\(\int \frac {1}{(a+\frac {c}{x^2}+\frac {b}{x}) x (d+e x)^2} \, dx\) [213]

Optimal result

Integrand size = 25, antiderivative size = 183 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x (d+e x)^2} \, dx=\frac {d}{\left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {\left (b c e^2+a d (b d-4 c e)\right ) \text {arctanh}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac {\left (a d^2-c e^2\right ) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}+\frac {\left (a d^2-c e^2\right ) \log \left (c+b x+a x^2\right )}{2 \left (a d^2-e (b d-c e)\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x (d+e x)^2} \, dx=\frac {\frac {2 d \left (a d^2+e (-b d+c e)\right )}{d+e x}-\frac {2 \left (b c e^2+a d (b d-4 c e)\right ) \arctan \left (\frac {b+2 a x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\left (-2 a d^2+2 c e^2\right ) \log (d+e x)+\left (a d^2-c e^2\right ) \log (c+x (b+a x))}{2 \left (a d^2+e (-b d+c e)\right )^2} \] Input:

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

Output:

((2*d*(a*d^2 + e*(-(b*d) + c*e)))/(d + e*x) - (2*(b*c*e^2 + a*d*(b*d - 4*c 
*e))*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (-2*a*d^ 
2 + 2*c*e^2)*Log[d + e*x] + (a*d^2 - c*e^2)*Log[c + x*(b + a*x)])/(2*(a*d^ 
2 + e*(-(b*d) + c*e))^2)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1893, 1200, 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.

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

\(\Big \downarrow \) 1893

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

d/((a*d^2 - b*d*e + c*e^2)*(d + e*x)) + ((b*c*e^2 + a*d*(b*d - 4*c*e))*Arc 
Tanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - 
c*e))^2) - ((a*d^2 - c*e^2)*Log[d + e*x])/(a*d^2 - e*(b*d - c*e))^2 + ((a* 
d^2 - c*e^2)*Log[c + b*x + a*x^2])/(2*(a*d^2 - e*(b*d - c*e))^2)
 

Defintions of rubi rules used

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1893

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) 
 + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c 
 + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && EqQ[mn 
, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\frac {\left (a^{2} d^{2}-a c \,e^{2}\right ) \ln \left (a \,x^{2}+b x +c \right )}{2 a}+\frac {2 \left (2 a c d e -b c \,e^{2}-\frac {\left (a^{2} d^{2}-a c \,e^{2}\right ) b}{2 a}\right ) \arctan \left (\frac {2 x a +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a \,d^{2}-b d e +c \,e^{2}\right )^{2}}+\frac {d}{\left (a \,d^{2}-b d e +c \,e^{2}\right ) \left (e x +d \right )}-\frac {\left (a \,d^{2}-c \,e^{2}\right ) \ln \left (e x +d \right )}{\left (a \,d^{2}-b d e +c \,e^{2}\right )^{2}}\)	\(187\)
risch	\(\frac {d}{\left (a \,d^{2}-b d e +c \,e^{2}\right ) \left (e x +d \right )}-\frac {\ln \left (e x +d \right ) a \,d^{2}}{a^{2} d^{4}-2 a b \,d^{3} e +2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 b c d \,e^{3}+c^{2} e^{4}}+\frac {\ln \left (e x +d \right ) c \,e^{2}}{a^{2} d^{4}-2 a b \,d^{3} e +2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 b c d \,e^{3}+c^{2} e^{4}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c \,d^{4}-a^{2} b^{2} d^{4}-8 a^{2} b c \,d^{3} e +8 a^{2} c^{2} d^{2} e^{2}+2 a \,b^{3} d^{3} e +2 a \,b^{2} c \,d^{2} e^{2}-8 a b \,c^{2} d \,e^{3}+4 a \,c^{3} e^{4}-b^{4} d^{2} e^{2}+2 b^{3} c d \,e^{3}-b^{2} c^{2} e^{4}\right ) \textit {\_Z}^{2}+\left (-4 a^{2} c \,d^{2}+a \,b^{2} d^{2}+4 a \,c^{2} e^{2}-b^{2} c \,e^{2}\right ) \textit {\_Z} +a c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (-2 a^{4} d^{6}+6 a^{3} b \,d^{5} e +2 a^{3} c \,d^{4} e^{2}-8 a^{2} b^{2} d^{4} e^{2}-4 a^{2} b c \,d^{3} e^{3}+10 a^{2} c^{2} d^{2} e^{4}+6 a \,b^{3} d^{3} e^{3}-2 a \,b^{2} c \,d^{2} e^{4}-10 a b \,c^{2} d \,e^{5}+6 a \,c^{3} e^{6}-2 b^{4} d^{2} e^{4}+4 b^{3} c d \,e^{5}-2 b^{2} c^{2} e^{6}\right ) \textit {\_R}^{2}+\left (-a^{3} d^{4}+2 a^{2} c \,d^{2} e^{2}+a \,b^{2} d^{2} e^{2}-4 a b d \,e^{3} c +3 a \,e^{4} c^{2}\right ) \textit {\_R} +a^{2} d^{2}\right ) x +\left (-a^{3} b \,d^{6}+8 a^{3} c \,d^{5} e +a^{2} b^{2} d^{5} e -19 a^{2} b c \,d^{4} e^{2}+16 e^{3} a^{2} c^{2} d^{3}+a \,b^{3} d^{4} e^{2}+10 a \,b^{2} c \,d^{3} e^{3}-19 a b \,c^{2} d^{2} e^{4}+8 e^{5} a \,c^{3} d -b^{4} d^{3} e^{3}+b^{3} c \,d^{2} e^{4}+b^{2} c^{2} d \,e^{5}-b \,c^{3} e^{6}\right ) \textit {\_R}^{2}+\left (-a^{2} b \,d^{4}+a \,b^{2} d^{3} e -b^{2} c d \,e^{3}+b \,c^{2} e^{4}\right ) \textit {\_R} +a c d e \right )\right )\)	\(743\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (177) = 354\).

Time = 2.88 (sec) , antiderivative size = 1059, normalized size of antiderivative = 5.79 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x (d+e x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(2*(a*b^2 - 4*a^2*c)*d^3 - 2*(b^3 - 4*a*b*c)*d^2*e + 2*(b^2*c - 4*a*c 
^2)*d*e^2 + (a*b*d^3 - 4*a*c*d^2*e + b*c*d*e^2 + (a*b*d^2*e - 4*a*c*d*e^2 
+ b*c*e^3)*x)*sqrt(b^2 - 4*a*c)*log((2*a^2*x^2 + 2*a*b*x + b^2 - 2*a*c + s 
qrt(b^2 - 4*a*c)*(2*a*x + b))/(a*x^2 + b*x + c)) + ((a*b^2 - 4*a^2*c)*d^3 
- (b^2*c - 4*a*c^2)*d*e^2 + ((a*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e 
^3)*x)*log(a*x^2 + b*x + c) - 2*((a*b^2 - 4*a^2*c)*d^3 - (b^2*c - 4*a*c^2) 
*d*e^2 + ((a*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(e*x + d) 
)/((a^2*b^2 - 4*a^3*c)*d^5 - 2*(a*b^3 - 4*a^2*b*c)*d^4*e + (b^4 - 2*a*b^2* 
c - 8*a^2*c^2)*d^3*e^2 - 2*(b^3*c - 4*a*b*c^2)*d^2*e^3 + (b^2*c^2 - 4*a*c^ 
3)*d*e^4 + ((a^2*b^2 - 4*a^3*c)*d^4*e - 2*(a*b^3 - 4*a^2*b*c)*d^3*e^2 + (b 
^4 - 2*a*b^2*c - 8*a^2*c^2)*d^2*e^3 - 2*(b^3*c - 4*a*b*c^2)*d*e^4 + (b^2*c 
^2 - 4*a*c^3)*e^5)*x), 1/2*(2*(a*b^2 - 4*a^2*c)*d^3 - 2*(b^3 - 4*a*b*c)*d^ 
2*e + 2*(b^2*c - 4*a*c^2)*d*e^2 + 2*(a*b*d^3 - 4*a*c*d^2*e + b*c*d*e^2 + ( 
a*b*d^2*e - 4*a*c*d*e^2 + b*c*e^3)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 
 + 4*a*c)*(2*a*x + b)/(b^2 - 4*a*c)) + ((a*b^2 - 4*a^2*c)*d^3 - (b^2*c - 4 
*a*c^2)*d*e^2 + ((a*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(a 
*x^2 + b*x + c) - 2*((a*b^2 - 4*a^2*c)*d^3 - (b^2*c - 4*a*c^2)*d*e^2 + ((a 
*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(e*x + d))/((a^2*b^2 
- 4*a^3*c)*d^5 - 2*(a*b^3 - 4*a^2*b*c)*d^4*e + (b^4 - 2*a*b^2*c - 8*a^2*c^ 
2)*d^3*e^2 - 2*(b^3*c - 4*a*b*c^2)*d^2*e^3 + (b^2*c^2 - 4*a*c^3)*d*e^4 ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(1/(a+c/x^2+b/x)/x/(e*x+d)^2,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(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x (d+e x)^2} \, dx=\frac {1}{2} \, e {\left (\frac {{\left (a d^{2} - c e^{2}\right )} \log \left (-a + \frac {2 \, a d}{e x + d} - \frac {a d^{2}}{{\left (e x + d\right )}^{2}} - \frac {b e}{e x + d} + \frac {b d e}{{\left (e x + d\right )}^{2}} - \frac {c e^{2}}{{\left (e x + d\right )}^{2}}\right )}{a^{2} d^{4} e - 2 \, a b d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, b c d e^{4} + c^{2} e^{5}} + \frac {2 \, d e}{{\left (a d^{2} e^{2} - b d e^{3} + c e^{4}\right )} {\left (e x + d\right )}} + \frac {2 \, {\left (a b d^{2} e - 4 \, a c d e^{2} + b c e^{3}\right )} \arctan \left (-\frac {2 \, a d - \frac {2 \, a d^{2}}{e x + d} - b e + \frac {2 \, b d e}{e x + d} - \frac {2 \, c e^{2}}{e x + d}}{\sqrt {-b^{2} + 4 \, a c} e}\right )}{{\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, b c d e^{3} + c^{2} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c} e^{2}}\right )} \] Input:

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

Output:

1/2*e*((a*d^2 - c*e^2)*log(-a + 2*a*d/(e*x + d) - a*d^2/(e*x + d)^2 - b*e/ 
(e*x + d) + b*d*e/(e*x + d)^2 - c*e^2/(e*x + d)^2)/(a^2*d^4*e - 2*a*b*d^3* 
e^2 + b^2*d^2*e^3 + 2*a*c*d^2*e^3 - 2*b*c*d*e^4 + c^2*e^5) + 2*d*e/((a*d^2 
*e^2 - b*d*e^3 + c*e^4)*(e*x + d)) + 2*(a*b*d^2*e - 4*a*c*d*e^2 + b*c*e^3) 
*arctan(-(2*a*d - 2*a*d^2/(e*x + d) - b*e + 2*b*d*e/(e*x + d) - 2*c*e^2/(e 
*x + d))/(sqrt(-b^2 + 4*a*c)*e))/((a^2*d^4 - 2*a*b*d^3*e + b^2*d^2*e^2 + 2 
*a*c*d^2*e^2 - 2*b*c*d*e^3 + c^2*e^4)*sqrt(-b^2 + 4*a*c)*e^2))
 

Mupad [B] (verification not implemented)

Time = 31.26 (sec) , antiderivative size = 1768, normalized size of antiderivative = 9.66 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x (d+e x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

d/((d + e*x)*(a*d^2 + c*e^2 - b*d*e)) - (log(56*a^3*b^2*c*d^4 - 96*a^4*c^2 
*d^4 - 96*a^2*c^4*e^4 - 8*b^4*c^2*e^4 - 8*a^2*b^4*d^4 + 56*a*b^2*c^3*e^4 - 
 4*a^3*b^3*d^4*x + 320*a^3*c^3*d^2*e^2 + 8*a*d^3*e*(b^2 - 4*a*c)^(5/2) - 8 
*c*d*e^3*(b^2 - 4*a*c)^(5/2) - 3*c*e^4*x*(b^2 - 4*a*c)^(5/2) - 8*b^5*c*e^4 
*x + 8*a^2*b*d^4*(b^2 - 4*a*c)^(3/2) - 8*b*c^2*e^4*(b^2 - 4*a*c)^(3/2) + 1 
2*a^3*d^4*x*(b^2 - 4*a*c)^(3/2) - 6*b*d*e^3*x*(b^2 - 4*a*c)^(5/2) + 16*a^4 
*b*c*d^4*x - 112*a^2*b^2*c^2*d^2*e^2 - 8*a*b^2*d^3*e*(b^2 - 4*a*c)^(3/2) + 
 8*b^2*c*d*e^3*(b^2 - 4*a*c)^(3/2) + 10*a*d^2*e^2*x*(b^2 - 4*a*c)^(5/2) - 
5*b^2*c*e^4*x*(b^2 - 4*a*c)^(3/2) + 6*b^3*d*e^3*x*(b^2 - 4*a*c)^(3/2) + 16 
*a*b^3*c^2*d*e^3 + 8*a*b^4*c*d^2*e^2 - 64*a^2*b*c^3*d*e^3 + 16*a^2*b^3*c*d 
^3*e - 64*a^3*b*c^2*d^3*e + 60*a*b^3*c^2*e^4*x - 112*a^2*b*c^3*e^4*x + 4*a 
*b^5*d^2*e^2*x - 8*a^2*b^4*d^3*e*x + 256*a^3*c^3*d*e^3*x - 256*a^4*c^2*d^3 
*e*x - 6*a*b^2*d^2*e^2*x*(b^2 - 4*a*c)^(3/2) - 160*a^2*b^2*c^2*d*e^3*x - 5 
6*a^2*b^3*c*d^2*e^2*x + 160*a^3*b*c^2*d^2*e^2*x + 24*a*b^4*c*d*e^3*x - 8*a 
^2*b*d^3*e*x*(b^2 - 4*a*c)^(3/2) + 96*a^3*b^2*c*d^3*e*x)*(b^2*((a*d^2)/2 - 
 (c*e^2)/2) - b*((a*d^2*(b^2 - 4*a*c)^(1/2))/2 + (c*e^2*(b^2 - 4*a*c)^(1/2 
))/2) - 2*a^2*c*d^2 + 2*a*c^2*e^2 + 2*a*c*d*e*(b^2 - 4*a*c)^(1/2)))/(4*a^3 
*c*d^4 + 4*a*c^3*e^4 - a^2*b^2*d^4 - b^2*c^2*e^4 - b^4*d^2*e^2 + 8*a^2*c^2 
*d^2*e^2 + 2*a*b^3*d^3*e + 2*b^3*c*d*e^3 - 8*a*b*c^2*d*e^3 - 8*a^2*b*c*d^3 
*e + 2*a*b^2*c*d^2*e^2) - (log(8*a^2*b^4*d^4 + 96*a^4*c^2*d^4 + 96*a^2*...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 865, normalized size of antiderivative = 4.73 \[ \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x (d+e x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b*d**3 - 2 
*sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*b*d**2*e*x + 8* 
sqrt(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*c*d**2*e + 8*sqr 
t(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*a*c*d*e**2*x - 2*sqrt 
(4*a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b*c*d*e**2 - 2*sqrt(4* 
a*c - b**2)*atan((2*a*x + b)/sqrt(4*a*c - b**2))*b*c*e**3*x + 4*log(a*x**2 
 + b*x + c)*a**2*c*d**3 + 4*log(a*x**2 + b*x + c)*a**2*c*d**2*e*x - log(a* 
x**2 + b*x + c)*a*b**2*d**3 - log(a*x**2 + b*x + c)*a*b**2*d**2*e*x - 4*lo 
g(a*x**2 + b*x + c)*a*c**2*d*e**2 - 4*log(a*x**2 + b*x + c)*a*c**2*e**3*x 
+ log(a*x**2 + b*x + c)*b**2*c*d*e**2 + log(a*x**2 + b*x + c)*b**2*c*e**3* 
x - 8*log(d + e*x)*a**2*c*d**3 - 8*log(d + e*x)*a**2*c*d**2*e*x + 2*log(d 
+ e*x)*a*b**2*d**3 + 2*log(d + e*x)*a*b**2*d**2*e*x + 8*log(d + e*x)*a*c** 
2*d*e**2 + 8*log(d + e*x)*a*c**2*e**3*x - 2*log(d + e*x)*b**2*c*d*e**2 - 2 
*log(d + e*x)*b**2*c*e**3*x - 8*a**2*c*d**2*e*x + 2*a*b**2*d**2*e*x + 8*a* 
b*c*d*e**2*x - 8*a*c**2*e**3*x - 2*b**3*d*e**2*x + 2*b**2*c*e**3*x)/(2*(4* 
a**3*c*d**5 + 4*a**3*c*d**4*e*x - a**2*b**2*d**5 - a**2*b**2*d**4*e*x - 8* 
a**2*b*c*d**4*e - 8*a**2*b*c*d**3*e**2*x + 8*a**2*c**2*d**3*e**2 + 8*a**2* 
c**2*d**2*e**3*x + 2*a*b**3*d**4*e + 2*a*b**3*d**3*e**2*x + 2*a*b**2*c*d** 
3*e**2 + 2*a*b**2*c*d**2*e**3*x - 8*a*b*c**2*d**2*e**3 - 8*a*b*c**2*d*e**4 
*x + 4*a*c**3*d*e**4 + 4*a*c**3*e**5*x - b**4*d**3*e**2 - b**4*d**2*e**...