\(\int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx\) [7]

Optimal result

Integrand size = 37, antiderivative size = 109 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {B \arctan \left (\frac {\sqrt {e} x \sqrt {-d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{e^{3/2}}-\frac {(B d+A e) \arctan \left (\frac {\sqrt {2} \sqrt {e} x \sqrt {-d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{\sqrt {2} d e^{3/2}} \] Output:

B*arctan(e^(1/2)*x*(e*x^2-d)^(1/2)/(-e^2*x^4+d^2)^(1/2))/e^(3/2)-1/2*(A*e+ 
B*d)*arctan(2^(1/2)*e^(1/2)*x*(e*x^2-d)^(1/2)/(-e^2*x^4+d^2)^(1/2))*2^(1/2 
)/d/e^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.90 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.38 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\frac {(B d+A e) \sqrt {-2 d+2 e x^2} \sqrt {d^2-e^2 x^4} \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {-d-e x^2}}\right )}{d \sqrt {-d-e x^2} \left (d-e x^2\right )}-2 i B \log \left (-2 i \sqrt {e} x-\frac {2 \sqrt {d^2-e^2 x^4}}{\sqrt {-d+e x^2}}\right )}{2 e^{3/2}} \] Input:

Integrate[(A + B*x^2)/(Sqrt[-d + e*x^2]*Sqrt[d^2 - e^2*x^4]),x]
 

Output:

(((B*d + A*e)*Sqrt[-2*d + 2*e*x^2]*Sqrt[d^2 - e^2*x^4]*ArcTan[(Sqrt[2]*Sqr 
t[e]*x)/Sqrt[-d - e*x^2]])/(d*Sqrt[-d - e*x^2]*(d - e*x^2)) - (2*I)*B*Log[ 
(-2*I)*Sqrt[e]*x - (2*Sqrt[d^2 - e^2*x^4])/Sqrt[-d + e*x^2]])/(2*e^(3/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1396, 25, 398, 224, 216, 291, 218}

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.

\(\displaystyle \int \frac {A+B x^2}{\sqrt {e x^2-d} \sqrt {d^2-e^2 x^4}} \, dx\)

\(\Big \downarrow \) 1396

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

Int[(A + B*x^2)/(Sqrt[-d + e*x^2]*Sqrt[d^2 - e^2*x^4]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 218

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

rule 1396

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x 
_Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d 
+ c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, 
x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* 
e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 1] && EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(479\) vs. \(2(89)=178\).

Time = 0.14 (sec) , antiderivative size = 480, normalized size of antiderivative = 4.40

method	result	size
default	\(\frac {\sqrt {-e^{2} x^{4}+d^{2}}\, \left (A \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {-d e}\right ) \left (e x +\sqrt {-d e}\right )}{e}}}\right ) \sqrt {2}\, \sqrt {-d}\, e -A \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}-d}}\right ) \sqrt {2}\, \sqrt {-d}\, e +A \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-\sqrt {d e}\, x -d \right )}{e x -\sqrt {d e}}\right ) e^{\frac {3}{2}} d -A \ln \left (\frac {2 e \left (\sqrt {d e}\, x +\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-d \right )}{e x +\sqrt {d e}}\right ) e^{\frac {3}{2}} d -B \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {-d e}\right ) \left (e x +\sqrt {-d e}\right )}{e}}}\right ) \sqrt {2}\, \sqrt {-d}\, d -B \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}-d}}\right ) \sqrt {2}\, \sqrt {-d}\, d +B \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-\sqrt {d e}\, x -d \right )}{e x -\sqrt {d e}}\right ) \sqrt {e}\, d^{2}-B \ln \left (\frac {2 e \left (\sqrt {d e}\, x +\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-d \right )}{e x +\sqrt {d e}}\right ) \sqrt {e}\, d^{2}\right ) \sqrt {2}}{2 \sqrt {e \,x^{2}-d}\, \sqrt {-e \,x^{2}-d}\, \left (\sqrt {-d e}-\sqrt {d e}\right ) \left (\sqrt {-d e}+\sqrt {d e}\right ) \sqrt {e}\, \sqrt {d e}\, \sqrt {-d}}\)	\(480\)

Input:

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

Output:

1/2*(-e^2*x^4+d^2)^(1/2)*(A*(d*e)^(1/2)*arctan(e^(1/2)*x/(1/e*(-e*x+(-d*e) 
^(1/2))*(e*x+(-d*e)^(1/2)))^(1/2))*2^(1/2)*(-d)^(1/2)*e-A*(d*e)^(1/2)*arct 
an(e^(1/2)*x/(-e*x^2-d)^(1/2))*2^(1/2)*(-d)^(1/2)*e+A*ln(2*e*(2^(1/2)*(-d) 
^(1/2)*(-e*x^2-d)^(1/2)-(d*e)^(1/2)*x-d)/(e*x-(d*e)^(1/2)))*e^(3/2)*d-A*ln 
(2*e*((d*e)^(1/2)*x+2^(1/2)*(-d)^(1/2)*(-e*x^2-d)^(1/2)-d)/(e*x+(d*e)^(1/2 
)))*e^(3/2)*d-B*(d*e)^(1/2)*arctan(e^(1/2)*x/(1/e*(-e*x+(-d*e)^(1/2))*(e*x 
+(-d*e)^(1/2)))^(1/2))*2^(1/2)*(-d)^(1/2)*d-B*(d*e)^(1/2)*arctan(e^(1/2)*x 
/(-e*x^2-d)^(1/2))*2^(1/2)*(-d)^(1/2)*d+B*ln(2*e*(2^(1/2)*(-d)^(1/2)*(-e*x 
^2-d)^(1/2)-(d*e)^(1/2)*x-d)/(e*x-(d*e)^(1/2)))*e^(1/2)*d^2-B*ln(2*e*((d*e 
)^(1/2)*x+2^(1/2)*(-d)^(1/2)*(-e*x^2-d)^(1/2)-d)/(e*x+(d*e)^(1/2)))*e^(1/2 
)*d^2)/(e*x^2-d)^(1/2)/(-e*x^2-d)^(1/2)/((-d*e)^(1/2)-(d*e)^(1/2))/((-d*e) 
^(1/2)+(d*e)^(1/2))*2^(1/2)/e^(1/2)/(d*e)^(1/2)/(-d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.83 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\left [-\frac {2 \, B d \sqrt {-e} \log \left (\frac {2 \, e^{2} x^{4} - d e x^{2} - 2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} - d} \sqrt {-e} x - d^{2}}{e x^{2} - d}\right ) + \sqrt {2} {\left (B d + A e\right )} \sqrt {-e} \log \left (-\frac {3 \, e^{2} x^{4} - 2 \, d e x^{2} + 2 \, \sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} - d} \sqrt {-e} x - d^{2}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right )}{4 \, d e^{2}}, -\frac {2 \, B d \sqrt {e} \arctan \left (\frac {\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} - d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right ) - \sqrt {2} {\left (B d + A e\right )} \sqrt {e} \arctan \left (\frac {\sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} - d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right )}{2 \, d e^{2}}\right ] \] Input:

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

Output:

[-1/4*(2*B*d*sqrt(-e)*log((2*e^2*x^4 - d*e*x^2 - 2*sqrt(-e^2*x^4 + d^2)*sq 
rt(e*x^2 - d)*sqrt(-e)*x - d^2)/(e*x^2 - d)) + sqrt(2)*(B*d + A*e)*sqrt(-e 
)*log(-(3*e^2*x^4 - 2*d*e*x^2 + 2*sqrt(2)*sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 
- d)*sqrt(-e)*x - d^2)/(e^2*x^4 - 2*d*e*x^2 + d^2)))/(d*e^2), -1/2*(2*B*d* 
sqrt(e)*arctan(sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 - d)*sqrt(e)*x/(e^2*x^4 - d 
^2)) - sqrt(2)*(B*d + A*e)*sqrt(e)*arctan(sqrt(2)*sqrt(-e^2*x^4 + d^2)*sqr 
t(e*x^2 - d)*sqrt(e)*x/(e^2*x^4 - d^2)))/(d*e^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=-\frac {B \log \left ({\left | -\sqrt {-e} x + \sqrt {-e x^{2} - d} \right |}\right )}{\sqrt {-e} e} - \frac {\sqrt {2} {\left (B d + A e\right )} \sqrt {-e} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} - d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} + 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} - d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} + 6 \, d \right |}}\right )}{4 \, e^{2} {\left | d \right |}} \] Input:

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

Output:

-B*log(abs(-sqrt(-e)*x + sqrt(-e*x^2 - d)))/(sqrt(-e)*e) - 1/4*sqrt(2)*(B* 
d + A*e)*sqrt(-e)*log(abs(2*(sqrt(-e)*x - sqrt(-e*x^2 - d))^2 - 4*sqrt(2)* 
abs(d) + 6*d)/abs(2*(sqrt(-e)*x - sqrt(-e*x^2 - d))^2 + 4*sqrt(2)*abs(d) + 
 6*d))/(e^2*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\int \frac {B\,x^2+A}{\sqrt {d^2-e^2\,x^4}\,\sqrt {e\,x^2-d}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.65 \[ \int \frac {A+B x^2}{\sqrt {-d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\sqrt {e}\, i \left (-4 \mathit {asinh} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right ) b d -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d \right )}{4 d \,e^{2}} \] Input:

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

Output:

(sqrt(e)*i*( - 4*asinh((sqrt(e)*x)/sqrt(d))*b*d - sqrt(2)*log((sqrt(d + e* 
x**2) - sqrt(d)*sqrt(2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e - sqrt(2)*log( 
(sqrt(d + e*x**2) - sqrt(d)*sqrt(2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d + 
sqrt(2)*log((sqrt(d + e*x**2) - sqrt(d)*sqrt(2) + sqrt(d) + sqrt(e)*x)/sqr 
t(d))*a*e + sqrt(2)*log((sqrt(d + e*x**2) - sqrt(d)*sqrt(2) + sqrt(d) + sq 
rt(e)*x)/sqrt(d))*b*d + sqrt(2)*log((sqrt(d + e*x**2) + sqrt(d)*sqrt(2) - 
sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e + sqrt(2)*log((sqrt(d + e*x**2) + sqrt(d 
)*sqrt(2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d - sqrt(2)*log((sqrt(d + e*x* 
*2) + sqrt(d)*sqrt(2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e - sqrt(2)*log((s 
qrt(d + e*x**2) + sqrt(d)*sqrt(2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d))/(4 
*d*e**2)