\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx\) [768]

Optimal result

Integrand size = 43, antiderivative size = 141 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {(3 i A+B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a \sqrt {c} f}-\frac {3 i A+B}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \] Output:

1/8*(3*I*A+B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2 
)/a/c^(1/2)/f-1/4*(3*I*A+B)/a/f/(c-I*c*tan(f*x+e))^(1/2)+1/2*(I*A-B)/a/f/( 
1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 2.41 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.80 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\frac {\sqrt {2} (3 i A+B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {c}}+\frac {-2 A+6 i B+(-6 i A-2 B) \tan (e+f x)}{(-i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}}{8 a f} \] Input:

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

Output:

((Sqrt[2]*((3*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c 
])])/Sqrt[c] + (-2*A + (6*I)*B + ((-6*I)*A - 2*B)*Tan[e + f*x])/((-I + Tan 
[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]))/(8*a*f)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3042, 4071, 27, 87, 61, 73, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4071

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} c^{3/2}}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\right )}{a f}\)

Input:

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

Output:

(c*((I*A - B)/(2*c*(1 + I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]) + ((3* 
A - I*B)*((I*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[ 
2]*c^(3/2)) - I/(c*Sqrt[c - I*c*Tan[e + f*x]])))/4))/(a*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 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 = 0.57 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {3 A}{2}-\frac {i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}-\frac {-i B +A}{4 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f a}\)	\(121\)
default	\(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {3 A}{2}-\frac {i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}-\frac {-i B +A}{4 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f a}\)	\(121\)

Input:

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x,method= 
_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (110) = 220\).

Time = 0.11 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.52 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} a c f \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{2} c f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{2} c f^{2}}} + 3 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f}\right ) - \sqrt {\frac {1}{2}} a c f \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{2} c f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{2} c f^{2}}} - 3 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f}\right ) - \sqrt {2} {\left (2 \, {\left (i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, A + B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a c f} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, 
algorithm="fricas")
 

Output:

1/8*(sqrt(1/2)*a*c*f*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^2*c*f^2))*e^(2*I*f*x 
 + 2*I*e)*log(1/2*(sqrt(2)*sqrt(1/2)*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt( 
c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^2*c*f^2)) + 
3*I*A + B)*e^(-I*f*x - I*e)/(a*f)) - sqrt(1/2)*a*c*f*sqrt(-(9*A^2 - 6*I*A* 
B - B^2)/(a^2*c*f^2))*e^(2*I*f*x + 2*I*e)*log(-1/2*(sqrt(2)*sqrt(1/2)*(a*f 
*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 
 - 6*I*A*B - B^2)/(a^2*c*f^2)) - 3*I*A - B)*e^(-I*f*x - I*e)/(a*f)) - sqrt 
(2)*(2*(I*A + B)*e^(4*I*f*x + 4*I*e) - (-I*A - 3*B)*e^(2*I*f*x + 2*I*e) - 
I*A + B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*c*f)
 

Sympy [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=- \frac {i \left (\int \frac {A}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \, {\left (\frac {\sqrt {2} {\left (3 \, A - i \, B\right )} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (3 \, A - i \, B\right )} c - 4 \, {\left (A - i \, B\right )} c^{2}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a c}\right )}}{16 \, c f} \] Input:

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, 
algorithm="maxima")
 

Output:

-1/16*I*(sqrt(2)*(3*A - I*B)*sqrt(c)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan 
(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a + 4*((- 
I*c*tan(f*x + e) + c)*(3*A - I*B)*c - 4*(A - I*B)*c^2)/((-I*c*tan(f*x + e) 
 + c)^(3/2)*a - 2*sqrt(-I*c*tan(f*x + e) + c)*a*c))/(c*f)
 

Giac [F(-2)]

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

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),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 = 6.24 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.50 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {B\,c-\frac {B\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{4}}{a\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-2\,a\,c\,f\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}+\frac {\frac {A\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4\,a\,f}-\frac {A\,c\,1{}\mathrm {i}}{a\,f}}{2\,c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{8\,a\,\sqrt {-c}\,f}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{8\,a\,\sqrt {c}\,f} \] Input:

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

Output:

(B*c - (B*(c - c*tan(e + f*x)*1i))/4)/(a*f*(c - c*tan(e + f*x)*1i)^(3/2) - 
 2*a*c*f*(c - c*tan(e + f*x)*1i)^(1/2)) + ((A*(c - c*tan(e + f*x)*1i)*3i)/ 
(4*a*f) - (A*c*1i)/(a*f))/(2*c*(c - c*tan(e + f*x)*1i)^(1/2) - (c - c*tan( 
e + f*x)*1i)^(3/2)) - (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/ 
2))/(2*(-c)^(1/2)))*3i)/(8*a*(-c)^(1/2)*f) + (2^(1/2)*B*atanh((2^(1/2)*(c 
- c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(8*a*c^(1/2)*f)
 

Reduce [F]

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

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

Output:

(sqrt(c)*(2*sqrt( - tan(e + f*x)*i + 1)*a*i - 2*sqrt( - tan(e + f*x)*i + 1 
)*b + int(sqrt( - tan(e + f*x)*i + 1)/(tan(e + f*x)**3*i + tan(e + f*x)**2 
 + tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*a*f - int(sqrt( - tan(e + f*x)*i 
 + 1)/(tan(e + f*x)**3*i + tan(e + f*x)**2 + tan(e + f*x)*i + 1),x)*tan(e 
+ f*x)**2*b*f*i + int(sqrt( - tan(e + f*x)*i + 1)/(tan(e + f*x)**3*i + tan 
(e + f*x)**2 + tan(e + f*x)*i + 1),x)*a*f - int(sqrt( - tan(e + f*x)*i + 1 
)/(tan(e + f*x)**3*i + tan(e + f*x)**2 + tan(e + f*x)*i + 1),x)*b*f*i + 9* 
int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3*i + tan 
(e + f*x)**2 + tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*a*f - int((sqrt( - t 
an(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3*i + tan(e + f*x)**2 + 
 tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*b*f*i + 9*int((sqrt( - tan(e + f*x 
)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3*i + tan(e + f*x)**2 + tan(e + f 
*x)*i + 1),x)*a*f - int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan 
(e + f*x)**3*i + tan(e + f*x)**2 + tan(e + f*x)*i + 1),x)*b*f*i + 8*int((s 
qrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**3*i + tan(e + f*x) 
**2 + tan(e + f*x)*i + 1),x)*tan(e + f*x)**2*a*f*i + 8*int((sqrt( - tan(e 
+ f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**3*i + tan(e + f*x)**2 + tan(e + 
 f*x)*i + 1),x)*a*f*i))/(2*a*c*f*(tan(e + f*x)**2 + 1))