\(\int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx\) [285]

Optimal result

Integrand size = 24, antiderivative size = 181 \[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}\right )}{e}-\frac {\sqrt {c} \text {arctanh}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}\right )}{d}-\frac {\sqrt {a d^2-e (b d-c e)} \text {arctanh}\left (\frac {2 a d-b e+\frac {b d-2 c e}{x}}{2 \sqrt {a d^2-e (b d-c e)} \sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}\right )}{d e} \] Output:

a^(1/2)*arctanh(1/2*(2*a+b/x)/a^(1/2)/(a+c/x^2+b/x)^(1/2))/e-c^(1/2)*arcta 
nh(1/2*(b+2*c/x)/c^(1/2)/(a+c/x^2+b/x)^(1/2))/d-(a*d^2-e*(b*d-c*e))^(1/2)* 
arctanh(1/2*(2*a*d-b*e+(b*d-2*c*e)/x)/(a*d^2-e*(b*d-c*e))^(1/2)/(a+c/x^2+b 
/x)^(1/2))/d/e
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\frac {x \sqrt {a+\frac {c+b x}{x^2}} \left (-2 \sqrt {-a d^2+b d e-c e^2} \arctan \left (\frac {\sqrt {a} (d+e x)-e \sqrt {c+x (b+a x)}}{\sqrt {-a d^2+b d e-c e^2}}\right )+2 \sqrt {c} e \text {arctanh}\left (\frac {\sqrt {a} x-\sqrt {c+x (b+a x)}}{\sqrt {c}}\right )-\sqrt {a} d \log \left (e \left (b+2 a x-2 \sqrt {a} \sqrt {c+x (b+a x)}\right )\right )\right )}{d e \sqrt {c+x (b+a x)}} \] Input:

Integrate[Sqrt[a + c/x^2 + b/x]/(d + e*x),x]
 

Output:

(x*Sqrt[a + (c + b*x)/x^2]*(-2*Sqrt[-(a*d^2) + b*d*e - c*e^2]*ArcTan[(Sqrt 
[a]*(d + e*x) - e*Sqrt[c + x*(b + a*x)])/Sqrt[-(a*d^2) + b*d*e - c*e^2]] + 
 2*Sqrt[c]*e*ArcTanh[(Sqrt[a]*x - Sqrt[c + x*(b + a*x)])/Sqrt[c]] - Sqrt[a 
]*d*Log[e*(b + 2*a*x - 2*Sqrt[a]*Sqrt[c + x*(b + a*x)])]))/(d*e*Sqrt[c + x 
*(b + a*x)])
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1773, 1802, 1270, 1154, 219, 1269, 1092, 219, 1154, 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.

\(\displaystyle \int \frac {\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}{d+e x} \, dx\)

\(\Big \downarrow \) 1773

\(\displaystyle \int \frac {\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}{x \left (\frac {d}{x}+e\right )}dx\)

\(\Big \downarrow \) 1802

\(\displaystyle -\int \frac {\sqrt {a+\frac {b}{x}+\frac {c}{x^2}} x}{\frac {d}{x}+e}d\frac {1}{x}\)

\(\Big \downarrow \) 1270

\(\displaystyle \frac {\int \frac {a d-b e-\frac {c e}{x}}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \left (\frac {d}{x}+e\right )}d\frac {1}{x}}{e}-\frac {a \int \frac {x}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}d\frac {1}{x}}{e}\)

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\int \frac {a d-b e-\frac {c e}{x}}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \left (\frac {d}{x}+e\right )}d\frac {1}{x}}{e}+\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {\frac {\left (a d^2-e (b d-c e)\right ) \int \frac {1}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \left (\frac {d}{x}+e\right )}d\frac {1}{x}}{d}-\frac {c e \int \frac {1}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}d\frac {1}{x}}{d}}{e}+\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {\left (a d^2-e (b d-c e)\right ) \int \frac {1}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \left (\frac {d}{x}+e\right )}d\frac {1}{x}}{d}-\frac {2 c e \int \frac {1}{4 c-\frac {1}{x^2}}d\frac {b+\frac {2 c}{x}}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}}{d}}{e}+\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\left (a d^2-e (b d-c e)\right ) \int \frac {1}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \left (\frac {d}{x}+e\right )}d\frac {1}{x}}{d}-\frac {\sqrt {c} e \text {arctanh}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{d}}{e}+\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {-\frac {2 \left (a d^2-e (b d-c e)\right ) \int \frac {1}{4 \left (a d^2-e (b d-c e)\right )-\frac {1}{x^2}}d\frac {2 a d-b e+\frac {b d-2 c e}{x}}{\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}}{d}-\frac {\sqrt {c} e \text {arctanh}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{d}}{e}+\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {\sqrt {a d^2-e (b d-c e)} \text {arctanh}\left (\frac {2 a d+\frac {b d-2 c e}{x}-b e}{2 \sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \sqrt {a d^2-e (b d-c e)}}\right )}{d}-\frac {\sqrt {c} e \text {arctanh}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{d}}{e}+\frac {\sqrt {a} \text {arctanh}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e}\)

Input:

Int[Sqrt[a + c/x^2 + b/x]/(d + e*x),x]
 

Output:

(Sqrt[a]*ArcTanh[(2*a + b/x)/(2*Sqrt[a]*Sqrt[a + c/x^2 + b/x])])/e + (-((S 
qrt[c]*e*ArcTanh[(b + (2*c)/x)/(2*Sqrt[c]*Sqrt[a + c/x^2 + b/x])])/d) - (S 
qrt[a*d^2 - e*(b*d - c*e)]*ArcTanh[(2*a*d - b*e + (b*d - 2*c*e)/x)/(2*Sqrt 
[a*d^2 - e*(b*d - c*e)]*Sqrt[a + c/x^2 + b/x])])/d)/e
 

Defintions of rubi rules used

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 1092

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

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 1270

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)/(((d_.) + (e_.)*(x_))*((f_.) + 
 (g_.)*(x_))), x_Symbol] :> Simp[(c*d^2 - b*d*e + a*e^2)/(e*(e*f - d*g)) 
Int[(a + b*x + c*x^2)^(p - 1)/(d + e*x), x], x] - Simp[1/(e*(e*f - d*g)) 
Int[Simp[c*d*f - b*e*f + a*e*g - c*(e*f - d*g)*x, x]*((a + b*x + c*x^2)^(p 
- 1)/(f + g*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[p] 
&& GtQ[p, 0]
 

rule 1773

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_) 
^(n2_.))^(p_.), x_Symbol] :> Int[((e + d*x^n)^q*(a + b*x^n + c*x^(2*n))^p)/ 
x^(n*q), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[n2, 2*n] && EqQ[mn, - 
n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])
 

rule 1802

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( 
e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1 
)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, 
c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(157)=314\).

Time = 0.13 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.18

method	result	size
default	\(\frac {\sqrt {\frac {a \,x^{2}+b x +c}{x^{2}}}\, x \left (a^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, e -2 a d x +b e x -b d +2 e c}{e x +d}\right ) d^{2}-\sqrt {a}\, \ln \left (\frac {2 c +b x +2 \sqrt {c}\, \sqrt {a \,x^{2}+b x +c}}{x}\right ) \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, \sqrt {c}\, e^{2}-\sqrt {a}\, \ln \left (\frac {2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, e -2 a d x +b e x -b d +2 e c}{e x +d}\right ) b d e +\sqrt {a}\, \ln \left (\frac {2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, e -2 a d x +b e x -b d +2 e c}{e x +d}\right ) c \,e^{2}+\ln \left (\frac {2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, a d e \right )}{\sqrt {a \,x^{2}+b x +c}\, d \,e^{2} \sqrt {a}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}}\)	\(395\)

Input:

int((a+c/x^2+b/x)^(1/2)/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

((a*x^2+b*x+c)/x^2)^(1/2)*x*(a^(3/2)*ln((2*(a*x^2+b*x+c)^(1/2)*((a*d^2-b*d 
*e+c*e^2)/e^2)^(1/2)*e-2*a*d*x+b*e*x-b*d+2*e*c)/(e*x+d))*d^2-a^(1/2)*ln((2 
*c+b*x+2*c^(1/2)*(a*x^2+b*x+c)^(1/2))/x)*((a*d^2-b*d*e+c*e^2)/e^2)^(1/2)*c 
^(1/2)*e^2-a^(1/2)*ln((2*(a*x^2+b*x+c)^(1/2)*((a*d^2-b*d*e+c*e^2)/e^2)^(1/ 
2)*e-2*a*d*x+b*e*x-b*d+2*e*c)/(e*x+d))*b*d*e+a^(1/2)*ln((2*(a*x^2+b*x+c)^( 
1/2)*((a*d^2-b*d*e+c*e^2)/e^2)^(1/2)*e-2*a*d*x+b*e*x-b*d+2*e*c)/(e*x+d))*c 
*e^2+ln(1/2*(2*(a*x^2+b*x+c)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d^2-b*d*e 
+c*e^2)/e^2)^(1/2)*a*d*e)/(a*x^2+b*x+c)^(1/2)/d/e^2/a^(1/2)/((a*d^2-b*d*e+ 
c*e^2)/e^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 39.16 (sec) , antiderivative size = 2411, normalized size of antiderivative = 13.32 \[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\text {Too large to display} \] Input:

integrate((a+c/x^2+b/x)^(1/2)/(e*x+d),x, algorithm="fricas")
 

Output:

[1/2*(sqrt(a)*d*log(-8*a^2*x^2 - 8*a*b*x - b^2 - 4*a*c - 4*(2*a*x^2 + b*x) 
*sqrt(a)*sqrt((a*x^2 + b*x + c)/x^2)) + sqrt(c)*e*log(-(8*b*c*x + (b^2 + 4 
*a*c)*x^2 + 8*c^2 - 4*(b*x^2 + 2*c*x)*sqrt(c)*sqrt((a*x^2 + b*x + c)/x^2)) 
/x^2) + sqrt(a*d^2 - b*d*e + c*e^2)*log((8*b*c*d*e - 8*c^2*e^2 - (b^2 + 4* 
a*c)*d^2 - (8*a^2*d^2 - 8*a*b*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 2*(4*a*b*d^2 
+ 4*b*c*e^2 - (3*b^2 + 4*a*c)*d*e)*x + 4*sqrt(a*d^2 - b*d*e + c*e^2)*((2*a 
*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(e^2*x^2 + 2 
*d*e*x + d^2)))/(d*e), -1/2*(2*sqrt(-a)*d*arctan(1/2*(2*a*x^2 + b*x)*sqrt( 
-a)*sqrt((a*x^2 + b*x + c)/x^2)/(a^2*x^2 + a*b*x + a*c)) - sqrt(c)*e*log(- 
(8*b*c*x + (b^2 + 4*a*c)*x^2 + 8*c^2 - 4*(b*x^2 + 2*c*x)*sqrt(c)*sqrt((a*x 
^2 + b*x + c)/x^2))/x^2) - sqrt(a*d^2 - b*d*e + c*e^2)*log((8*b*c*d*e - 8* 
c^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*a^2*d^2 - 8*a*b*d*e + (b^2 + 4*a*c)*e^2)* 
x^2 - 2*(4*a*b*d^2 + 4*b*c*e^2 - (3*b^2 + 4*a*c)*d*e)*x + 4*sqrt(a*d^2 - b 
*d*e + c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c) 
/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e), 1/2*(sqrt(a)*d*log(-8*a^2*x^2 - 
8*a*b*x - b^2 - 4*a*c - 4*(2*a*x^2 + b*x)*sqrt(a)*sqrt((a*x^2 + b*x + c)/x 
^2)) + sqrt(c)*e*log(-(8*b*c*x + (b^2 + 4*a*c)*x^2 + 8*c^2 - 4*(b*x^2 + 2* 
c*x)*sqrt(c)*sqrt((a*x^2 + b*x + c)/x^2))/x^2) - 2*sqrt(-a*d^2 + b*d*e - c 
*e^2)*arctan(-1/2*sqrt(-a*d^2 + b*d*e - c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 
 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2)/(a*c*d^2 - b*c*d*e + c^2*e^2 + (...
 

Sympy [F]

\[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\int \frac {\sqrt {a + \frac {b}{x} + \frac {c}{x^{2}}}}{d + e x}\, dx \] Input:

integrate((a+c/x**2+b/x)**(1/2)/(e*x+d),x)
 

Output:

Integral(sqrt(a + b/x + c/x**2)/(d + e*x), x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\int { \frac {\sqrt {a + \frac {b}{x} + \frac {c}{x^{2}}}}{e x + d} \,d x } \] Input:

integrate((a+c/x^2+b/x)^(1/2)/(e*x+d),x, algorithm="maxima")
 

Output:

integrate(sqrt(a + b/x + c/x^2)/(e*x + d), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+c/x^2+b/x)^(1/2)/(e*x+d),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\int \frac {\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}{d+e\,x} \,d x \] Input:

int((a + b/x + c/x^2)^(1/2)/(d + e*x),x)
 

Output:

int((a + b/x + c/x^2)^(1/2)/(d + e*x), x)
 

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx=\frac {\sqrt {a \,d^{2}-b d e +c \,e^{2}}\, \mathrm {log}\left (-2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a \,d^{2}-b d e +c \,e^{2}}+2 a d x +b d -b e x -2 c e \right )-\sqrt {a \,d^{2}-b d e +c \,e^{2}}\, \mathrm {log}\left (e x +d \right )+\sqrt {a}\, \mathrm {log}\left (-2 \sqrt {a}\, \sqrt {a \,x^{2}+b x +c}-2 a x -b \right ) d +\sqrt {c}\, \mathrm {log}\left (2 \sqrt {c}\, \sqrt {a \,x^{2}+b x +c}-b x -2 c \right ) e -\sqrt {c}\, \mathrm {log}\left (x \right ) e}{d e} \] Input:

int((a+c/x^2+b/x)^(1/2)/(e*x+d),x)
 

Output:

(sqrt(a*d**2 - b*d*e + c*e**2)*log( - 2*sqrt(a*x**2 + b*x + c)*sqrt(a*d**2 
 - b*d*e + c*e**2) + 2*a*d*x + b*d - b*e*x - 2*c*e) - sqrt(a*d**2 - b*d*e 
+ c*e**2)*log(d + e*x) + sqrt(a)*log( - 2*sqrt(a)*sqrt(a*x**2 + b*x + c) - 
 2*a*x - b)*d + sqrt(c)*log(2*sqrt(c)*sqrt(a*x**2 + b*x + c) - b*x - 2*c)* 
e - sqrt(c)*log(x)*e)/(d*e)