13{}\mathrm {i}\right )\,4{}\mathrm {i}}{105\,c^4\,f}+\frac {a^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (3\,A+B\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{35\,c^4\,f}+\frac {a^3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (A+B\,13{}\mathrm {i}\right )\,2{}\mathrm {i}}{105\,c^4\,f}+\frac {a^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{14\,c^4\,f}-\frac {a^3\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (A+B\,13{}\mathrm {i}\right )\,1{}\mathrm {i}}{210\,c^4\,f}\right ) \] Input:

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1 
i)^(7/2),x)
 

Output:

-(c - (c*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^3*(A + B*13i)*4i)/(105*c 
^4*f) + (a^3*exp(e*6i + f*x*6i)*(3*A + B*4i)*1i)/(35*c^4*f) + (a^3*exp(e*2 
i + f*x*2i)*(A + B*13i)*2i)/(105*c^4*f) + (a^3*exp(e*8i + f*x*8i)*(A - B*1 
i)*1i)/(14*c^4*f) - (a^3*exp(e*4i + f*x*4i)*(A + B*13i)*1i)/(210*c^4*f))
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\text {too large to display} \] Input:

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x)
 

Output:

(sqrt(c)*a**3*(66*sqrt( - tan(e + f*x)*i + 1)*a*i - 16*sqrt( - tan(e + f*x 
)*i + 1)*b - 12*int(( - sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**5*i - 
3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i - 2*tan(e + f*x)**2 - 3*tan(e + f* 
x)*i + 1),x)*tan(e + f*x)**2*a*f + 18*int(( - sqrt( - tan(e + f*x)*i + 1)) 
/(tan(e + f*x)**5*i - 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i - 2*tan(e + 
f*x)**2 - 3*tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*b*f*i - 12*int(( - sqrt 
( - tan(e + f*x)*i + 1))/(tan(e + f*x)**5*i - 3*tan(e + f*x)**4 - 2*tan(e 
+ f*x)**3*i - 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*a*f + 18*int(( 
- sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**5*i - 3*tan(e + f*x)**4 - 2* 
tan(e + f*x)**3*i - 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*b*f*i - 3 
6*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**5*i 
- 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i - 2*tan(e + f*x)**2 - 3*tan(e + 
f*x)*i + 1),x)*tan(e + f*x)**2*a*f + 54*int(( - sqrt( - tan(e + f*x)*i + 1 
)*tan(e + f*x)**4)/(tan(e + f*x)**5*i - 3*tan(e + f*x)**4 - 2*tan(e + f*x) 
**3*i - 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*b*f*i 
 - 36*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)** 
5*i - 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i - 2*tan(e + f*x)**2 - 3*tan( 
e + f*x)*i + 1),x)*a*f + 54*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f* 
x)**4)/(tan(e + f*x)**5*i - 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i - 2*ta 
n(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*b*f*i + 144*int(( - sqrt( - ta...