\(\int \frac {1}{x (d+e x) (a+b x+c x^2)^{5/2}} \, dx\) [50]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 515 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \left (c \left (b^2-2 a c\right ) d-b^3 e+3 a b c e+c \left (b c d-b^2 e+2 a c e\right ) x\right )}{3 a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \left (3 b^6 d e^2+b^4 c d \left (3 c d^2-16 a e^2\right )+24 a^2 c^3 d \left (c d^2+2 a e^2\right )-2 a b^2 c^2 d \left (11 c d^2+4 a e^2\right )-4 a^2 b c^2 e \left (14 c d^2+17 a e^2\right )+a b^3 c e \left (44 c d^2+43 a e^2\right )-6 b^5 \left (c d^2 e+a e^3\right )+c \left (3 b^5 d e^2+b^3 c d \left (3 c d^2-14 a e^2\right )-4 a b c^2 d \left (5 c d^2+4 a e^2\right )-8 a^2 c^2 e \left (2 c d^2+5 a e^2\right )+2 a b^2 c e \left (20 c d^2+19 a e^2\right )-6 b^4 \left (c d^2 e+a e^3\right )\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2 \sqrt {a+b x+c x^2}}-\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{5/2} d}-\frac {e^5 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{d \left (c d^2-b d e+a e^2\right )^{5/2}} \] Output:

2/3*(c*(-2*a*c+b^2)*d-b^3*e+3*a*b*c*e+c*(2*a*c*e-b^2*e+b*c*d)*x)/a/(-4*a*c 
+b^2)/(a*e^2-b*d*e+c*d^2)/(c*x^2+b*x+a)^(3/2)+2/3*(3*b^6*d*e^2+b^4*c*d*(-1 
6*a*e^2+3*c*d^2)+24*a^2*c^3*d*(2*a*e^2+c*d^2)-2*a*b^2*c^2*d*(4*a*e^2+11*c* 
d^2)-4*a^2*b*c^2*e*(17*a*e^2+14*c*d^2)+a*b^3*c*e*(43*a*e^2+44*c*d^2)-6*b^5 
*(a*e^3+c*d^2*e)+c*(3*b^5*d*e^2+b^3*c*d*(-14*a*e^2+3*c*d^2)-4*a*b*c^2*d*(4 
*a*e^2+5*c*d^2)-8*a^2*c^2*e*(5*a*e^2+2*c*d^2)+2*a*b^2*c*e*(19*a*e^2+20*c*d 
^2)-6*b^4*(a*e^3+c*d^2*e))*x)/a^2/(-4*a*c+b^2)^2/(a*e^2-b*d*e+c*d^2)^2/(c* 
x^2+b*x+a)^(1/2)-arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(5/2 
)/d-e^5*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/( 
c*x^2+b*x+a)^(1/2))/d/(a*e^2-b*d*e+c*d^2)^(5/2)
 

Mathematica [A] (verified)

Time = 11.66 (sec) , antiderivative size = 459, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {\frac {2 \left (b^2-2 a c+b c x\right )}{a (a+x (b+c x))^{3/2}}+\frac {2 \left (\left (b^2-6 a c\right ) \left (3 b^2-4 a c\right )+b c \left (3 b^2-20 a c\right ) x\right )}{a^2 \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)}}+\frac {2 e \left (-b^2 e+2 c (a e+c d x)+b c (d-e x)\right )}{\left (c d^2+e (-b d+a e)\right ) (a+x (b+c x))^{3/2}}-\frac {2 e \left (3 b^4 e^3+b^3 c e^2 (d+3 e x)+8 c^2 \left (3 a^2 e^3+2 c^2 d^3 x+5 a c d e^2 x\right )+4 b c^2 \left (2 c d^2 (d-3 e x)+5 a e^2 (d-e x)\right )+2 b^2 c e \left (-11 a e^2+c d (-6 d+e x)\right )\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right )^2 \sqrt {a+x (b+c x)}}-\frac {3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{a^{5/2}}+\frac {3 \left (b^2-4 a c\right ) e^5 \text {arctanh}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{\left (c d^2+e (-b d+a e)\right )^{5/2}}}{3 \left (b^2-4 a c\right ) d} \] Input:

Integrate[1/(x*(d + e*x)*(a + b*x + c*x^2)^(5/2)),x]
 

Output:

((2*(b^2 - 2*a*c + b*c*x))/(a*(a + x*(b + c*x))^(3/2)) + (2*((b^2 - 6*a*c) 
*(3*b^2 - 4*a*c) + b*c*(3*b^2 - 20*a*c)*x))/(a^2*(b^2 - 4*a*c)*Sqrt[a + x* 
(b + c*x)]) + (2*e*(-(b^2*e) + 2*c*(a*e + c*d*x) + b*c*(d - e*x)))/((c*d^2 
 + e*(-(b*d) + a*e))*(a + x*(b + c*x))^(3/2)) - (2*e*(3*b^4*e^3 + b^3*c*e^ 
2*(d + 3*e*x) + 8*c^2*(3*a^2*e^3 + 2*c^2*d^3*x + 5*a*c*d*e^2*x) + 4*b*c^2* 
(2*c*d^2*(d - 3*e*x) + 5*a*e^2*(d - e*x)) + 2*b^2*c*e*(-11*a*e^2 + c*d*(-6 
*d + e*x))))/((b^2 - 4*a*c)*(c*d^2 + e*(-(b*d) + a*e))^2*Sqrt[a + x*(b + c 
*x)]) - (3*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c* 
x)])])/a^(5/2) + (3*(b^2 - 4*a*c)*e^5*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + 
b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(c*d^2 + 
 e*(-(b*d) + a*e))^(5/2))/(3*(b^2 - 4*a*c)*d)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 477, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1289, 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.

\(\displaystyle \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1289

\(\displaystyle \int \left (\frac {1}{d x \left (a+b x+c x^2\right )^{5/2}}-\frac {e}{d (d+e x) \left (a+b x+c x^2\right )^{5/2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{5/2} d}+\frac {2 \left (b c x \left (3 b^2-20 a c\right )+\left (b^2-6 a c\right ) \left (3 b^2-4 a c\right )\right )}{3 a^2 d \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {e^5 \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{d \left (a e^2-b d e+c d^2\right )^{5/2}}+\frac {2 e \left (-c x (2 c d-b e) \left (-4 c e (2 b d-5 a e)-3 b^2 e^2+8 c^2 d^2\right )-\left (2 a c e+b^2 (-e)+b c d\right ) \left (-4 c e (b d-3 a e)-3 b^2 e^2+8 c^2 d^2\right )+4 a c e (2 c d-b e)^2\right )}{3 d \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2}+\frac {2 e \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{3 a d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

Input:

Int[1/(x*(d + e*x)*(a + b*x + c*x^2)^(5/2)),x]
 

Output:

(2*(b^2 - 2*a*c + b*c*x))/(3*a*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) + 
(2*e*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*d*(c* 
d^2 - b*d*e + a*e^2)*(a + b*x + c*x^2)^(3/2)) + (2*((b^2 - 6*a*c)*(3*b^2 - 
 4*a*c) + b*c*(3*b^2 - 20*a*c)*x))/(3*a^2*(b^2 - 4*a*c)^2*d*Sqrt[a + b*x + 
 c*x^2]) + (2*e*(4*a*c*e*(2*c*d - b*e)^2 - (b*c*d - b^2*e + 2*a*c*e)*(8*c^ 
2*d^2 - 3*b^2*e^2 - 4*c*e*(b*d - 3*a*e)) - c*(2*c*d - b*e)*(8*c^2*d^2 - 3* 
b^2*e^2 - 4*c*e*(2*b*d - 5*a*e))*x))/(3*(b^2 - 4*a*c)^2*d*(c*d^2 - b*d*e + 
 a*e^2)^2*Sqrt[a + b*x + c*x^2]) - ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + 
 b*x + c*x^2])]/(a^(5/2)*d) - (e^5*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x) 
/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(d*(c*d^2 - b*d*e 
 + a*e^2)^(5/2))
 

Defintions of rubi rules used

rule 1289
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( 
IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 953, normalized size of antiderivative = 1.85

method result size
default \(\frac {\frac {1}{3 a \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 a}+\frac {\frac {1}{a \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{a^{\frac {3}{2}}}}{a}}{d}-\frac {\frac {e^{2}}{3 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}-\frac {\left (b e -2 c d \right ) e \left (\frac {\frac {4 c \left (x +\frac {d}{e}\right )}{3}+\frac {2 \left (b e -2 c d \right )}{3 e}}{\left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}+\frac {16 c \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{3 {\left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right )}^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {e^{2} \left (\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{a \,e^{2}-b d e +c \,d^{2}}}{d}\) \(953\)

Input:

int(1/x/(e*x+d)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/3/a/(c*x^2+b*x+a)^(3/2)-1/2*b/a*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b 
*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2))+1/a*(1/a/( 
c*x^2+b*x+a)^(1/2)-b/a*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-1/a^(3/2) 
*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)))-1/d*(1/3/(a*e^2-b*d*e+c*d 
^2)*e^2/(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)- 
1/2*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(2/3*(2*c*(x+d/e)+(b*e-2*c*d)/e)/(4* 
c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/(c*(x+d/e)^2+(b*e-2*c*d)/e*(x 
+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+16/3*c/(4*c*(a*e^2-b*d*e+c*d^2)/e^2-( 
b*e-2*c*d)^2/e^2)^2*(2*c*(x+d/e)+(b*e-2*c*d)/e)/(c*(x+d/e)^2+(b*e-2*c*d)/e 
*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))+1/(a*e^2-b*d*e+c*d^2)*e^2*(1/(a*e 
^2-b*d*e+c*d^2)*e^2/(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2) 
/e^2)^(1/2)-(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(2*c*(x+d/e)+(b*e-2*c*d)/e)/ 
(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/(c*(x+d/e)^2+(b*e-2*c*d)/e 
*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-1/(a*e^2-b*d*e+c*d^2)*e^2/((a*e^2- 
b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e 
)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a* 
e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4174 vs. \(2 (492) = 984\).

Time = 19.05 (sec) , antiderivative size = 16795, normalized size of antiderivative = 32.61 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x/(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (d + e x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/x/(e*x+d)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Integral(1/(x*(d + e*x)*(a + b*x + c*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} {\left (e x + d\right )} x} \,d x } \] Input:

integrate(1/x/(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((c*x^2 + b*x + a)^(5/2)*(e*x + d)*x), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/x/(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,\left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int(1/(x*(d + e*x)*(a + b*x + c*x^2)^(5/2)),x)
 

Output:

int(1/(x*(d + e*x)*(a + b*x + c*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 6.13 (sec) , antiderivative size = 13522, normalized size of antiderivative = 26.26 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(1/x/(e*x+d)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

(48*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e 
**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**7*c**2*e**5 - 24 
*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 
 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**6*b**2*c*e**5 + 96* 
sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 
- b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**6*b*c**2*e**5*x + 96 
*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 
 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**6*c**3*e**5*x**2 + 
3*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e** 
2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**5*b**4*e**5 - 48*s 
qrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - 
 b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**5*b**3*c*e**5*x + 96* 
sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 
- b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**5*b*c**3*e**5*x**3 + 
 48*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e 
**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**5*c**4*e**5*x**4 
 + 6*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a* 
e**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**4*b**5*e**5*x - 
 18*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e 
**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**4*b**4*c*e**5...