\(\int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} (1+x^2)} \, dx\) [140]

Optimal result

Integrand size = 43, antiderivative size = 135 \[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\frac {\sqrt {2} \sqrt {A} \sqrt {B} \arctan \left (\frac {A}{B}+\frac {\sqrt {A} \sqrt {e \left (\frac {A}{B}-\frac {B}{A}+2 x\right )}}{\sqrt {B} \sqrt {e}}\right )}{\sqrt {e}}-\frac {\sqrt {2} \sqrt {A} \sqrt {B} \arctan \left (\frac {A}{B}-\frac {\sqrt {A} \sqrt {\left (\frac {A}{B}-\frac {B}{A}\right ) e+2 e x}}{\sqrt {B} \sqrt {e}}\right )}{\sqrt {e}} \] Output:

2^(1/2)*A^(1/2)*B^(1/2)*arctan(A/B+A^(1/2)*(e*(A/B-B/A+2*x))^(1/2)/B^(1/2) 
/e^(1/2))/e^(1/2)-2^(1/2)*A^(1/2)*B^(1/2)*arctan(A/B-A^(1/2)*((A/B-B/A)*e+ 
2*e*x)^(1/2)/B^(1/2)/e^(1/2))/e^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=-\frac {i \sqrt {2} \sqrt {A} \sqrt {B} \sqrt {\frac {A}{B}-\frac {B}{A}+2 x} \left (\text {arctanh}\left (\frac {\sqrt {A} \sqrt {B} \sqrt {\frac {A}{B}-\frac {B}{A}+2 x}}{A-i B}\right )-\text {arctanh}\left (\frac {\sqrt {A} \sqrt {B} \sqrt {\frac {A}{B}-\frac {B}{A}+2 x}}{A+i B}\right )\right )}{\sqrt {e \left (\frac {A}{B}-\frac {B}{A}+2 x\right )}} \] Input:

Integrate[(A + B*x)/(Sqrt[(A^2*e - B^2*e)/(2*A*B) + e*x]*(1 + x^2)),x]
 

Output:

((-I)*Sqrt[2]*Sqrt[A]*Sqrt[B]*Sqrt[A/B - B/A + 2*x]*(ArcTanh[(Sqrt[A]*Sqrt 
[B]*Sqrt[A/B - B/A + 2*x])/(A - I*B)] - ArcTanh[(Sqrt[A]*Sqrt[B]*Sqrt[A/B 
- B/A + 2*x])/(A + I*B)]))/Sqrt[e*(A/B - B/A + 2*x)]
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.46, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {654, 27, 1475, 1083, 217}

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.

\(\displaystyle \int \frac {A+B x}{\left (x^2+1\right ) \sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x}} \, dx\)

\(\Big \downarrow \) 654

\(\displaystyle 2 \int \frac {2 \left (\left (A^2+B^2\right ) e+2 A B \left (\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e\right )\right )}{A \left (\frac {\left (A^2+B^2\right )^2 e^2}{A^2 B^2}-\frac {4 \left (A^2-B^2\right ) \left (\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e\right ) e}{A B}+4 \left (\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e\right )^2\right )}d\sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \int \frac {\left (A^2+B^2\right ) e+2 A B \left (\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e\right )}{\frac {\left (A^2+B^2\right )^2 e^2}{A^2 B^2}-\frac {4 \left (A^2-B^2\right ) \left (\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e\right ) e}{A B}+4 \left (\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e\right )^2}d\sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e}}{A}\)

\(\Big \downarrow \) 1475

\(\displaystyle \frac {4 \left (\frac {1}{4} A B \int \frac {1}{\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+\frac {\left (A^2+B^2\right ) e}{2 A B}+x e-\frac {\sqrt {2} \sqrt {A} \sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e} \sqrt {e}}{\sqrt {B}}}d\sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e}+\frac {1}{4} A B \int \frac {1}{\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+\frac {\left (A^2+B^2\right ) e}{2 A B}+x e+\frac {\sqrt {2} \sqrt {A} \sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e} \sqrt {e}}{\sqrt {B}}}d\sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e}\right )}{A}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {4 \left (-\frac {1}{2} A B \int \frac {1}{-\frac {2 B e}{A}-\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e-x e}d\left (2 \sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e}-\frac {\sqrt {2} \sqrt {A} \sqrt {e}}{\sqrt {B}}\right )-\frac {1}{2} A B \int \frac {1}{-\frac {2 B e}{A}-\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e-x e}d\left (\frac {\sqrt {2} \sqrt {A} \sqrt {e}}{\sqrt {B}}+2 \sqrt {\frac {1}{2} \left (\frac {A}{B}-\frac {B}{A}\right ) e+x e}\right )\right )}{A}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {4 \left (\frac {A^{3/2} \sqrt {B} \arctan \left (\frac {\sqrt {A} \left (2 \sqrt {\frac {1}{2} e \left (\frac {A}{B}-\frac {B}{A}\right )+e x}-\frac {\sqrt {2} \sqrt {A} \sqrt {e}}{\sqrt {B}}\right )}{\sqrt {2} \sqrt {B} \sqrt {e}}\right )}{2 \sqrt {2} \sqrt {e}}+\frac {A^{3/2} \sqrt {B} \arctan \left (\frac {\sqrt {A} \left (2 \sqrt {\frac {1}{2} e \left (\frac {A}{B}-\frac {B}{A}\right )+e x}+\frac {\sqrt {2} \sqrt {A} \sqrt {e}}{\sqrt {B}}\right )}{\sqrt {2} \sqrt {B} \sqrt {e}}\right )}{2 \sqrt {2} \sqrt {e}}\right )}{A}\)

Input:

Int[(A + B*x)/(Sqrt[(A^2*e - B^2*e)/(2*A*B) + e*x]*(1 + x^2)),x]
 

Output:

(4*((A^(3/2)*Sqrt[B]*ArcTan[(Sqrt[A]*(-((Sqrt[2]*Sqrt[A]*Sqrt[e])/Sqrt[B]) 
 + 2*Sqrt[((A/B - B/A)*e)/2 + e*x]))/(Sqrt[2]*Sqrt[B]*Sqrt[e])])/(2*Sqrt[2 
]*Sqrt[e]) + (A^(3/2)*Sqrt[B]*ArcTan[(Sqrt[A]*((Sqrt[2]*Sqrt[A]*Sqrt[e])/S 
qrt[B] + 2*Sqrt[((A/B - B/A)*e)/2 + e*x]))/(Sqrt[2]*Sqrt[B]*Sqrt[e])])/(2* 
Sqrt[2]*Sqrt[e])))/A
 

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 217

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

rule 654

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), 
x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d* 
x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1475

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^ 
2, x], x], x] + Simp[e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F 
reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && 
 (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] 
, 0]))
 

Maple [A] (verified)

Time = 8.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {A B \sqrt {2}\, \left (\arctan \left (\frac {\sqrt {\frac {e \left (2 x A B +A^{2}-B^{2}\right )}{A B}}\, A B -\sqrt {A^{3} B e}}{\sqrt {A e B}\, B}\right )+\arctan \left (\frac {\sqrt {\frac {e \left (2 x A B +A^{2}-B^{2}\right )}{A B}}\, A B +\sqrt {A^{3} B e}}{\sqrt {A e B}\, B}\right )\right )}{\sqrt {A e B}}\)	\(110\)
default	\(2 \sqrt {2}\, B^{2} A \left (\frac {\arctan \left (\frac {2 A B \sqrt {2 e x +\frac {e \left (A^{2}-B^{2}\right )}{A B}}+2 \sqrt {A^{3} B e}}{2 B \sqrt {A e B}}\right )}{2 B \sqrt {A e B}}+\frac {\arctan \left (\frac {2 A B \sqrt {2 e x +\frac {e \left (A^{2}-B^{2}\right )}{A B}}-2 \sqrt {A^{3} B e}}{2 B \sqrt {A e B}}\right )}{2 B \sqrt {A e B}}\right )\)	\(135\)
derivativedivides	\(4 B^{2} A \left (\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 \sqrt {4 e x +\frac {2 A^{2} e -2 B^{2} e}{A B}}\, A B +2 \sqrt {2}\, \sqrt {A^{3} B e}\right ) \sqrt {2}}{4 B \sqrt {A e B}}\right )}{4 B \sqrt {A e B}}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 \sqrt {4 e x +\frac {2 A^{2} e -2 B^{2} e}{A B}}\, A B -2 \sqrt {2}\, \sqrt {A^{3} B e}\right ) \sqrt {2}}{4 B \sqrt {A e B}}\right )}{4 B \sqrt {A e B}}\right )\)	\(156\)

Input:

int(2*(B*x+A)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2)/(x^2+1),x,method=_RETURNVE 
RBOSE)
 

Output:

A*B*2^(1/2)*(arctan(((e*(2*A*B*x+A^2-B^2)/A/B)^(1/2)*A*B-(A^3*B*e)^(1/2))/ 
(A*e*B)^(1/2)/B)+arctan(((e*(2*A*B*x+A^2-B^2)/A/B)^(1/2)*A*B+(A^3*B*e)^(1/ 
2))/(A*e*B)^(1/2)/B))/(A*e*B)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\left [\frac {1}{2} \, \sqrt {2} \sqrt {-\frac {A B}{e}} \log \left (\frac {A^{2} x^{2} - 4 \, A B x - A^{2} + 2 \, B^{2} + 2 \, {\left (A x - B\right )} \sqrt {-\frac {A B}{e}} \sqrt {\frac {2 \, A B e x + {\left (A^{2} - B^{2}\right )} e}{A B}}}{x^{2} + 1}\right ), \sqrt {2} \sqrt {\frac {A B}{e}} \arctan \left (\frac {{\left (A x - B\right )} \sqrt {\frac {A B}{e}} \sqrt {\frac {2 \, A B e x + {\left (A^{2} - B^{2}\right )} e}{A B}}}{2 \, A B x + A^{2} - B^{2}}\right )\right ] \] Input:

integrate(2*(B*x+A)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2)/(x^2+1),x, algorithm 
="fricas")
 

Output:

[1/2*sqrt(2)*sqrt(-A*B/e)*log((A^2*x^2 - 4*A*B*x - A^2 + 2*B^2 + 2*(A*x - 
B)*sqrt(-A*B/e)*sqrt((2*A*B*e*x + (A^2 - B^2)*e)/(A*B)))/(x^2 + 1)), sqrt( 
2)*sqrt(A*B/e)*arctan((A*x - B)*sqrt(A*B/e)*sqrt((2*A*B*e*x + (A^2 - B^2)* 
e)/(A*B))/(2*A*B*x + A^2 - B^2))]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\sqrt {2} \left (\int \frac {A}{x^{2} \sqrt {\frac {A e}{B} + 2 e x - \frac {B e}{A}} + \sqrt {\frac {A e}{B} + 2 e x - \frac {B e}{A}}}\, dx + \int \frac {B x}{x^{2} \sqrt {\frac {A e}{B} + 2 e x - \frac {B e}{A}} + \sqrt {\frac {A e}{B} + 2 e x - \frac {B e}{A}}}\, dx\right ) \] Input:

integrate(2*(B*x+A)/(2*(A**2*e-B**2*e)/A/B+4*e*x)**(1/2)/(x**2+1),x)
 

Output:

sqrt(2)*(Integral(A/(x**2*sqrt(A*e/B + 2*e*x - B*e/A) + sqrt(A*e/B + 2*e*x 
 - B*e/A)), x) + Integral(B*x/(x**2*sqrt(A*e/B + 2*e*x - B*e/A) + sqrt(A*e 
/B + 2*e*x - B*e/A)), x))
 

Maxima [F]

\[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\int { \frac {2 \, {\left (B x + A\right )}}{\sqrt {4 \, e x + \frac {2 \, {\left (A^{2} e - B^{2} e\right )}}{A B}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(2*(B*x+A)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2)/(x^2+1),x, algorithm 
="maxima")
 

Output:

2*integrate((B*x + A)/(sqrt(4*e*x + 2*(A^2*e - B^2*e)/(A*B))*(x^2 + 1)), x 
)
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 13.79 (sec) , antiderivative size = 18107, normalized size of antiderivative = 134.13 \[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate(2*(B*x+A)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2)/(x^2+1),x, algorithm 
="giac")
 

Output:

1/2*sqrt(2)*(6*A*B*e^2*abs(A)*abs(B)*cos(1/2*real_part(arccos(A^3*B*e/abs( 
A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e))))^2*cosh(1/2*imag_par 
t(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e))) 
)^3*sin(1/2*real_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs( 
A^3*B*e + A*B^3*e)))) - 2*A*B*e^2*abs(A)*abs(B)*cosh(1/2*imag_part(arccos( 
A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e))))^3*sin(1 
/2*real_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + 
 A*B^3*e))))^3 - 18*A*B*e^2*abs(A)*abs(B)*cos(1/2*real_part(arccos(A^3*B*e 
/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e))))^2*cosh(1/2*ima 
g_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3 
*e))))^2*sin(1/2*real_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e 
/abs(A^3*B*e + A*B^3*e))))*sinh(1/2*imag_part(arccos(A^3*B*e/abs(A^3*B*e + 
 A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e)))) + 6*A*B*e^2*abs(A)*abs(B)*co 
sh(1/2*imag_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B 
*e + A*B^3*e))))^2*sin(1/2*real_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) 
 - A*B^3*e/abs(A^3*B*e + A*B^3*e))))^3*sinh(1/2*imag_part(arccos(A^3*B*e/a 
bs(A^3*B*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e)))) + 18*A*B*e^2*abs 
(A)*abs(B)*cos(1/2*real_part(arccos(A^3*B*e/abs(A^3*B*e + A*B^3*e) - A*B^3 
*e/abs(A^3*B*e + A*B^3*e))))^2*cosh(1/2*imag_part(arccos(A^3*B*e/abs(A^3*B 
*e + A*B^3*e) - A*B^3*e/abs(A^3*B*e + A*B^3*e))))*sin(1/2*real_part(arc...
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.53 \[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\frac {\sqrt {2}\,\sqrt {A}\,\sqrt {B}\,\left (\mathrm {atan}\left (\frac {A^{3/2}\,{\left (4\,e\,x+\frac {2\,A^2\,e-2\,B^2\,e}{A\,B}\right )}^{3/2}\,\sqrt {2\,B}}{8\,e^{3/2}\,\left (A^2+B^2\right )}-\frac {A^{5/2}\,\sqrt {8\,e\,x+\frac {2\,\left (2\,A^2\,e-2\,B^2\,e\right )}{A\,B}}}{4\,\sqrt {B}\,\sqrt {e}\,\left (A^2+B^2\right )}+\frac {3\,\sqrt {2}\,\sqrt {A}\,B^{3/2}\,\sqrt {4\,e\,x+\frac {2\,A^2\,e-2\,B^2\,e}{A\,B}}}{4\,\sqrt {e}\,\left (A^2+B^2\right )}\right )+\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {A}\,\sqrt {4\,e\,x+\frac {2\,A^2\,e-2\,B^2\,e}{A\,B}}}{4\,\sqrt {B}\,\sqrt {e}}\right )\right )}{\sqrt {e}} \] Input:

int((2*A + 2*B*x)/((x^2 + 1)*(4*e*x + (2*A^2*e - 2*B^2*e)/(A*B))^(1/2)),x)
 

Output:

(2^(1/2)*A^(1/2)*B^(1/2)*(atan((A^(3/2)*(4*e*x + (2*A^2*e - 2*B^2*e)/(A*B) 
)^(3/2)*(2*B)^(1/2))/(8*e^(3/2)*(A^2 + B^2)) - (A^(5/2)*(8*e*x + (2*(2*A^2 
*e - 2*B^2*e))/(A*B))^(1/2))/(4*B^(1/2)*e^(1/2)*(A^2 + B^2)) + (3*2^(1/2)* 
A^(1/2)*B^(3/2)*(4*e*x + (2*A^2*e - 2*B^2*e)/(A*B))^(1/2))/(4*e^(1/2)*(A^2 
 + B^2))) + atan((2^(1/2)*A^(1/2)*(4*e*x + (2*A^2*e - 2*B^2*e)/(A*B))^(1/2 
))/(4*B^(1/2)*e^(1/2)))))/e^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.44 \[ \int \frac {A+B x}{\sqrt {\frac {A^2 e-B^2 e}{2 A B}+e x} \left (1+x^2\right )} \, dx=\frac {\sqrt {e}\, \sqrt {b}\, \sqrt {a}\, \sqrt {2}\, \left (\mathit {atan} \left (\frac {\sqrt {2 a b x +a^{2}-b^{2}}-a}{b}\right )+\mathit {atan} \left (\frac {\sqrt {2 a b x +a^{2}-b^{2}}+a}{b}\right )\right )}{e} \] Input:

int(2*(B*x+A)/(2*(A^2*e-B^2*e)/A/B+4*e*x)^(1/2)/(x^2+1),x)
 

Output:

(sqrt(e)*sqrt(b)*sqrt(a)*sqrt(2)*(atan((sqrt(a**2 + 2*a*b*x - b**2) - a)/b 
) + atan((sqrt(a**2 + 2*a*b*x - b**2) + a)/b)))/e