Integrand size = 22, antiderivative size = 215 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\frac {(2 c d-b e) (b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{8 \left (c d^2-b d e+a e^2\right )^2 (d+e x)^2}-\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{16 \left (c d^2-b d e+a e^2\right )^{5/2}} \] Output:
1/8*(-b*e+2*c*d)*(b*d-2*a*e+(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(1/2)/(a*e^2-b*d *e+c*d^2)^2/(e*x+d)^2-1/3*e*(c*x^2+b*x+a)^(3/2)/(a*e^2-b*d*e+c*d^2)/(e*x+d )^3-1/16*(-4*a*c+b^2)*(-b*e+2*c*d)*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))/(a*e^2-b*d*e+c*d^2)^(5/2)
Time = 10.36 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\frac {-\frac {2 e (a+x (b+c x))^{3/2}}{(d+e x)^3}+3 (2 c d-b e) \left (\frac {\sqrt {a+x (b+c x)} (-2 a e+2 c d x+b (d-e x))}{4 \left (c d^2+e (-b d+a e)\right ) (d+e x)^2}+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{8 \left (c d^2+e (-b d+a e)\right )^{3/2}}\right )}{6 \left (c d^2+e (-b d+a e)\right )} \] Input:
Integrate[Sqrt[a + b*x + c*x^2]/(d + e*x)^4,x]
Output:
((-2*e*(a + x*(b + c*x))^(3/2))/(d + e*x)^3 + 3*(2*c*d - b*e)*((Sqrt[a + x *(b + c*x)]*(-2*a*e + 2*c*d*x + b*(d - e*x)))/(4*(c*d^2 + e*(-(b*d) + a*e) )*(d + e*x)^2) + ((b^2 - 4*a*c)*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)])])/(8*(c*d^2 + e* (-(b*d) + a*e))^(3/2))))/(6*(c*d^2 + e*(-(b*d) + a*e)))
Time = 0.35 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1157, 1152, 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.
\(\displaystyle \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 1157 |
\(\displaystyle \frac {(2 c d-b e) \int \frac {\sqrt {c x^2+b x+a}}{(d+e x)^3}dx}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {\sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{4 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{8 \left (a e^2-b d e+c d^2\right )}\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 \left (c d^2-b e d+a e^2\right )-\frac {(b d-2 a e+(2 c d-b e) x)^2}{c x^2+b x+a}}d\left (-\frac {b d-2 a e+(2 c d-b e) x}{\sqrt {c x^2+b x+a}}\right )}{4 \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{4 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(2 c d-b e) \left (\frac {\sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{4 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{8 \left (a e^2-b d e+c d^2\right )^{3/2}}\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e \left (a+b x+c x^2\right )^{3/2}}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}\) |
Input:
Int[Sqrt[a + b*x + c*x^2]/(d + e*x)^4,x]
Output:
-1/3*(e*(a + b*x + c*x^2)^(3/2))/((c*d^2 - b*d*e + a*e^2)*(d + e*x)^3) + ( (2*c*d - b*e)*(((b*d - 2*a*e + (2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(4* (c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) - ((b^2 - 4*a*c)*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])] )/(8*(c*d^2 - b*d*e + a*e^2)^(3/2))))/(2*(c*d^2 - b*d*e + a*e^2))
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])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b *x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a *c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
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]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d ^2 - b*d*e + a*e^2))), x] + Simp[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && EqQ[m + 2*p + 3, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1252\) vs. \(2(197)=394\).
Time = 1.34 (sec) , antiderivative size = 1253, normalized size of antiderivative = 5.83
Input:
int((c*x^2+b*x+a)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
1/e^4*(-1/3/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^3*(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 )*(-1/2/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/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)-1/4*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(- 1/(a*e^2-b*d*e+c*d^2)*e^2/(x+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)^(3/2)+1/2*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^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)+1/2*(b*e-2*c*d)/e *ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(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))/c^(1/2)-(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)))+2*c/(a*e^2-b*d*e+c*d^2)*e^2*(1/4*( 2*c*(x+d/e)+(b*e-2*c*d)/e)/c*(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)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/c ^(3/2)*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(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))))+1/2*c/(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)+1/2*(b*e -2*c*d)/e*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(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))/c^(1/2)-(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*...
Leaf count of result is larger than twice the leaf count of optimal. 954 vs. \(2 (197) = 394\).
Time = 2.31 (sec) , antiderivative size = 1950, normalized size of antiderivative = 9.07 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\text {Too large to display} \] Input:
integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)^4,x, algorithm="fricas")
Output:
[1/96*(3*(2*(b^2*c - 4*a*c^2)*d^4 - (b^3 - 4*a*b*c)*d^3*e + (2*(b^2*c - 4* a*c^2)*d*e^3 - (b^3 - 4*a*b*c)*e^4)*x^3 + 3*(2*(b^2*c - 4*a*c^2)*d^2*e^2 - (b^3 - 4*a*b*c)*d*e^3)*x^2 + 3*(2*(b^2*c - 4*a*c^2)*d^3*e - (b^3 - 4*a*b* c)*d^2*e^2)*x)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b ^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 + 4*sqrt (c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e) *x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e* x + d^2)) + 4*(6*b*c^2*d^5 + 22*a^2*b*d*e^4 - 8*a^3*e^5 - (9*b^2*c + 20*a* c^2)*d^4*e + (3*b^3 + 40*a*b*c)*d^3*e^2 - (17*a*b^2 + 28*a^2*c)*d^2*e^3 + (4*c^3*d^4*e - 8*b*c^2*d^3*e^2 + (7*b^2*c - 4*a*c^2)*d^2*e^3 - (3*b^3 - 4* a*b*c)*d*e^4 + (3*a*b^2 - 8*a^2*c)*e^5)*x^2 + 2*(6*c^3*d^5 - 13*b*c^2*d^4* e + 11*b^2*c*d^3*e^2 - a^2*b*e^5 - 2*(2*b^3 + a*b*c)*d^2*e^3 + (5*a*b^2 - 6*a^2*c)*d*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d^9 - 3*b*c^2*d^8*e - 3*a^2 *b*d^4*e^5 + a^3*d^3*e^6 + 3*(b^2*c + a*c^2)*d^7*e^2 - (b^3 + 6*a*b*c)*d^6 *e^3 + 3*(a*b^2 + a^2*c)*d^5*e^4 + (c^3*d^6*e^3 - 3*b*c^2*d^5*e^4 - 3*a^2* b*d*e^8 + a^3*e^9 + 3*(b^2*c + a*c^2)*d^4*e^5 - (b^3 + 6*a*b*c)*d^3*e^6 + 3*(a*b^2 + a^2*c)*d^2*e^7)*x^3 + 3*(c^3*d^7*e^2 - 3*b*c^2*d^6*e^3 - 3*a^2* b*d^2*e^7 + a^3*d*e^8 + 3*(b^2*c + a*c^2)*d^5*e^4 - (b^3 + 6*a*b*c)*d^4*e^ 5 + 3*(a*b^2 + a^2*c)*d^3*e^6)*x^2 + 3*(c^3*d^8*e - 3*b*c^2*d^7*e^2 - 3*a^ 2*b*d^3*e^6 + a^3*d^2*e^7 + 3*(b^2*c + a*c^2)*d^6*e^3 - (b^3 + 6*a*b*c)...
\[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\int \frac {\sqrt {a + b x + c x^{2}}}{\left (d + e x\right )^{4}}\, dx \] Input:
integrate((c*x**2+b*x+a)**(1/2)/(e*x+d)**4,x)
Output:
Integral(sqrt(a + b*x + c*x**2)/(d + e*x)**4, x)
Exception generated. \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)^4,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-b*d*e>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 1992 vs. \(2 (197) = 394\).
Time = 0.45 (sec) , antiderivative size = 1992, normalized size of antiderivative = 9.27 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\text {Too large to display} \] Input:
integrate((c*x^2+b*x+a)^(1/2)/(e*x+d)^4,x, algorithm="giac")
Output:
-1/8*(2*b^2*c*d - 8*a*c^2*d - b^3*e + 4*a*b*c*e)*arctan(-((sqrt(c)*x - sqr t(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*sqrt (-c*d^2 + b*d*e - a*e^2)) + 1/24*(6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5* b^2*c*d*e^4 - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a*c^2*d*e^4 - 3*(sq rt(c)*x - sqrt(c*x^2 + b*x + a))^5*b^3*e^5 + 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a*b*c*e^5 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*c^(7/2)*d ^4*e - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*b*c^(5/2)*d^3*e^2 + 78*(sq rt(c)*x - sqrt(c*x^2 + b*x + a))^4*b^2*c^(3/2)*d^2*e^3 - 24*(sqrt(c)*x - s qrt(c*x^2 + b*x + a))^4*a*c^(5/2)*d^2*e^3 - 15*(sqrt(c)*x - sqrt(c*x^2 + b *x + a))^4*b^3*sqrt(c)*d*e^4 - 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a* b*c^(3/2)*d*e^4 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^2*c^(3/2)*e^5 + 32*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c^4*d^5 + 16*(sqrt(c)*x - sqrt (c*x^2 + b*x + a))^3*b*c^3*d^4*e - 84*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^ 3*b^2*c^2*d^3*e^2 - 112*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c^3*d^3*e^ 2 + 74*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^3*c*d^2*e^3 + 120*(sqrt(c)* x - sqrt(c*x^2 + b*x + a))^3*a*b*c^2*d^2*e^3 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^4*d*e^4 - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^2*c *d*e^4 + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*c^2*d*e^4 + 8*(sqrt( c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^3*e^5 + 48*(sqrt(c)*x - sqrt(c*x^2 ...
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (d+e\,x\right )}^4} \,d x \] Input:
int((a + b*x + c*x^2)^(1/2)/(d + e*x)^4,x)
Output:
int((a + b*x + c*x^2)^(1/2)/(d + e*x)^4, x)
Time = 1.16 (sec) , antiderivative size = 2879, normalized size of antiderivative = 13.39 \[ \int \frac {\sqrt {a+b x+c x^2}}{(d+e x)^4} \, dx =\text {Too large to display} \] Input:
int((c*x^2+b*x+a)^(1/2)/(e*x+d)^4,x)
Output:
( - 12*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*b*c*d**3*e - 36*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*b*c*d**2*e**2*x - 36*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*b*c*d*e**3*x**2 - 12*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*b*c*e**4*x**3 + 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*c**2*d**4 + 72*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*c**2*d**3*e*x + 72*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*c**2*d**2*e**2*x**2 + 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*c**2*d*e**3*x**3 + 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)*b**3*d**3*e + 9*sqrt(a*e**2 - b*d*e + c*d**2)*log(2*sq rt(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)*b**3*d**2*e**2*x + 9*sqrt(a*e**2 - b*d*e + c*d**2)*log(2*sqrt...